2016SS
2016SS
Andreas Kriegl
      email:andreas.kriegl@univie.ac.at
0O 0O
          
0     /E            / sN        / / sN /E   /0
                        OO              OO
          
0     /E       / sN ×sN /E Q̃     / / Q̃    /0
                       O                O
                        ?              ?
                       sO             sO
                       0              0
This is the script for my lecture course during the summer semester 2016. It
can be downloaded at http://www.mat.univie.ac.at/∼kriegl/Skripten/2016SS.pdf
Many of the proofs are taken from Meise and Vogt’s book [MV92] and I will give
detailed references to it, but also to Jarchow’s book [Jar81].
As prerequiste the user is assumed to be familiar with basic functional analysis
(for Banach spaces) and the basics of locally convex theory as presented in lecture
courses on higher functional analysis. I will refer to my script [Kri14] for these
results.
The main focus is on Fréchet spaces and additional topological properties for them.
Leading examples of Fréchet spaces will be the Köthe sequences spaces and in
particular the power series spaces with the space s of rapidly decreasing sequences
as most relevant member. We will have to consider several of these properties also
for general locally convex spaces, in particular, since the strong dual of Fréchet
spaces is rarely Fréchet.
Our discussion will start with properties of locally convex spaces which are pre-
served by the formation of inductive or projective limits. And we will then consider
what is inherited by the strong dual. Then we consider how properties of con-
tinuous linear maps translate into properties of the adjoint mappings using short
exact sequences. And we will introduce topological properties which garantee the
splitting of such sequences. These and further properties will also play a role in
determining situations where continuous linear mappings are locally bounded and
for characterizing the subspaces and the quotients of s.
I will put online a detailed list of the treated sections at the end of the semester
under http://www.mat.univie.ac.at/∼kriegl/LVA-2016-SS.html.
Obviously the attentive reader will find misprints and even errors. Thus I kindly
ask to inform me about such - future generations of students will appreciate the
corrections.
4. Duality                                                            57
  Spaces of (linear) functions                                        57
  Completeness of dual spaces                                         58
  Barrelledness and bornologicity of dual spaces                      60
  Duals of Fréchet spaces                                           64
  Duals of Köthe sequence spaces                                     66
  Semi-reflexivity and stronger conditions on dual spaces             68
  Dual morphisms                                                      84
  Splitting sequences                                                 99
  Locally bounded linear mappings                                    116
  The subspaces and the quotients of s                               126
Bibliography 129
Index 131
A projective limit is called reduced, iff all projections prj : lim F → F(j) have
                                                                 ←−
dense image. By replacing F(j) with the closure F̄(j) of the image of prj (lim F)
                                                                              ←−
in F(j) we get that lim F equals lim F̄, which is a reduced projective limit. Note
                        ←−            ←−
that F̄(j  j 0 ) is then a well defined continuous linear mapping with dense image.
As closed subspace in the product the limit of complete lcs is complete. Recall,
that an lcs E is complete iff every Cauchy-net (i.e. x : (I, ) → E satisfying
∀p ∈ P ∀ε > 0 ∃i ∈ I ∀i0 , i00  i: p(xi0 − xi00 ) < ε) converges in E.
Fréchet spaces are Baire spaces, hence the closed graph theorem (cf. [Kri14,
4.3.1], [MV92, 24.31 p.270], [Jar81, 5.4.1 p.92]) and the open mapping theorem
(cf. [Kri14, 4.3.5], [MV92, 24.30 p.270], [Jar81, 5.5.2 p.95]) hold for linear maps
between Fréchet spaces.
Proof. The trace of the countable 0-neighborhoodbasis (or countable many semi-
norms) is a 0-neighborhoodbasis (are the generating seminorms) of the subspace.
The quotient seminorms q̃(x + F ) := inf{y ∈ F : q(x + y)} are a basis of semi-
norms on the quotient, see [Kri14, 3.3.3]. And since Cauchy-sequences can be lifted
along the quotient mapping (see [Kri14, 3.5.3]) the quotient is (sequentially-)com-
plete as well.
We obviously get a countable basis of seminorms for the product of countable many
Fréchet spaces, and since limits of complete spaces are complete, such a limit is a
Fréchet space.
as subspace of λ∞ (A).
1.10 Remark.
1.11 Lemma. Köthe sequence spaces as limits (See [MV92, 27.2 p.307]).
The Köthe sequence space λq (A) is isomorphic to lim F, where the functor F on
                                                                                0
(A, ≥) is given by F(a) := `q and F(a ≥ a0 ) : F(a) → F(a0 ) is given by x 7→ aa x,
where                      ( 0
                             an
                  a0             for an 6= 0,
                     : n 7→ an
                  a         0    for an = 0 (and hence a0n = 0).
In particular, if A is countable, then λq (A) and c0 (A) are Fréchet spaces (See
[MV92, 27.1 p.307]).
Obviously, the coproduct R(Na ) is dense in `p (a) for p < ∞ and hence also R(N) ⊆ E,
since R(N) ⊆ E/ ker k ka ⊆ `p (a). ⇒ Ea := (E/ ker k ka )∼ ∼       = `p (a) for 1 ≤ p < ∞
          ∼
(resp. Ea = c0 (a)). By completeness E = lima Ea .
                                            ←−
                               Ea ∼= `p (a) ∼
                                               ·a   / `p (carr a)      / `p
                                77             =
                          |carr a
             E = λp (A)                          |carr a0
                                                                               0
                                                                            · aa           · aa
                                                                                               0
                        |carr a0        ''                                              
                                    E   a0
                                             ∼
                                             = `p (a0 )
                                                            ·a0   / `p (carr a0 )     / `p
                                                             ∼
                                                             =
                                        ∼ `p and c0 (A) =
    1. If A is a singleton, then λp (A) =                ∼ c0 .
    2. Let A := {en : n ∈ N}, where en are the standard unit vectors in RN . Then
       λ∞ (A) = λp (A) = c0 (A) = RN for all p ∈ [1, ∞]. Note that we can equally
       take {χF = max{ek : k ∈ F } : F ⊆ N is finite} instead of A.
       Let A = RN be the set of all real sequences (ak )k . Then λ∞ (A) = K(N) :=
    3. `
          j∈N K (cf. [Kri14, 3.6.1]): Suppose there is an x ∈ λ (A) with carr(x)
                                                                   ∞
       In fact, the power series n an z converges for all |z| < R iff {an rn : n ∈ N}
                                         n
                                P
       is bounded (equivalently, absolutely summable) for all r < R.
    6. If A = {n 7→ rn : 0 < r < 1} = {n 7→ es n : s < 0} then λ∞ (A) = λ1 (A) =
       H(D), the space of holomorphic functions on the unit disk [MV92,
       29.4.3 p.340].
                                         X∞         
                                x 7→ z 7→     xn z n
                                             n=0
It is the power series space λ0 (α) for α(n) := n (See [MV92, 29.4.2 p.340])
                                1       1
    7. For 1 ≤ p < ∞ and        q   +   p   = 1 we have λ1 (`p ) = (`q , σ(`q , `p ) as lcs:
                                                       1
       (⊇) By the Hölder inequality kxkλy = kx · yk`1 ≤ kxk`q · kyk`p < ∞ for all
       y ∈ `p and x ∈ `q .
       (⊆) Let x ∈ KN be such that kx·yk1 < ∞ for all y ∈ `p . Then the linear map
       y 7→ Px · y, `p → `1 has closed graph and thus is continuous. Consequently,
       y 7→ n xn · yn is a continuous linear functional, hence x ∈ (`p )∗ = `q (see
       [Kri14, 5.3.1]).
       λ1 (`∞ ) = (`1 , σ(`1 , `∞ )): For (⊆) choose y = 1.
       λ1 (c0 ) = (`1 , σ(`P
                           1
                             , c0 )): Suppose x ∈ λ1 (c0 ) \ `1 , choose k 7→ nk strictly
                              nk+1
       increasing with j=n         k +1
                                        |xj | ≥ k and yj := k1 for nn < j ≤ nk+1 . Then
                    P
       kx · yk`1 ≥ k 1 = ∞.
                 (R) ∼
                                                            R                      
               ∞                                              1   π          −ikt
Proof. (1) C2π       = s via Fourier-coefficients f 7→ 2π            f (t) e      dt      ,
                                                    Rπ
                                                                 −π                   k∈Z
cf. [Kri07b, 5.4.5] and 1.26.3 : Let ck (f ) := 2π −π f (t) e−ikt dt. Then ck (f 0 ) =
                                                  1
                                                         R
polynomials with respect to the inner product hf |giρ := R f (t)g(t) ρ(t) dt. Ob-
viously L2ρ (R) ∼
                = L2 (R) via f 7→ ρ f . Gram-Schmidt orthonormalization applied
                                 √
to the monomials t 7→ tn gives an orthonormal basis ( √ n1 √ Hn )n∈N , where Hn
                                                                           2 n! π
are the Hermite polynomials (cf. [Kri07b, 6.3.9 p.118]), which can also be
obtained recursively H0 := 1, Hn+1 (t) := 2t Hn (t) − 2n Hn−1 (t):
From the recursion we get Hn0 = 2n Hn−1 by induction. In fact H00 = 0, H1 (t) = 2t,
H10 = 2H0 , and hence
            0
                                      0
         Hn+1   = 2 id Hn − 2n Hn−1 = 2 Hn + 2 id Hn0 − 2n Hn−1    0
               ∞
               X                             ∞
                                             X
        ⇒            nm |hf |hn i|2 ≤ 2−m          |hAm           2
                                                      + f |hn+m i| ≤ 2
                                                                      −m
                                                                         kAm    2
                                                                           + f kL2 (R) < ∞,
               n=0                           n=0
hence S → s, f 7→ (hf |hn i)n≥0 isPcontinuous and obviously injective.
It is also onto: Let a ∈ s. Then n an hn converges in S, since
                                     X     √                  p
        − h0k + id ·hk = A1+ (hk ) =   21/2 n hhk |hn−1 i hn = 2(k + 1) hk+1
                                             n≥1
                r               r                   r          r
               n         n+1                          n          n+1
    ⇒ h0n =       hn−1 −      hn+1 and id ·hn =         hn−1 +       hn+1 .
                2          2                          2           2
      ∞
           (R) := f ∈ C ∞ (R) : f (x) = 0 ∀x ∈
                                             / [a, b] ∼
                   
(3) C[a,b]                                            =s
(See [MV92, 29.5.3 p.342]):
W.l.o.g. −a = b = π/2.
    Φ : S(R) → C[−π/2,π/2]
                ∞
                           (R), Φ(f )(t) := f (tan(t)) ∀|t| < π/2 is an iso, since
                  p
                                                          (R, R)
                 X    g̃j,p                            ∞
    Φ(f )(p) =           j+p
                             f (j) ◦ tan with g̃j,p ∈ C2π
                 j=1
                     cos
    | tan(t)k f (j) (tan(t))| ≤ sup |xk f (j) (x)| =: Ck,j < ∞ ∀|t| < π/2
                                   x∈R
                     1
  Since tan(x) ∼          for x near ±π/2 we get |Φ(f )(p) (t)| → 0 for t → ±π/2.
                   cos(x)
And the inverse mapping is given by f 7→ f ◦ arctan using analogous arguments:
                                         qn (s)
arctan0 (s) = 1+s
               1
                 2 ⇒ arctan
                             (n)
                                 (s) = (1+s  2 )n with deg(qn ) ≤ n − 1. Thus
                                  p
                                 X   tk gj,p (t) (j)
        tk Φ−1 (f )(p) (t) =                    f (arctan(t)) with deg(gj,p ) ≤ n − 1,
                                 j=1
                                     (1 + t2 )n
and
                                            n+k−1                 
   tn+k−1 f (j) (arctan(t)) = tan ±(π/2 − s)         f (j) ±(π/2 − s)
                                                              
                            = (± cot(s))n+k−1 f (j) ±(π/2 − s)
                                         n+k−1 (j)
                                ±s cos(s)          f (±(π/2 − s))
                            =                                      → 0 for s & 0.
                                  sin(s)               sn+k−1
Now the result follows since S(R) ∼
                                  = s by 2 .
     ∞         ∼ s (See [MV92, 29.5.4 p.343]):
(4) C ([a, b]) =
W.l.o.g. −a = b = 1.
                    Φ : f 7→ f ◦ cos, C ∞ ([−1, 1]) → C2π,even
                                                       ∞       ∼
                                                               =s
is continuous and injective. It is also onto, since
                                                    g0                  g0
f := g ◦ arccos ∈ C([−1, 1]) ∩ C ∞ (]−1, 1[), f 0 = −  ◦ arccos, and           ∞
                                                                           ∈ C2π,even .
                                                   sin                 sin
                                    ∞                  ∞
Note that via Fourier-coefficents C2π,even = {f ∈ C2π     : f (x) = f (−x)} = {f ∈
   ∞                       ∼               P                  
C2π : cn (f ) = c−n (f )} = s, via x 7→      n≥0 an cos(n x)    ← (an )n . Thus s →
C ∞ ([−1, 1]) is given by a 7→
                                P                           P
                                 n∈N an cos(n arccos t) =      n∈N an Tn , where Tn :
t 7→ cos(n arccos t) are the Tschebyscheff(=Chebyshev) polynomials.
                                       X
                ∀a ∃a0 ∃C > 0 ∀x :           |xj | kej · ak`p ≤ C kx · a0 k`p ,
                                        j
                                                   | {z }
                                                        =|aj |
Thus
                              k+p
                              X                                        k+p
                                                                       X
                   ∀k, p :           x∞
                                      j ej          ≤ 2ε +                       ξj (xm ) ej        .
                                                n                                               n
                             j=k+1                                 j=k+1
Since j ξj (xm ) ej converges in E, the sequence j x∞
       P                                              P
                                                           j ej is Cauchy, hence con-
                         ∞
verges to some x∞ := j=0 x∞                         ∞      ∞
                      P
                             j  e j ∈ E with ξ j (x   ) = xj , since (ej ) is a Schauder-
                                                    0
                                                  m       ∞
basis. By the inequality above, we have that x → x with respect to τ .
1.20 H(DR ) has (z k )k∈N as absolute basis (See [MV92, 27.27 p.323]).
Let DR := {z ∈ C : |z| < R} be the disk with radius 0 < R ≤ ∞. Taylor
                       P (k)
development f (z) = k f k!(0) z k shows that (z 7→ z k )k∈N is a Schauder-basis of
H(DR ). This is even an absolute basis: kf kr := sup{|f (z)| : |z| ≤ r} for r < R is a
basis of seminorms and ∀f ∈ H(DR ) ∀r < r0 < R :
   X f (j) (0)                             Z                       ∞  j
                        [Kri11, 3.30] X 1          f (z)          X    r
                kz j kr ==========                       dz r j
                                                                =            kf kr0 .
     j
           j!                         j
                                        2πi |z|=r0 z j+1
                                                                  j=0
                                                                       r0
1.21 The Fréchet spaces with absolute basis are the spaces λ1 (A)
(See [MV92, 27.26 p.323], [Jar81, 14.7.8 p.314]).
For Fréchet space E we have: ∃A countable: E ∼
                                              = λ1 (A) ⇔ E has an absolute basis.
Proof. (⇒) The standard basis (ej )j∈N is obviously an absolute basis of λ1 (A).
(⇐) (See [MV92, 27.25 p.322]) Let (ej )j be an absolute basis of E and consider
the Köthe matrix A := (j 7→ kej kp )p∈N . Then ξ : E → KN , x 7→ (ξj (x))j is linear.
(ej )j absolute basis ⇒
                                     X
                      ∀p ∃p0 ∃C ∀x :     |ξj (x)| kej kp ≤ Ckxkp0
                                               j
             1                       1
⇒ ξ(x) ∈ λ (A) and ξ : E → λ (A) continuous and injective.
Claim: ξ is onto λ1 (A):
                                                   n+k
                                                   X                      n+k
                                                                          X
                 y = (yj )j ∈ λ1 (A) ⇒                    yj ej       ≤           |yj | kej kp
                                                                  p
                                               j=n+1                      j=n+1
y ∈ λ1 (A)       P                                                                 P
=     ⇒ n 7→
=======            j≤n yj ej   Cauchy in E ⇒ converges to x :=                          j   yj ej with ξ(x) = y,
             open map.thm.
i.e. ξ onto. =         ⇒ ξ is isomorphism.
             ===========
Proof.
() y ∈ λ∗ :
                       ∞
                       X                              ∞
                                                     X            X ∞
        ∀x ∈ λ : x =         xj ej ⇒ y(x) = y                xj ej =     xj y(ej ) .
                       j=0                               j=0         j=0
                                                                            | {z }
                                                                                        =:yj
                                                            X
                  kεk∞ ≤ 1 ⇒ ε · x ∈ λ, hence                     xj yj converges absolutely.
                                                              j
Counter-example.
Let A = c0 . By 1.15.7 we have λ := λ1 (A) = (`1 , σ(`1 , A)), and hence λ∗ = A =
c0 , whereas λ1 (λ) = λ1 (`1 ) = `∞ .
Proof. Let first 1 < p < ∞ and y ∈ λ∗ . We assume first, that carr y ⊆ carr a.
Then
          1.23                        1.22     X               X
   kyk∗a =
         ===== ky|Ua k∞ = sup |y(x)| =
                                     ===== sup   xj yj = sup         xj yj
                                                                         j∈N
                                        x∈Ua                   x∈Ua                         x∈Ua j∈carr a
                  n         X                 yj                                   o      `q = (`p )∗   y
             = sup                 xj aj ·       : (xj aj )j∈carr a        `p
                                                                                ≤ 1 ========                     ,
                        j∈carr a
                                              aj                                                        a   `q
1.25 Theorem. Equality of λp (A) for various p (See [MV92, 27.16 p.315]).
                                                     0
       1. ∃ 1 ≤ p 6= p0 ≤ ∞: λp (A) = λp (A) as lcs;
                                                     0
 ⇔ 2. ∀ 1 ≤ p 6= p0 ≤ ∞: λp (A) = λp (A) as lcs;
 ⇔ 3. ∀a ∈ A ∃a0 ∈ A:              a
                                   a0   ∈ `1 .
( 2 ⇒ 1 ) is trivial.
( 1 ⇒ 3 ) For p0 = ∞ we get λ∞ (A) = c0 (A), since `p ⊆ c0 ⊆ `∞ .
                                                                                      0
                                                               p                  λp (A)
                       ⇒ ∀a ∃a0 ≥ a ∃C > 0 : k kλa                 (A)
                                                                         ≤ C k ka0
     [Kri14, 1.3.3,1.3.7]                                  0                     p0
                ⇒ ∀a ∃a0 ≥ a ∃C > 0 : Uap0 := {x : kxkλa0 ≤ 1} ⊆ C Uap
  ===============
                   1.24
                      ⇒ ∀y ∈ h(Uap )o ilin.sp : kyk∗a0 ,p0 ≤ Ckyk∗a,p .
                                             
                  =
                  ====
           Hölder, 1.22                                        0
          ==========
          =         ⇒η · a ∈ (Uap )o ⊆ C (Uap0 )o , since ka · xk`p ≤ 1 ∀x ∈ Uap .
                 1.24 X                   0
                                 0 aj q 0 1/q
                ====
                =   ⇒      |ηj |q 0            = kη · a|U p0 k∞ ≤ C
                        j
                                   aj                     a0
                           0
                  ξ := η q
                              nX       aj q0           0                      0
                                                                                   o      0
                  ====== ⇒ sup   |ξj | 0      : ξ ∈ `q/q , kξk`q/q0 = (kηk`q )q ≤ 1 ≤ C q
                               j
                                       aj
        q                       q0                0
t :=   q−q 0 ,   i.e.   1
                        t   +   q    = 1, (`q/q )∗ = `t
                          a q 0         X aj q0 t   a q0                                    t
                   1.15.7                                                                                    0
                                     t
                       ⇒ 0
                  ======          ∈ ` and            =                                                  ≤ C q t.
                          a               j
                                             a0j         a0                                   `t
                                Hölder inductive
                                                      X a(0)
                                                         j
                                                                          Y a(k)
                                                                        X d−1 j
                                            ⇒
                                =============               (d)
                                                                    =              (k+1)
                                                                                             < ∞.
                                                      j    aj           j   k=0   aj
       1. Let 0 < α = (αn )n % ∞, R ∈ [0, ∞), p ∈ [1, ∞]. Then λpR (α) ∼  p
                                                                       = λ0 (α).
       2. Let R ∈ {0, ∞}. Then λ1R (α) = λ1R (β) ⇔ ∃C ≥ 1: C1 α ≤ β ≤ Cα.
                         < ∞ ⇒ λpR (α) × λpR (α) ∼
               α                                     p
       3. sup 2j+1αj                              = λR (α) for R ∈ {0, +∞} and p ∈
          [1, ∞]. In particular, s × s ∼= s and s(Z) ∼
                                                     = s(N).
                                                           < ∞ ⇔ sup ln
            1          p
                                                    P αj              j
       4. λ∞ (α) = λ∞ (α) ∀p ∈ [1, ∞] ⇔ ∃r < 1:       jr                αj : j < ∞.
By the open-mapping theorem λ1R (α) = λ1R (γ) = λ1R (β) as Fréchet spaces.
                       1.24                                       λ1 (α)                     λ1 (β)
                      ⇒ ∀r ∃s > r ∃C > 0 : k kr R
                   ====
                   =                                                          ≤ C k ks R
     [Kri14, 1.3.3,1.3.7]
               ⇒ ∀r ∃s > r ∃C > 0 : Usβ ⊆ C Urα
 ===============
               ⇒ ∀y ∈ h(Urα )o ilin.sp : kyk∗s,β ≤ Ckyk∗r,α
                                      
                         1.25                               P a
 4 λ1∞ (α) = λp∞ (α) ⇐      ⇒ ∀r ∃s(> r) : j (er−s )αj = j aj,r
                                             P
                        ====                                      j,s
                                                                      <∞⇔
          r−s
                     P αj
⇔ ∃q = e      < 1 : j q < ∞ ⇔ ∃δ > 0 : δ ln j ≤ αj :
(⇐) q αj ≤ e(r−s)δ ln j = j δ(r−s) ≤ j −2 , provided s > r + 2δ .
(⇒) ln  j                              ln jn
     αj unbounded ⇒ ∀n ∃jn : αjn ≥ n, w.l.o.g. jn+1 ≥ 2jn ≥ 8. Then for
     −x
q = e with x > 0 we have
                X                  X           jn
                                               X                     X
                         q αj =                       e−x αj ≥                (jn − jn−1 ) e−x αjn
                                      n j=jn−1 +1                         n
                                                                              |    {z   }
                   j
                                                                                    ≥jn /2
                                                                                  ln( j2n )− n
                                                                                             x
                                                                     X                                        X
                                                                                               ln(jn )
                                                                 ≥            e                          ≥            1,
                                                                          n                                   n≥2x
                                x/n                              1−x/n
since ln( j2n ) ≥ ln(jn ), or equivalently jn                                 ≥ 41/2 = 2.
                                   1.25                                                                                aj,r
 5 λ10 (α) = λp0 (α) ⇐====⇒ ∀r < 0 ∃ (r <)s < 0 :                                               r−s αj
                                                                                      P                        P
                                                                                         j (e      )     =           j aj,s   < ∞ ⇔
∀q = er−s < 1 : j q αj < ∞ ⇔ limj→∞ ln    j
                  P
                                         αj = 0:
                       ln j
(⇐) limj→∞              αj    = 0 ⇒ ∀x ∃N ∀j ≥ N : ln j    x
                                                    αj < 2 ⇒
                                 X           X       2
                                                             X 1
                                    e−x αj ≤    e−x x ln j =      <∞
                                                               j2
                                   j≥N                   j≥N                        j≥N
                       ln j                                       ln jn
(⇒) limj→∞              αj     6= 0 ⇒ ∃δ > 0 ∀n ∃jn :              αjn        ≥ δ, w.l.o.g. jn+1 ≥ 2jn ≥ 8. Then
              −x                       δ
for q := e         with x :=           2    we have
                              jn
                                                                                                              jn
                                                                                                                   )− x
X              X              X                     X                                            X                                    X
      q αj =                          e−x αj ≥            (jn − jn−1 ) e−x αjn ≥                       eln(    2      δ ln(jn )   ≥       1,
                n j=jn−1 +1                          n
                                                          |    {z   }                             n                                   n
 j
                                                               ≥jn /2
                    1/2                       1−1/2
since ln( j2n ) ≥ ln(jn ), or equivalently jn         ≥ 41/2 = 2.
A sequence (xn )n∈N in an lcs is called Mackey convergent towards x∞ iff there
exists a sequence λn → ∞ in R with {λn (xn − x∞ ) : n ∈ N} being bounded.
                                                                                   (k)
Proof. Let (pk )k∈N be a basis of seminorms. Since for each k the sequence µn :=
pk (xn − x∞ ) → 0 for n → ∞ we find another sequence 0 6= µ∞               k  ∞
                                                            n → 0 with {µn /µn :
n ∈ N} bounded for each k (See [Kri14, 2.1.6]). Then λn := 1/µk has the required
                                                                ∞
property.
2.3 Corollary (See [MV92, 23.23 p.253], [Kri14, 4.1.11], [Kri14, 4.2.4] ).
Fréchet spaces are ultrabornological, hence bornological, barrelled and infrabarrelled.
Proof. Metrizable lcs are bornological (See [Kri14, 2.1.7], [MV92, 24.13 p.264]),
since any bounded linear mapping f on them is (sequentially) continuous: Let xn →
x∞ , then by 2.2 there are λn → ∞ with n 7→ f (λn (xn −x∞ )) = λn (f (xn )−f (x∞ ))
bounded, hence f (xn ) → f (x∞ ). Completeness implies now that the space is even
ultrabornological.
2.4 Colimits.
Let F : J → lcs be a functor from a partially ordered set (J, )op = (J, ≺) or even
from a small category J into that of locally convex spaces. The colimit colim F
of F (See [Kri08, 3.25]) is then given as quotient of the coproduct (direct sum,
cf. [Kri14, 3.6.1])
           a          n      Y                                            o
              F(j) := x ∈       F(j) : xj = 0 for all but finitely many j
            j                  j
                                                                                   `
with the final locally convex structure with respect to the inclusions Ej ,→ j F(j)
(whose continuous seminorms are those which restricted to each summand F(j)
are seminorms of F(j)), where we factor out the congruence relation generated
                     0
x(j) ∼ (F(f )(x))(j ) for every j ≺ j 0 (morphism f : j → j 0 in J ), where x(j)
denotes the point with j-th coordinate x ∈ F(j) and all other coordinates equal to
0. Since the topology on this quotient need not be Hausdorff, one has to factor out
the closure of {0} in addition, i.e. the intersection of the kernels of all its seminorms.
In the particular case, where J op    F = (J, ) is directed, the first (not necessarily
Hausdorff) quotient is given by j F(j)/ ∼, where x1 ∈ F(j1 ) is equivalent to
x2 ∈ F(j2 ) iff for some j  j1 , j2 : F(j1 ≺ j)(x1 ) = F(j2 ≺ j)(x2 ). In this case the
colimit is also called inductive limit (See [Jar81, 4.5 p.82]) and denoted lim F.
                                                                                 −→
An inductive limits is called reduced, iff all ιj : F(j) → lim F are injective.
                                                                    −→
By replacing F(j) with the image F̃(j) of ιj in lim F supplied with its quotient
                                                      −→
structure, we get that lim F equals lim F̃, which is a reduced inductive limit (See
                         −→             −→
[Jar81, 4.5.2 p.82]). Note for this that F̃(j ≺ j 0 ) is then a well defined injective
continuous linear mapping.
An even more restricted situation is, when J = (N, ≤), i.e. we have an inductive
limit of a sequence of spaces (the steps of the limit). The inductive limit of a
sequence of Fréchet-spaces (a so-called (LF)-space) is almost never a Fréchet space
(See [Kri14, 4.1.13]): Strict inductive limits of sequences (i.e. En is a closed
topological subspace in En+1 for each n), which are not finally constant, can not
be Baire spaces and hence are not Fréchet; And, more generally, by [Jar81, 12.4.4
p.259] a metrizable space with a countable base of bornology has to be normed, in
particular this is valid for (locally) complete (LB)-spaces (See [Flo73, 5.5 p.73]),
i.e. inductive limits of a sequenceSof Banach spaces. Even more generally, if all Fn
and F∞ := limn Fn (hence F∞ = n∈N ιn (Fn )) are Fréchet, then by Grothendieck’s
             −→
                             E
                                  f
                                         / S fn (En )      /F
                                            n O
                            f˜
                                            % ?
                           ∃Em            / fm (Em )
                                   ∃fm
            (1)     (0)
similarly En into Ek ).
Proof. ( 1 ) b ∈ λ∞ (A) ⇔ ∀a: b · a bounded (by Ca > 0), i.e. ∀j: |bj | ≤ Ca /aj ⇔
∀j: |bj | ≤ inf a Ca /aj .
( 2 ) Let A := {a(k) : k ∈ N} with k 7→ a(k) increasing. For each k ∈ N choose
               (k)          (k)         (k)                                     (k)
Ck > k max{1, a0 , . . . , ak }. ⇒ Ck /aj ≥ k for all k ≥ j. ⇒ bj := inf k Ck /aj =
             (k)
mink Ck /aj        > 0 and b ∈ λ∞ (A).
( 3 ) By ( 2 ) there is a b0 ∈ λ∞ (A) with b0j > 0 forall j. For b ∈ λ∞ (A) also
b00 : j 7→ max{|bj |, b0j } is in λ∞ (A) and satisfies b00 ≥ |b| and ∀j : b00j ≥ b0j > 0.
Proof. b ∈ λ∞ (A) ⇒ Bbp ⊆ λp (A) bounded, since x ∈ Bbp ⇒ carr x ⊆ carr b and
∀a ∈ A: kxka = kx · ak`p = kx · 1b · b · ak`p ≤ kx/bk`p · kb · ak∞ ≤ 1 · kb · ak∞ .
Conversely, let B ⊆ λ∞ (A) be bounded, i.e. ∀a ∈ A ∃Ca > 0 ∀x ∈ B: kx·ak`∞ ≤ Ca .
Let bj := inf{ Caja : a ∈ A}, which is < ∞, since aj > 0 for some a. Then b ∈ λ∞ (A),
since |bj aj | ≤ Ca for all a ∈ A and j ∈ N. Furthermore, since |xj Cja | ≤ 1 for all
                                                                       a
x ∈ B.
Since c0 (A) is a subspace of λ∞ (A), this works for c0 (A) as well.
Now for 1 ≤ p < ∞ and A = {a(k) : k ∈ N} countable: Let B ⊆ λp (A) be bounded,
                                                           2.9.1
                                       ⇒ b := inf k 2k+1 Ck /a(k) ∈ λ∞ (A).
i.e. ∀k ∃Ck > 0 ∀x ∈ B : kxkk ≤ Ck =====
                       (k)                           (k)
      1         aj      X aj
         = sup k+1    ≤          ⇒
      bj    k 2    Ck    2k+1 Ck
                                        k
                    X xa(k)                              X kxa(k) k`p               X kxkk     X 1
      kx/bk`p     ≤                                  ≤                          ≤            ≤       =1
                     2k+1 Ck                    `p               2k+1 C   k          2k+1 Ck    2k+1
                       k                                   k                          k           k
  ⇒x∈     Bbp ,    i.e. B ⊆   Bbp .
2.11 Counter-example.
Let A := `p for 1 ≤ p < ∞. Then λ∞ (A) := {x ∈ KN : ∀y ∈ `p : kx · yk`∞ < ∞} is
the linear space `∞ :
(⊇) x ∈ `∞ , y ∈ `p ⊆ `∞ ⇒ kx · yk`∞ ≤ kxk`∞ · kyk`∞ .
(⊆) Suppose x ∈ λ∞ (A) is unbounded ⇒ ∃jn (W.l.o.g. strictly increasing) with
|xjn | ≥ n 2n .                   (
                                   2−n for j = jn
                           yj :=                   .
                                   0     otherwise
                                                                       / λ∞ (A).
Then kyk`p ≤ kyk`1 = n 21n < ∞, but kx · yk`∞ ≥ |xjn yjn | ≥ n, i.e. x ∈
                       P
Proof. 2.9.3 ⇒ ∀b ∈ λ∞ (A) ∃b0 ∈ λ∞ (A) ∀j: 0 6= b0j ≥ |bj | ⇒ Bbp ⊆ Bbp0 and
λp (A)B p0 ∼
           = `p .
       b
In the following sections we consider classes of locally convex spaces which are
invariant under the formation of limits, i.e. various completeness conditions, semi-
reflexivity, Montel spaces, Schwartz spaces, and nuclear spaces. And we characterize
those Köthe sequence spaces having these properties.
3.1 Completeness.
For lcs E we consider the following completeness conditions:
    •   E is called complete iff evevry Cauchy net (or Cauchy filter) converges.
    •   E is called quasi complete iff every closed bounded subset is complete.
    •   E is called sequentially complete iff Cauchy sequences converge.
    •   E is called locally complete (or Mackey-complete) iff EB is a Banach
        space for every closed absolutely convex bounded subset B ⊆ E.
One obviously has the implications:
     complete ⇒ quasi-complete ⇒ sequentially complete ⇒ locally complete.
For metrizable spaces all 4 conditions are equivalent (See [Kri14, 2.2.2]). Each
of these completeness properties is inherited by closed subspaces ([Kri14, 3.1.4]),
products ([Kri14, 3.2.1]), and coproducts ([Kri14, 3.6.1]) (See [Jar81, 3.2.5 p.59],
[Jar81, 3.2.6 p.59], [Jar81, 6.6.7 p.111]).
The completion (i.e. reflector) of any lcs E is given by the space of all linear func-
tionals on E ∗ , whose restrictions to equicontinuous subsets are σ(E ∗ , E)-continuous,
supplied with the topology of uniform convergence on the equicontinuous subsets,
see [Kri14, 5.5.7].
Proof. (⇐) If T has such a representation, then the finite sums define finite
dimensional operators which converge to T .
(⇒) Since any compact T : E → F induces a compact injective operator T :
(ker T )⊥ → T (E) with dense image, we may assume that T is injective. Now we
consider the positive compact operator T ∗ T . Its eigenvalues are all non-zero, since
T ∗ T x = 0 implies kT xk2 = hT x, T xi = hT ∗ T x, xi = 0. By [Kri07b, 6.5.4] there is
an orthonormal P sequences   of Eigen-vectors en with Eigen-value 0 6= λ2n → 0 such
        ∗
that T T x = n λn hen , xien . Let fn := λ1n T en . Then a simple direct calculation
                     2
                                                          P
shows that the fn are orthonormal. Note that x = n hen , xien . Otherwise the
compact positive operator T ∗ T restricted to the orthogonal complement {ek : k}⊥
would have a unit Eigen-vector e with positive P Eigen-value λ. Which is impossible
by definition of the ek . So we obtain T x = n hen , xiλn fn .
Another way to prove this is to use the polar decomposition T = U |T |, see [Kri14,
7.24], where U is a partial isometry and |T | a positive and also compact operator.
The spectral P theorem for |T | gives an orthonormal family en and λ ∈ c0 , such
that T x = k λk hek , xiek . Applying U to this equation, shows that we may take
fk := U ek .
Proof. ( 1 ⇒ 2 ) Let
                                               (
                                                   xj /dj   for |dj | ≥ ε,
                   Tε : ` → `,    Tε (x)j :=
                                                   0        elsewhere.
Proof (See [MV92, Aufgabe 16.(3) p.392]). Note that kDk = kdk`∞ = sup{|di | :
i ∈ N} since kD(x)k = kd · xk`2 ≤ kdk`∞ · kxk`2 and D(e(k) ) = dk e(k) .
Thus an (D) ≤ kDn k = sup{|dk | : k ≥ n}, where Dn is the diagonal operator with
entries d · χ[n,∞) with dim((D − Dn )(`2 )) = n. Conversely, let dim T (`2 ) ≤ n. Then
      Pn
∃y = i=0 yi ei with kyk`2 = 1 and T (y) = 0. Thus kD − T k`2 ≥ k(D − T )yk`2 =
             Pn
kDyk`2 = ( i=0 |di yi |2 )1/2 ≥ min{|di | : i ≤ n} kyk`2 = |dn |.
     (1) λr (α) ∼
                = λr (β);
⇔ (2) λr (α) = λr (β) as lcs;
⇔ (3) λr (α) = λr (β) as sets;
⇔ (4) ∃C > 0 ∃n0 ∈ N ∀n ≥ n0 :          1
                                        C αn    ≤ βn ≤ Cαn .
Proof. (4 ⇒ 3) is obvious.
(3 ⇒ 2) apply the closed graph theorem using that convergence in λr implies
coordinatewise convergence.
(2 ⇒ 1) is obvious.
(1 ⇒ 4) Let Λα
             s := λr (α)s := λr (α)/ ker k ks for s < r.
                                   λr (β)
                                                   B    / λr (α)         A   / λr (β)
                       }}                   {{              ιss
                                                                  0           ##             !!
                Λβt0
                            B̃      / Λα0                                       / Λβs   Ã    / Λβt
                                       s                                                        8
                                                                  0
                                                            ιtt
Proof. By 3.14 λ∞ (β) has (DN). Indirectly, suppose λ0 (α) has (DN), i.e.
                ∃τ < 0 ∀t < 0 ∃T < 0 ∃C > 0 : kxk2t ≤ Ckxkτ kxkT .
                                                                              1
x := ej ⇒ e2tαj ≤ C eτ αj +T αj ≤ C eτ αj ⇒ 2t ≤                              αj   ln(C) + τ , limj αj = +∞ ⇒
t ≤ τ , a contradiction.
Reflexive spaces
Proof. Since every (F) space is (infra-)barrelled by 2.3 the result follows from
3.18 .
3.20 λp (A) ist reflexive for 1 < p < ∞ (See [MV92, 27.3 p.307]).
                     1.13 , 3.16
Proof. `p reflexive =       ⇒ λp reflexive.
                    =========
Montel spaces
E ∗ = Ẽ ∗ into which all polars U o for 0-nbhds U in E (or the completion Ẽ) with
their compact topology continuously embed and τc (E ∗ , Ẽ) is the topology of uniform
convergence on compact subsets of the completion Ẽ.
Proof. Since for 0-nbhds U in E (or Ẽ) the polar U o is σ(E ∗ , Ẽ) compact and
even τpc (E ∗ , Ẽ) = τc (E ∗ , Ẽ) compact by 3.4 , we have γ ≥ τc (E ∗ , Ẽ) ≥ σ(E ∗ , Ẽ).
By Grothendieck’s completion result (See [Kri14, 5.5.7]) Ẽ = (E ∗ , γ)∗ , hence γ
is compatible with the duality (E ∗ , Ẽ), i.e. coincides with the topology of uniform
convergence on the closed equicontinuous subsets in (E ∗ , γ)∗ = Ẽ (see [Kri14,
5.4.11]). Let C be set of these subsets. All of them are compact for τpc (Ẽ, (E ∗ , γ))
by 3.4 . The identity (Ẽ, τpc (Ẽ, (E ∗ , γ))) → Ẽ is continuous, since Ẽ carries
the topology of uniform convergence on the equicontinuous subsets (polars U o ) in
Ẽ ∗ = E ∗ and polars U o are γ-compact by definition. Thus the sets in C are compact
in Ẽ. Hence τc (E ∗ , Ẽ) ≥ γ.
                                       2.9.3                                      2.10
Proof. ( 4 ⇒ 3 ) b ∈ λ∞ (A) =   ⇒ W.l.o.g. bj > 0 for all j =====⇒ B := {x :
                            =====
                                                   3.17
kx/bk`1 ≤ 1} is bounded in E := λ1 (A) =  ⇒ B is weakly relatively compact in E
                                       ====
⇒ ∀k : ιk ◦ ιB : EB  E  Ek weakly compact. Define (for ` := `1 )
                             R : ` → EB ,        x 7→ b · x,
                             S : Ek → `,         [x] 7→ x · a(k) , and
                             D : ` → `,        x 7→ b · a(k) · x.
( 3 ⇒ 2 ) By 3.18 and 2.3 we have to show that bounded sets B in λp (A) are
relatively compact. W.l.o.g. B = Bbp with b ∈ λ∞ (A) by 2.10 . λ∞ (A) = c0 (A) ⇒
D from above (with ` := `p for p < ∞ and ` := c0 for p = ∞) is compact by 3.10
⇒ ∀k : ιk ◦ ιB compact ⇒ B relatively compact (cf. the proof of 3.31 ).
( 2 ⇒ 1 ) trivial.
( 2 ⇒ 5 ) since normed Montel spaces are finite dimensional by 3.23 and 4.171 .
Schwartz spaces
3.30 Lemma (See [MV92, 24.17 p.265], [Kri07a, 6.7], [Jar81, 17.1.7 p.370]).
An lcs is Schwartz iff for every continuous linear T : E → F into a normed space
F there exists a 0-neighborhood in E with precompact image in F .
3.36 Example of (FM), but not Schwartz (See [MV92, 27.21 p.319]).
                              (
                                (ki)k for j < k
A := {a : k ∈ N} with ai,j :=
       (k)             (k)
                                                ⇒ λp (A) is (F), (M), not (S)
                                kj    for j ≥ k
Claim: I is finite:
k := n + 1, j ≥ n + 1 ⇒
                                                          j
                                    (n)   (n+1)         n
                          εn+1 ≤ ai,j /ai,j      =
                                                      n+1
⇒ ∃j0 : I ⊆ N × {1, . . . , j0 }. Let 1 ≤ j ≤ j0 , k > max{j, n}, (i, j) ∈ I ⇒
                                        (ni)n   nn n−k
                      (n) (k)
                     ai,j /ai,j = (ki)k = kk i
                                                        for j < n(< k)
               εk ≤
                      (n) (k)           nj
                     
                         ai,j /ai,j = (ki)  k            for (k >)j ≥ n
⇒ I ∩ (N × {j}) is finite.
 3.28.6
     ⇒ λp (A) Montel (for all 1 ≤ p ≤ ∞).
======
Tensor products
In this section we introduce the projective tensor product as universal solution for
linearizing bilinear continuous maps and the injective tensor product as subspace
of te space of all bounded linear (or bilinear) operators. of locally convex spaces
in order to define nuclearity. Nuclear spaces are then deinied as those locally
convex spaces, where these to tensor product functors coincide. And we use these
tensor products to obtain descriptions for various types of vector valued summable
sequences.
                               E×F
                                          ⊗
                                                 / E⊗F
                                                            ∃!f˜ linear
                                  ∀f bilinear         % 
                                                      ∀G
The linear space E ⊗ F can be obtained as subspace of L(E, F ; K)∗ (the dual of
the bilinear forms) generated by the image of ⊗ : E × F → E ⊗ F ⊆ L(E, F ; K)∗
given by (x, y) 7→ ev(x,y) (See [Kri07a, 3.1 p.50]).
For locally convex spaces the solution of the corresponding universal problem for
(bi)linear continuous mappings is called projective tensor product E ⊗π F ,
it is the linear spaces E ⊗ F supplied with the finest locally convex topology for
which ⊗ : E × F → E ⊗ F is continuous. This topology exists since the union of
locally convex topologies is locally convex and E × F → E ⊗ F is continuous for the
weak topology on E ⊗ F generated by those linear functionals which correspond
to continuous bi-linear functionals on E × F . It has the universal property, since
the inverse image of a locally convex topology under a linear mapping T̃ is again a
locally convex topology, such that ⊗ is continuous, provided the associated bilinear
mapping T is continuous.
The space E ⊗πPF is Hausdorff, since the set E ∗ × F ∗ separates points in E ⊗ F :
Let 0 6= z =      k xk ⊗ yk be given. By replacing linear dependent xk by the
corresponding linear combinations and using bilinearity of ⊗, we may assume that
                                                                  P
Proof. Let z ∈ P       λ · (U ⊗ V ) with λ > 0. Then  P z = λ λk (uk ⊗ vk ) with uk ∈
P, vk ∈ V and
U                         k |λ
                             Pk | = 1. Hence z =         xk ⊗ vk , where xk = λλk uk , and
   k p U (x k )·pV (v k ) ≤     λ|λ k | = λ. Taking the infimum of all λ shows that pU ⊗V (z)
is greater or equal to the infimum on the right side.
Proof. Every compact set K in the Fréchet space E ⊗    ˆ π F is contained in the closed
                                                   ˆ π F by 3.6 . For this 0-sequence
absolutely convex hull of a 0-sequence (zk )k in E ⊗
we can choose kn strictly increasing, such that zk ∈ Un ⊗ Vn for all k ≥ kn , where
(Un )n and (Vn )n are countable 0-neighborhood bases of the topology    P of E and F .
For kn ≤ k < kn+1 we can choose finite (disjoint) sets Nk ⊆ N and j∈Nk |λj | = 1,
                                      P                                         F
xj ∈ Un and yj ∈ Vn such that zk = j∈Nk λj xj ⊗ yj . Let A := {xj : j ∈ k Nk }
                     F
and B := {yj : j ∈ k Nk }. These are formed by two sequences converging to 0,
and hence are precompact. Furthermore, each z ∈ K can be written as
                              ∞
                              X               X X
                         z=         µk zk =            µk λj xj ⊗ yj
                              k=0             k j∈Nk
      P                          P P                   P       P
with k |µk | ≤ 1 and hence k j∈Nk |µk λj | = k |µk | j∈Nk |λj | ≤ 1. From
this it easily follows that the series on the right hand side converges (even Mackey)
and hence z is contained in the closed absolutely convex hull of A ⊗ B.
                                          0
                                            finite : F 0 ∩P
                                                                    P
       (⇒) Otherwise, ∃U ∀F  P finite ∃F P                F = ∅ and n∈F
                                                                      P 0 xn ∈/ 2U
       (∃F1 , F2 ⊇ F : 4U 63 n∈F2 xn − n∈F1 xn = n∈F2 \F1 xn − n∈F1 \F2 xn ,
       now take F 0 := F2 \P F1 or F 0 := F1 \ F2 ). Since n xn is Cauchy, there is
                                                            P
                               n2
       some n0 such that n=n          xn ∈ U for all n2 ≥ n1 ≥ n0 . Let F0 := {n ∈
       N : n ≤ n0 } and F0 a corresponding set. We construct nk , Fk , and Fk0 6= ∅
                                    1
                           0
       where
       P      Fk+1 \ Fk = {n : nkP< n ≤ nk+1 } with nP                          k+1 ≥ nk ≥ n0 , so
          n∈Fk+1 \Fk  x n ∈ U , wheras        n∈Fk  0 x n   ∈/   2U    , hence    n∈Fk00 xn ∈
                                                                                            / U . The
       elements inP  the  sequence F0 , F0
                                          0
                                            , F 00
                                                0  ,  F1
                                                        0
                                                          , F 00
                                                              1  , . . . define a permutation  σ of N
       for which n xσ(n) is not Cauchy.
    • `1 [F ] := L(c0 , F ), the space of scalarly absolutely summable se-
      quences in F (See [Kri07a, 4.9] and [Jar81, 19.4.3 p.427]): Since the
      standard unit vectors ek generate a dense subspace in c0 every f ∈ L(c0 , F )
      is uniquely determined by its values       P fk := f∗ (ek ). Moreover, f is contin-
      uous=bounded iff {(y ∗ ◦ f )(x) =             j∈N xj y (fj ) : x ∈ c0 , kxk∞ ≤ 1} is
                                ∗     ∗            ∗
      bounded for each y ∈ F , i.e. {(xj y (fj ))j : x ∈ c0 , kxk∞ ≤ 1} is bounded
      in `1 , i.e. (y ∗ (fj ))j ∈ λ1 (c0 ) = `1 by 1.15.7 , i.e. (fj )j is scalarly absolutely
      summable.
This can be extended 1 < q < ∞:
Proof. We first show that the natural mapping `1c ⊗π F → `1c (N, F ), x ⊗ y 7→
(xj y)j∈N , is an isomorphism, where `1c is the dense subspace in `1 of finite sequences
and `1c (N, F ) the analogous
                       P (k) subspace    in `1 (N, F ). Since Rk ⊗π F ∼   = F k we have a
                                  (k)     1
bijection. Let z = k x ⊗ y            ∈ `c ⊗ F and p be a seminorm of F . For the
corresponding norm p̃ of `1c (N, F ) we have
                                             !
            X            X     X (k)             X X (k)
                                         (k)
   p̃(z) :=     p(zj ) =    p      xj y        ≤          |xj | p(y (k) ) ≤
              j            j             k                   j       k
                                                             (k)
                                                 XX                              X
                                             ≤             |xj | p(y (k) )   =       kx(k) k`1 · p(y (k) ),
                                                   k   j                         k
Taking the infimum of the right side over all representations of z shows that p̃ ≤ pπ ,
where pπ is projective tensor norm formed from k k`1 and p, see 3.38 .
Conversely    each z = (zj )j ∈ `1c (N, F ) can be written as image of the finite sum
                                                               1
P
  j ej ⊗ zj , where ej denotes the standard unit vector in ` . Thus we have for the
                π
tensor norm p that
                              X                      X
                     pπ (z) ≤     kej k`1 · p(zj ) =   p(zj ) = p̃(z)
                                     j                           j
L(E, F ) (x 7→ x∗ (x)y)
                                    3.43                    X               X       
                                     = p̃(x − y) + p
                                  ====
                                  =                                  yk −          yk ≤ ε + ε,
                                                              k∈K2          k∈K1
                            P
which shows that K 7→             k∈K      xk is a Cauchy-net.
Since `1 [F ] ∼
              = L(c0 , F ), it is complete for complete F .
Proof. By 3.48 we have that `1 ⊗ε F ∼        = c00 ⊗ε F embeds into L(c0 , F ), the space
of scalarly absolutely summable sequences. Obviously λ ⊗ y ∈ `1 ⊗ F is contained
in `1 {F } ⊆ `1 hF i. We show that `1c ⊗ F = K(N) ⊗ F ∼      = F (N) is dense in `1 hF i
with respect to the structure inherited from ` [F ]. So let x ∈ `1 hF i and consider
                                                    1
xn := x|[0,...,n−1] ∈ F n ⊆ F (N) ⊆ `1 [F ]. We             n          1
                                             P claim that x → x in ` [F ]: Let p be a
continuous seminorm on F . Since K 7→ k∈K xk is Cauchy, we have for K = R:
                                                                 m
             3.43      nX                                 o     nX                                 o
p̃(x − xn ) =
            ====
               = sup              |y ∗ (xk )| : |y ∗ | ≤ p = sup   |y ∗ (xk )| : |y ∗ | ≤ p, m ≥ n
                           k≥n                                         k=n
                n X                              X                                  o
           = sup    y ∗ (xk ) +                        y ∗ (xk ) : |y ∗ | ≤ p, m ≥ n
                     m≥k≥n                     m≥k≥n
                    y ∗ (xk )>0               y ∗ (xk )<0
                n X                      o     n X           o
           ≤ sup y ∗ xk : |y ∗ | ≤ p, m ≥ n + sup y ∗ xk : . . .
                         m≥k≥n                                                      m≥k≥n
                        y ∗ (xk )>0                                                y ∗ (xk )<0
                  n X                                    o
           ≤ 2 sup p  xk : K 0 finite, K 0 ∩ [0, n − 1] = ∅ ≤ 2ε
                           k∈K 0
for n sufficiently
                P large. In the complex case we have to make a more involved
estimation for k>n |y ∗ (xk )|. Let P := {z ∈ C : <z > 0 and − <z < =z ≤ <z}.
For every z 6= 0 there is a unique j ∈ {0, 1, 2, 3} with ij z ∈ P . Then |z| ≤ 2<(ij z) ≤
2|z|. Thus we can split the sum into 4 parts corresponding to j ∈ {0, 1, 2, 3}, where
Operator ideals
All these classes are operator ideals, since for A, B ∈ L they are closed under
T 7→ A ◦ T ◦ B. For approximable this follows from an+m (R ◦ S) ≤ an (R) · am (S),
see 3.53 below, for the others from S∗ (`p {E}) ⊆ `p ({F }) and S∗ (`p [E]) ⊆ `p [F ]
(since `q (N, ) and L(`, ) are obviously functorial).
Proof. By 3.40 the elements of E ∗ ⊗  ˆ π F are those of the from                       λn x∗n ⊗ yn with
                                                                                  P
                                                                                    n
x∗ , y bounded sequences and λ ∈ `1 .
It can be shown that the converse is valid as well, see [Jar81, 20.2.3 p.454].
hence T ∈ Ap .
Proof. Let e1 , . . . , en be an algebraic basis of E. For the weakly compact unit ball
K of E ∗ we consider the continuous map f : K n → K, (x∗1 , . . . , x∗n ) 7→ | det(x∗j (ei ))|.
Let (x∗1 , . . . , x∗n ) be a point where it attains its maximum. Since the ei are linearly
independent this maximum is positive. Hence there is a unique solution with xj ∈ E
of the equations
                                  X
                                     x∗j (ei )xj = ei for 1 ≤ i ≤ n.
                                     j
Thus | det(yj∗ (xi ))| ≤ 1. Choosing yj∗ = x∗j for all j 6= k shows that |yk∗ (xk )| ≤ 1
and hence kxk k ≤ 1. From 1 = x∗j (xj ) ≤ kx∗j k kxj k we conclude that kxj k = 1 =
kx∗j k.
Proof. See [Jar81,   P19.8.5 p.442]. Let T ∈ A1 (E, F ). We have to show that it can
be written as T = n λn x∗n ⊗ yn with x∗n ∈ oE ∗ , yn ∈ oF and λ ∈ `1 .
Let ε > 0. Choose Tn with dim Tn (E) ≤ 2n and kT − Tn k ≤ (1 + ε) a2n (T ). Let
Dn := Tn+1 − Tn . Then dn := dim Dn (E) ≤ 3 · 2n and since an (T ) → 0 we have
                               P∞                             P∞ Pdn
kT − Tn k → 0, hence T = n=0 Dn . By 3.56 we have T = n=0 j=1              λn,j x∗n,j ⊗
yn,j , with x∗n,j ∈ oE ∗ , yn,j ∈ oF and 0 ≤ λn,j ≤ kDn k. We estimate as follows
        dn
       XX                 X                        X
                 λn,j ≤        dn kDn k ≤ 3            2n (kTn+1 − T k + kTn − T k)
        n j=1             n                        n
                              X
                                    2n (1 + ε) a2n+1 (T ) + a2n (T )
                                                                       
                       ≤3·
                               n
                              X                                            X
                       ≤3·          2n+1 (1 + ε) a2n (T ) ≤ 22 3 (1 + ε)        2n−1 a2n (T )
                               n                                            n
                                           X
                          2
                       ≤ 2 3 (1 + ε)           an (T )    (since an (T ) is decreasing)
                                           n
A map T : E → F between Banach spaces is nuclear iff there are continuous linear
operators S : E → `∞ and R : `1 → F such that T factors as diagonal operator
D : `∞ → `1 with diagonal d ∈ `1 , i.e.
                                        E
                                                T       /F
                                                          O
                                        S                     R
                                            
                                        `∞              / `1
                                                D
Proof. (⇒) Let T be represented by k dk x∗k ⊗yk P  with kx∗k kE ∗ ≤ 1, kyk kF ≤ 1 and
                                    P
     1                  ∗
d ∈ ` . Then S(x) := (xk (x))k and R((µk )k ) := k µk yk define linear operators
S : E → `∞ and R : `1 → F of norm ≤ 1 and T = R ◦ D ◦ S, where D : `∞ → `1
denotes the diagonal operator, with diagonal (dk )k .
(⇐) Since the nuclear operators form an ideal, it is enough to show that such
                                                  ∞   1
diagonal operators D : (µk )k 7→ (d
                                  Pk µk )k ,∗D : ` → ` are nuclear, which is clear
since they can be represented by k dk xk ⊗ yk , where x∗k := ek ∈ `1 ⊆ (`∞ )∗ and
yk := ek ∈ `1 .
Proof. Let T be a nuclear mapping. Since the compact mappings form an ideal,
we may assume by 3.59 that T is a diagonal-operator `∞ → `1 with absolutely
summable
P           diagonal (λk )k . Such an operator is compact, since the finite sub-sums
  k≤n λ k e k ⊗ ek define finite dimensional operators, which converge to T uniformly
on the unit-ball of `∞ .
                                    p
As in 3.42 one can show that ` {F } is complete (see [Jar81, 19.4.1 p.426]). For
p > 1 it is however not isomorphic to `p ⊗      ˆ π F . Otherwise we would obtain for
                 ˆ π `p = `p {`p } = `p (N × N), which is not the case..
E = `p , that `p ⊗
On `p [E] we consider the operator norm of L(`q , E) (see 3.43 ):
                             nX               1/p                       o
            k(xk )k kε := sup      |x∗ (xk )|p      : x∗ ∈ E ∗ , kx∗ k ≤ 1 .
                                    k
It is obvious, that the inclusion `p {E} → `p [E] is a contraction (i.e. has norm ≤ 1).
for all finite sequences (x )k . The smallest such R is the norm of T∗ : `p [E] →
                                     (k)
Proof.
(⇒) By 3.61 we have that T∗ is continuous, and hence the required property holds
with R := kT∗ k and all (even the infinite) sequences in `p [E].
                                                                            (k) p 1/p
                                                                Pm               
(⇐) For x = (x(k) )k ∈ `p [E] we have k(T x(k) )k kπ = supm         k=1 kT x   k      ≤
R · k(x(k) )k≤m kε ≤ R · k(x(k) )k kε < ∞ and hence (T x(k) )k ∈ `p {F }.
(N1 ⊆ S1 ) by 3.59 , since any diagonal operator D : `∞ → `1 with diagonal d ∈ `1
is 1-summing:
  m
                                             (k)                                   (k)
  X                    XX                               X                X                                             X
        kDx(k) k`1 =                     |dj xj | =              |dj |        |xj | ≤ kdk`1 sup                             | evj (x(k) )|
                                                                                                               j
  k=1                      k     j                          j             k                                            k
                                        nX                                       o
                   ≤ kdk`1           sup   |x∗ (x(k) )| : x∗ ∈ (`∞ )∗ , kx∗ k ≤ 1 .
                                             k
Also Np ⊆ Nq can be shown under the same assumption, see [Jar81, 19.7.5 p.437].
                                                                               1         1            1
Proof. Let T ∈ Sp and let r ≥ 0 be given by                                    r   +     q    =       p.   Let λk := kT xk kq/r .
                       r/q
Then kT xk k = λk and hence kT (λk xk )kp = kλk T (xk )kp = λpk · kT xk kp =
          q
kT xk kp( r +1) = kT xk kq and so the Hölder’s inequality (cf. the proof of 3.53 )
shows that
   X             1/p X                  1/p
         kT xk kq     =      kT (λk xk )kp      = (T (λk xk ))k π
       k                             k
                                                                                                      X                           1/p
                           ≤ kT kSp · (λk xk )k                  ε
                                                                     = kT kSp · sup                            |x∗ (λpk xk )|p
                                                                                       kx∗ k≤1             k
                                             X             1/r                   X                          1/q
                                                                                               ∗
                           ≤ kT kSp ·                 λrk            · sup                   |x (xk )|q
                                                 k                    kx∗ k≤1          k
                                             X                      1/r                    X                            1/q
                           ≤ kT kSp ·                 kT xk kq                · sup                    |x∗ (xk )|q
                                                 k                            kx∗ k≤1             k
              P                     1/r        P                      1/p−1/q
Dividing by       k   kT xk kq              =          k    kT xk kq                         gives
                   X                       1/q                                                          1/q
                               kT xk kq              ≤ kT kSp · sup                    |x∗ (xk )|q                 .
                           k                                             kx∗ k≤1
Thus T ∈ Sq by 3.62 .
Proof. Note that the right hand side is nothing else but M · kδ(x)kp , where
δ : E  C(oE ∗ ).
(⇐) If µ is a probability measure (i.e. µ(oE ∗ ) = 1) with that property, then
X                 Z     X                                  nX                        o
   kT xk kp ≤ M p          |x∗ (xk )|p dµ(x∗ ) ≤ M p · sup    |x∗ (xk )|p : x∗ ∈ oE ∗ .
 k                   oE ∗   k                                               k
So T ∈ Sp by 3.62 .
(⇒) Let T ∈ Sp (E, F ). For every finite sequence x = (x1 , . . . , xm ) in E let fx ∈
C(oE ∗ ) be defined by
                       X               X            X                                 
 fx (x∗ ) := kT kpSp ·   |x∗ (xi )|p −   kT xi kp =    kT kpSp · |x∗ (xi )|p − kT xi kp .
                      i                          i                      i
                                 (N)
The set B := {fx : x ∈ E } is convex in C(oE ∗ ). In fact let x and y be two
finite sequences in E and λ + µ = 1 with λ ≥ 0 and µ ≥ 0. Let z be the sequence
obtained by appending µ1/p y to λ1/p x. Then
                              X                                
          (λfx + µfy )(x∗ ) =   λ kT kpSp |x∗ (xi )|p − kT xi kp +
                                    i
                                            X                                
                                        +    µ kT kpSp |x∗ (yj )|p − kT yj kp
                                             j
                                  X
                                =  (kT kpSp )|x∗ (λ1/p xi )|p − kT (λ1/p xi )kp +
                                    i
                                            X
                                        +    (kT kpSp )|x∗ (µ1/p yj )|p − kT (µ1/p yj )kp
                                             j
                                  X
                                =  (kT kpSp )|x∗ (zk )|p − kT (zk )kp = fz (x∗ ).
                                   k
By 3.62 we have that supx∗ ∈oE ∗ fx (x∗ ) ≥ 0. Thus the open set A := {f ∈
C(oE ∗ ) : supx∗ ∈oE ∗ f (x) < 0} is disjoint from B. So by the consequence [Kri07b,
7.2.1] of Hahn-Banach there exists a regular Borel measure µ on oE ∗ and a constant
α such that hµ, f i < α ≤ hµ, gi for all f ∈ A and g ∈ B. Since 0 ∈ B we have α ≤ 0.
Since A contains the constant negative functions we have α = 0 and µ(oE ∗ ) > 0.
Without loss of generality we may assume kµk = 1. Hence for every x ∈ E we have
                                 Z                               
                  0 ≤ hµ, fx i =        kT kpSp |x∗ (x)|p − kT xkp dµ(x∗ )
                                  oE ∗
                          p
and thus kT xk ≤ kT kSp · oE ∗ |x∗ (x)|p dµ(x∗ ).
                p
                               R
                                             E
                                                            T     /F
                                                                    O
                                          
                                                          i   / L2 (µ)
                                        C(K)
Proof. (⇐) It is enough to show that the canonical mapping ι : C(K) → L2 (µ) is
absolutely 2-summing. So let δx be the point measure at x. Then for finitely many
fk ∈ C(K) we have
          X                 Z X                   Z X
              kι(fk )k2`2 =     |fk (x)|2 dµ(x) =      |δx (fk )|2 dµ(x)
                     k                      K       k                                        K      k
                                                                 nX                                       o
                                        ≤ µ(K) · sup                        |ν(fk )| : ν ∈ C(K)∗ , kνk ≤ 1 ,
                                                                                       2
                                                 E
                                                                            T                 /F
                                                 _                                           < O
                                                                                                    R
                                                             S         ##              R
                                                                            H q              /H
                                                δ
                                                                                       1        O
                                                                                                    P
                                                                                            "
                                            C(oE ∗ )
                                                                            ι              / L2 (µ)
where H denotes the closure of the image of ι◦δ in L2 (µ). The operator T factorizes
via a continuous linear operator R : H → F , since kT xk ≤ M · kι(δ(x))k`2 for some
M > 0. Using the ortho-projection P : L2 (µ) → H we get the factorization
R ◦ P ◦ (ι ◦ δ) = R ◦ ι ◦ δ = R ◦ S = T .
Proof.
For orthonormal families ek and fk and T ∈ S2 we have by 3.61
                     X                  1/2                                   X                         1/2
                              kT ek k2          ≤ kT kS2 · sup                             |hx, ek i|2                  ≤ kT kS2 .
                         k                                         kxk≤1           k
3.67 Overview. One has the following inclusions for 1 < p < q < ∞:
                               3.57
                                                    / N 1  
                                                                                3.62
                                                                                                        / S1
       A 1                                              _                                                 _
          _
obvious                      [Jar81, 19.7.5 p.437]                                                                   3.63
                                                                                                        
                                                     N p 
                                                                 [Jar81, 19.7.8 p.438]
       A p                                                                                             / Sp
          _                                              _                                                      _
obvious                      [Jar81, 19.7.5 p.437]                                                                   3.63
                                                                                                        
                                                     Nq 
                                                                 [Jar81, 19.7.8 p.438]
       Aq                                                                                               / Sq
For Hilbert spaces one has the following results for 1 < p < ∞:
                                              
                                       N 1 
              [Jar81, 20.2.5 p.456]
       A1                                                                / S1
          _                               _
                                                                             [Jar81, 20.5.1 p.467]
             [Jar81, 20.5.1 p.467]              [Jar81, 20.5.1 p.467]
       A2                             Np                                  Sp
          _                                                                 _
                  [Jar81, 20.5.1 p.467]
                                                                           
      A∞                               N∞                                / S∞
Nuclear spaces
In this section we characterize nuclear spaces in several ways and we prove their
inheritance properties We show that the nuclear (Fréchet) spaces are exactly the
(closed) subspaces of products of (countable many) copies of s.
3.68 Definition.
A linear mapping T : E → F between lcs is called nuclear operator (See [Jar81,
17.3 p.376], [Kri07a, 5.6]) iff there exist {an : n ∈ N} ⊆ E ∗ equicontinuous, B a
Banach disk, bn ∈ B, and λ ∈ `1 with
                                      ∞
                                      X
                            Tx =            λn an (x) bn for all x ∈ E.
                                      n=1
This is exactly the case, iff there is an absolutely convex 0-neighborhood U ⊆ E and
a Banach disk B ⊆ F , such that T factors over a nuclear mapping T̃ : E    cU → FB ,
i.e.
                                             E
                                                     T    /F
                                                            O
                                              
                                             E
                                             fU           / FB
                                                     T̃
The nuclear mappings form an ideal: For composition from the left side with some
R replace bn by R(bn ), and from the right side replace an by an ◦ R = R∗ (an ) (Note
that an ∈ U o ⇒ R∗ (an ) ∈ (R−1 (U ))o ).
Note, that as for any bifunctor we denote with T ⊗ G the morphism T ⊗ idG .
                                                      o
                             P
Proof. We may represent T =   n λn an ⊗ bn with an ∈ U for some 0-nbhd.
U and bn ∈ B, a Banach-disk. Let V ⊆ F and W ⊆ G be 0-nbhds and let
                                      Pk
ρ := sup{qV (b) : b ∈ B}. For w =         j=1   xj ⊗ zj ∈ E ⊗ G we get
                           k
                           X                         X
            (T ⊗ G)(w) =         T (xj ) ⊗ zj =            λn an (xj ) bn ⊗ zj
                           j=1                       j,n
                           X                X
                      =          λ n bn ⊗        an (xj ) zj and hence
                           n                j
                           ∞
                            X           k
                                        X            
     πV,W (T ⊗ G)(w) = πV,W    λ n bn ⊗    an (xj )zj
                                    n=1                   j=1
                           ∞
                           X                           k
                                                      nX                                         o
                      ≤          |λn | qV (bn ) sup              an (xj )z ∗ (zj ) : z ∗ ∈ W o
                           n=1                             j=1
                            ∞
                            X                   n   k
                                                    X                                                o
                      ≤ρ           |λn | sup              x∗ (xj )z ∗ (zj ) : x∗ ∈ U o , z ∗ ∈ W o
                            n=1                     j=1
                      ≤ ρ kλk`1 εU,W (w).
Proof. We give the proof for 1 ≤ p ≤ 2 only. For the general case one needs in
addition that Sp ◦ Sq ⊆ Sr (see [Jar81, 19.10.3 p.446]) and N1 ⊆ Np ⊆ Sp (see
[Jar81, 19.7.5 p.437] and [Jar81, 19.7.8 p.438]).
( 1 ⇒ 2 ⇒ 3 ) and ( 5 ⇒ 4 ⇒ 3 ) are obvious by 3.42 and 3.50 .
( 3 ⇒ 6 ) From ( 3 ) we obtain that `1 hẼi ∼
                                            = `1 {Ẽ}. Thus for every U ⊆ E there
exists a V ⊆ E and a δ > 0 such that πU ≤ δ εV , where
                                          X
                            πU (xk )k :=      pU (xk )
                                                      k
( 6 ⇔ 7 ⇔ 8 ) Now let us show that for all mentioned ideals it is the same to
assume that the connecting mappings belong to them.
In fact, we have A1 ⊆ N1 ⊆ S1 ⊆ Sp ⊆ S2 for 1 ≤ p ≤ 2 by 3.57 , 3.62 , 3.63 .
The composite of three S2 maps belongs to A2 , since the following diagram shows
that it factors over a map between Hilbert spaces (see 3.65 ) of class S2 ⊆ A2 (by
 3.66 ):
                 E3
                                   S2
                                                    / E2     S2
                                                                  / E1               S2
                                                                                               / E0
                                                     ;                                          ;
                                 #                          S2 ⊆A2
                                                                                    #
                                L2 (µ3 )                                         / L2 (µ1 )
Since (A2 )2 ⊆ A1 by 3.53 we have that (S2 )6 ⊆ A1 . Now choose for a given
seminorm p successively p6 ≥ p5 ≥ · · · ≥ p1 ≥ p such that the connecting maps all
belong to S2 . Then the connecting mapping E  gp6 → Ep belongs to A1 .
                                                     f
Proof. Since `1 {E} and `1 hEi are Fréchet spaces it follows from the closed graph
theorem that the identity is a homeomorphism.
Proof.
                                      Q
( 1 ) A typical seminorm on E :=    Ei is of the form p : x 7→ maxi∈A pi (xi ), where
                                          i
A is finite and pi are seminorms on Ei . Obviously Efp = Q       ^
                                                             i∈A (Ei )pi . For every pi
we can find a seminorm qi ≥ pi such that the canonical mapping (^            ^
                                                                    Ei )qi → (Ei )pi
                                                      
                                       Fq           / Fp
Since the bottom arrow is precompact, the same is true for the top arrow.
Now for nuclear spaces. The corresponding proof will not work for nuclear map-
pings, but for absolutely summing mappings, since the ideal S1 is obviously injec-
tive, i.e. if T : E → F1 ,→ F belongs to S1 and F1 is a closed subspace of F , then
T : E → F1 belongs to S1 , since `1 {F1 } = `1 {F } ∩ F1N .
( 3 ) First for`
               Schwartz spaces. Recall that a basis of seminorms on a countable co-
product E = k Ek is given by supk pk , where the pk run through the seminorms of
Ek and supk pk : (xk )k 7→ supk pk (xk ). By assumption we can find seminorms qk ≥
pk such that the connecting map Tk : (Ek )qk → (Ek )pk is precompact. Furthermore
we may assume that its norm is less than 21k , by replacing qk with       k
                                                                       ` 2 kTk kq∼
                                                                                 k.
Now the following diagram shows that we get a natural bijection k (Ek )pk =
 `                                               `
( k Ek )supk pk which is an isometry if we supply k (Ek )pk with the norm (xk )k 7→
sup{pk (xk ) : k}, and analogously for the qk .
                                                         `
                  ker(supk pk )                             k ker(p )
                       _                                      _ k
                   `                                                                        ` 
                      k   Ek                                                                   k   Ek
                           supk pk                                         w
                                                                               `        pk
                                             '
                                                 Ro                     Rg
                                                        sup                    k
                                             7
                                                                `
                                                                k
                                                                                              
                                                         ∼
               ( k Ek )supk pk o                                                    /
                `                                        =                              `
                                                                                             k (E  k )pk
                                                                                                 
                  N                  /E                                                        //F
                                                                    π
p̃◦π p̃
                                                                    '
                                                                    7   R xe
                                                      p̃◦π                              p̃
                                                                                                  
                                                                    ∼
                                     Ep̃◦π /                                                   / / Fp̃
                                                                    =
Another argumentation for the same result would be an application of the isomorphy-
theorem F/ ker p̃ ∼
                  = (E/N )/(ker(p̃ ◦ π)/N ) ∼
                                            = E/ ker(p̃ ◦ π).
Hence we have the diagram: Note that connecting morphisms are always quotient
              Fq̃       / Fp̃ maps, since the projections E → Eq are. So the di-
            == O           O    agonal arrow is open, since it is up to the vertical
             ∼
             =          ∼
                        =       isomorphism the connecting map Eq → Eq̃◦π . Hence
                                the image of the unit ball in Eq is a 0-nbhd in Fq̃
        / / Eq̃◦π   / / Ep̃◦π .
 Eq                             whose image is precompact in Ep̃◦π ∼
                                                                   = Fp̃ .
Now in order that this proof works also for nuclear spaces, we can use the following:
It is enough to consider the situation, where E  E1 → F is nuclear, E  E1 is a
quotient map and E a Hilbert space (by 3.72 ). But then the sequence E2  E 
E1 splits, where E2 is the kernel of the quotient map E  E1 , and hence E1 → F
can be written as E1 ,→ E  E1 → F and thus is nuclear.
                                                                     
( 5 ) Use that E  fq = E   f̃ ,
                            q̃            ker
                                            _ q        E ∩ ker q̃       / ker q̃
                                                                                 _
where q̃ denotes the unique                                                      
extension of q to a seminorm               E                                 / / Ẽ
on Ẽ.
                                                          ) u
                                                          6Rh
                                                      q                q̃
                                               
                                                      q                q̃
                                                                                      
                                                                         / / Ẽq̃
                                           Eq
Proof. ( 2 ⇒ 3 ) λ∞ (A) (N) ⇒ λ∞ (A) is (S) by 3.60 and hence (M) by 3.31
⇒ λ∞ (A) = c0 (A) by 3.28 .
( 1 ⇒ 6 ) and ( 3 ⇒ 6 ) follows for p < ∞ from 3.58 for the diagonal operators
D∼ = ιm     p                           ∞
      k on ` resp. c0 , hence also for λ (A) = c0 (A).
( 2 ⇒ 1 ) obvious.
( 4 ⇔ 5 ⇔ 6 ) follows from 1.25
It was open for a long time whether all (NF) spaces have a Schauder-basis. The
first counter-example was given in [MZ74], see [Jar81, 21.10.9 p.516] for a sim-
pler counter-example. Rather recently it was shown in [DV00] that the complete
ultra-bornological nuclear space C ω (R, R) of real-analytic functions does not have
a Schauder-basis.
 3.74 .
                                                P p(e )
(⇐) For any continuous seminorm p choose p0 with j p0 (ejj ) < ∞. By 1.19 there
exists a p00 and C > 0 such that
                                   ∀x ∀j : |ξj (x)| p0 (ej ) ≤ C p00 (x),
where ξj are the coefficient functionals. Then ξj factors (for p0 (ej ) 6= 0) over
ιp00 : E  Ep00 to a ξ˜j ∈ (Ep00 )∗ . Thus D : Ep00 → Ep , x 7→ j=0 ξ˜j (x) ιp (ej ), is a
                                                               P∞
nuclear mapping, since
    ∞                            ∞                                                 ∞
    X                            X                                                X   p(ej )
           kξ˜j k kιp (ej )k =         sup{|ξj (x)| : p00 (x) ≤ 1} p(ej ) ≤ C                  < ∞.
     j=0                         j=0                                              j=0
                                                                                      p0 (ej )
                                              00
Thus the connecting mapping ιpp is nuclear, since it equals D:
                                         ∞
                                         X                              ∞
                                                                        X
                                                   ˜
                                                            
                   (D ◦ ι )(x) =
                          p00                      ξj ιp (x) ιp (ej ) =
                                                        00                ξj (x) ιp (ej )
                                         j=0                               j=0
                                               ∞
                                              X                     00
                                       = ιp           ξj (x) ej = (ιpp ◦ ιp00 )(x).
                                               j=0
                                               log(n)
       1. λ∞ (α) is nuclear ⇔ supn               αn     < ∞.
       2. λ0 (α) is nuclear ⇔          limn log(n)
                                              αn       = 0.
                                          3.74
                                                                                  e−tαn < ∞
                                                                          P
Proof. ( 1 ) λ∞ (α) nuclear ⇐  ⇒ ∃t > 0 : C :=
                            ====                                              n
                                n
                    −tαn
                                X                        log(n)   log(C)     log(C)
 (⇒)       ⇒ ne            ≤          e−tαj ≤ C ⇒               ≤        +t≤        + t =: D.
                                j=1
                                                           αn       αn         α0
                 log(n)              1  X
 (⇐)       sup          ≤ D ⇒ e−Dαn ≤ ⇒   e−2Dαn < ∞
            n      αn                n  n
(2)
                            3.74                 X
     λ0 (α) is nuclear ⇐  ⇒ ∀t > 0 :
                       ====                             e−tαn < ∞, now proceed as in ( 1 ).
                                                  n
Proof.
                3.70.8 , 3.53
                      ⇒ ∀k > 0 ∃pk ≥ p, cont. Hilbert SN : ιppk ∈ A1/k (Epk , Ep )
     E (N ) ===========
           As in 3.33 :         (Ep )∗ ∼
                                       = EU∗ o , (Epk )∗ ∼
                                                         = EU∗ ko with Uk := {x : pk (x) ≤ 1}
                                                      j=1
                                          (k)
                                und     (ej )j   ONB in EU∗ o , da ιk inj.
                                                                    (j)
Let (ẽn )n be the diagonal-enumeration of (ei )i,j , drop recursively those which
are linearly dependent on ealier ones, and apply Gram-Schmidt to obtain an ONB
                      C2           2     (2k C)2   Ck
                 ≤       2k
                            ke n k U o ≤     k
                                                 =: k for all large n
                    (jn )                  n       n
      ⇒ (en )n fast falling in E ∗ .
Proof. (⇒)
     E (N ) ⇒ ∃(pi )i∈I basis of Hilbert SN, let Ui := {x : pi (x) ≤ 1}
          3.79
           ⇒ ∀i ∃(ein )n fast falling ONB in EU∗ io
        ====
        =
             ⇒ ∀k ∃Vk : {nk ein : n} ⊆ (Vk )o , i.e. ∀x ∈ Vk : sup |nk ein (x)| ≤ 1
                                                                               n
             ⇒ fi : E → s, x 7→         (ein (x))n ,   is continuous
                                             I
             ⇒ f := (fi )i∈I : E → s is continuous.
     ∀x ∈ E : evx = διi (x) is continuous on the Hilbert space EU∗ io = (EUi )∗
[Kri07b, 6.2.9]
           ⇒ ∃x∗ ∈ EU∗ io ∀y ∗ ∈ EU∗ io : hx∗ , y ∗ i = evx (y ∗ ) = y ∗ (x)
============
                           X                 X
           ⇒ kfi (x)k20 :=     |ein (x)|2 =         |hx∗ , ein i|2 = kx∗ k2E ∗ o = k evx k2 = pi (x)2
                                                                                     U
                                                                                         i
                                   n                   n
                                                                  I
             ⇒ f is an embedding onto f (E) ⊆ s .
                          3.73.1 , 3.73.2
(⇐) s (N), E ,→ sI =          ⇒ E (N).
                   ============
3.81 Nuclear Fréchet spaces (See [MV92, 29.9 p.346], [Jar81, 21.7.3 p.502]).
E is (NF) ⇔ E is isomorphic to a closed linear subspace of sN .
3.82 Remark.
Note that we have the following implications under the assumption on the bottom
of the arrow:
formed in cbs), since then B = ιj (Bj ) with Bj ⊆ Ej bounded for some j and hence
B o ⊇ ((ιj )∗ )−1 (Bjo ), but not in general.
Note, that the representation E = limB EB of a bornological space is such a regular
                                            −→
inductive limit.
Note furthermore, that if ( )∗ is supplied with the bornology of equicontinuous sets,
then (colimj Ej )∗ = limj Ej∗ in cbs: In fact, let U o be a typical bounded set in E ∗ ,
i.e. U ⊆ E := colim Ej a 0-nbhd.QThen each Uj := (ιj )−1 (U ) ⊆ Ej is Q      a 0-nbhd
and the image of U o in limj Ej∗ ⊆ j Ej∗ is contained in the bounded set j (Uj )o ,
since |(ιj )∗ (u∗ )(uj )| = |u∗ (ιj (uj ))| ≤ 1 for u∗ ∈ U and uj ∈ Uj .
We want to consider inheritance with respect to L or L. If E 6= {0}, then F is
a topological direct summand in L(E, F ) and in L(E, F ): In fact, let 0 6= x ∈ E
and x∗ ∈ E ∗ with x∗ (x) = 1. Then ι : K → E, λ 7→ λ · x has x∗ : E → K as
                                       = L(K, F ) → L(E, F ) has L(ι, F ) : L(E, F ) →
left-inverse, and hence L(x∗ , F ) : F ∼
L(K, F ) ∼= F as left-inverse and the same works for L. And similary, F ∗ = L(F, K)
is a topological direct summand in L(F, E), via L(F, ι) with left-inverse L(F, x∗ )
and the same way F 0 = L(F, K) is a topological direct summand in L(F, E). Thus
in order to show some topological property for L(E, F ) it is reasonable to assume
the property for F and for E ∗ . Consequently a first step in answering this question
is to consider inheritance with respect to ( )∗ .
In this section we consider completeness conditions for the (strong) dual and we
introduce the classes of infra-c0 -barrelled and of c0 -barrelled space in this connec-
tion.
Recall the Banach Steinhaus Theorem [Kri07b, 5.2.6], by which L(E, F ) is sequen-
tially complete if E is barrelled and F is sequentially complete:
Let (fn )n be a Cauchy-sequence in L(E, F ). Then (fn )n∈N is Cauchy pointwise,
hence pointwise convergent to some function f∞ : E → F , which is continuous by
the Banach Steinhaus Theorem. For each bounded B ⊆ E and closed absolutely
convex 0-nbhd U ⊆ F there exists an n with (fn0 − fn00 )(B) ⊆ U for n0 , n00 ≥ n.
Taking for each x ∈ B the pointwise limit for n00 → ∞ yields (fn0 − f∞ )(x) ∈ U .
Thus fn → f∞ in L(E, F ).
It is σ(`1 , c0 )-σ(E, E ∗ )-continuous, since x∗ ◦ T = (x∗ (xn ))n∈N ∈ c0 ⊆ (`1 )∗ for each
x∗ ∈ E ∗ . Since o`1 is σ(`1 , c0 )-compact, its image A0 is σ(E, E ∗ )-compact, abso-
lute convex, and contains the xn . Hence their absolutely convex hull is relatively
compact for σ(E, E ∗ ).
( 3 ⇒ 2 ) Let xn be a Mackey-0-sequence. By 3 the σ(E, E ∗ )-closure C of the
absolutely convex hull of {xn : n ∈ N} is σ(E, E ∗ )-compact and hence σ(E, E ∗ )-
complete. Since closed absolutely convex sets in E are σ(E, E ∗ )-closed, C is even
complete in E by the next lemma 4.3 . Since {xn : n ∈ N} ∪ {0} is compact, its
closed absolutely convex hull is precompact (by the proof of [Kri07b, 6.4.3]) and
thus compact by completeness of C.
( 2 ⇒ 1 ) Suppose there is a closed absolutely convex bounded set B, such that EB
is not complete. Choose x̃ ∈ E      fB \ EB and iteratively construct a sequence (xi )i∈N
in EB such that
                                            n
                                           X             1
                                      x̃ −     xi    ≤ n+2
                                           i=1
                                                  B    3
                   P∞
and hence x̃ = i=1 xi . Now let yn := 2n xn ∈ EB and observe that
                                  Xn                n−1
                                                     X           2 n+1
                         n
             kyn kB ≤ 2       x̃ −     xi    + x̃ −      xi      ≤        → 0.
                                   i=1
                                           B
                                                     i=1
                                                             B      3
                P∞ −n
Hence x̃ =         n=1 2     yn is in the closure of the absolutely convex hull of the
(Mackey-)0-sequence (yn ) in the Banach space E           fB . Consider the initial topol-
ogy τ with respect to the inclusion ι : EB  E. Since B is closed in E, it is closed
        0
for τ 0 , thus (EB , k kB ) has a basis of τ 0 -closed sets. By the lemma 4.4 below the
extension ι̃ : E  fB  Ẽ is injective. Since ι̃(x̃) = P∞ 2−n yn is in the (by 2 )
                                                               n=1
compact closure of the absolutely convex hull of {yn : n ∈ N} ⊆ B in E, we get
x̃ ∈ E ∩ B ⊆ EB , a contradiction.
Proof. Cf. the proof of the corollary to 3.17 : Let (xi )i be a τ -Cauchy net,
which is τ 0 -convergent to x∞ , and let U ∈ U. Thus there exists an i such that
xi0 − xi00 ∈ U for all i0 , i00  i. For fixed i0 , the net i00 7→ xi0 − xi00 ∈ U is τ -Cauchy
and τ 0 -convergent to xi0 − x∞ . Since U is τ 0 -closed we get xi0 − x∞ ∈ U , i.e. (xi )i
is τ -convergent to x∞ .
Proof. Let x̃ ∈ ker T̃ ⊆ Ẽ. Thus there exists a net (xi )i in E convergent to x̃ in
Ẽ and hence T (xi ) = T̃ (xi ) → T̃ (x̃) = 0. Thus xi → 0 with respect to τ 0 and then
x̃ = τ - limi→∞ xi = 0 by 4.3 .
For any functor F we have a natural morphism F(lim Xi ) → lim F(Xi ) by the
universal property of the right side.
                                                                    F (Xα )
             XT i
                         Xα
                                     / Xj         F(Xi )                            / F(Xj )
                                       J            T a                                = J
                                                                  pri        prj
                                                                 limk F(Xk )
                pri                 prj             F (pri )           O            F (prj )
                                                                        !
limk Xk F(limk Xk )
4.9 Lemma. The dual of reduced projective limits (See [Kri07a, 3.27]).
The functor ( )∗ : lcs → cbsop preserves reduced projective limits, where E ∗ is again
considered with the bornology of equicontinuous sets.
have dense image the dual cone pr∗i : Ei∗  E ∗ consists of injective mappings only.
Let x∗ ∈ E ∗ be given. Then there has to exist an i and a 0-neighborhood Ui ⊆ Ei
with x∗ (pr−1i (Ui )) ⊆ D := {λ ∈ K : |λ| ≤ 1}. In particular x (ker pri |E ) = 0 and
                                                                 ∗
hence there exists a linear xi : pri (E) → R with x = xi ◦ pri = pr∗i (x∗i ). Since
                                ∗                       ∗     ∗
in Ui ⊆ (Ei ) on the closure Ei of pri (E). Thus the union of all images pr∗i ((Ei )∗ )
      o        ∗
Remark.
We have shown the following implications, where the dotted ones are valid under
In this section we describe the property (DF), which the strong duals of Fréchet
spaces have, and which garantees in turn that their strong dual is Fréchet.
theorem.
Thus W := n Bno ⊆ E ∗ satisfies W = (Wo )o and absorbs each Uio , hence Wo is
              T
bounded and thus (Wo )o = W ⊆ V is a 0-nbhd. in E ∗ . This shows infra-countably-
barrelledness. Since E ∗ is complete, countably-barrelledness follows by 4.14 .
We will give an example (of a non-distinguished λ1 (A)) in 4.25 for which these
equivalent conditions are not satisfied.
In [Jar81, 13.4.2 p.279] it is shown that for metrizable E the bornologification
β(E ∗ , E)born of β(E ∗ , E) is β(E ∗ , E ∗∗ ).
4.17 Definition. (DF)-spaces (See [Jar81, 12.4.1 p.257], [MV92, 25.6 p.279]).
An lcs E is called (DF)-space, iff it has a countable base of the bounded sets
and is infra-countably-barrelled (see 4.12 ), i.e. every bornivorous subsets which is
the intersection of countable many closed absolutely convex 0-neighborhoods is a
0-neighborhood.
An lcs E is called (df)-space iff it has a countable base of its bornology and is
infra-c0 -barrelled.
4.18 Proposition.
Proof. ( 1 ) This is 4.15 , since for the bornological space E the dual E ∗ = E 0 is
complete, and a countable basis of the bornology is given by the family Uno , where
{Un : n ∈ N} is a 0-nbhd basis of E.
( 2 ) By assumption a (DF)-space E has a countable base {Bn : n ∈ N} of bornology
and hence (E ∗ , β(E ∗ , E)) a countable 0-nbhd basis {Bno : n ∈ N}, so is metrizable.
Let (x∗n )n be Cauchy in E ∗ . ThenT x∗n converges pointwise to some linear x∗∞ : E →
K. Let Vn := {xn }o and V∞ := n∈N Vn . Since (x∗n )n is Cauchy, it is bounded,
                  ∗
In this section we describe the duals of Köthe sequence spaces and characterize
reflexivity (and the Montel property) of λ∞ (A) (and of c0 (A)). We also give an
example of a Köthe sequence space, whose strong dual fails to be (infra-)barrelled.
                                           P∞
Proof. By 1.22 we have λ∗ ∼
                                                     
                          = λ1 (λ) via x 7→ j=0 xj yj ← y.
By 2.10 the sets Bbp := {x : kx/bk`p ≤ 1} (resp. Bbo : Bb∞ ∩ c0 (A)) for b ∈ λ∞ (A)
(w.l.o.g. ∀j : bj > 0) form a basis of the bornology on λp (A) (resp. c0 (A)). Let
y ∈ λ1 (λ) ∼
           = λ∗ and 1q + p1 := 1, then by 1.23 the Minkowski-functional p(Bbp )o is
given by
                      ∞             nXx                        o
                                          j
                     X
 sup |y(x)| = sup        xj yj = sup        bj yj : kx/bk`p ≤ 1 = ky · bk`q =: kykb .
x∈Bbp          x∈Bbp j=0               j
                                         bj
4.21 c0 (A)∗∗ ∼
              = λ∞ (A) (See [MV92, 27.14 p.314]).
Proof. By 4.20 the family (k kb )b∈λ∞ (A) is a basis of seminorms for c0 (A)∗ and
                 = y ∈ KN : kykb :=
                   n                                              o
         c0 (A)∗ ∼
                                     X
                                          |yj bj | < ∞ ∀b ∈ λ∞ (A) . ⇒
                                            j∈N
             ∞
                              yj bj is in c0 (A)∗∗ .
                          P
(⊇) ∀b ∈ λ (A): y 7→       j∈N
i.e. x ∈ λ∞ (A).
Thus c0 (A)∗∗ = λ∞ (A) as linear spaces and, by the closed graph theorem, also as
lcs.
For 1 < p < ∞ the space λp (A) is reflexive by 3.20 and thus distinguished by
 4.10 . What about λ1 (A)?
        ∀ε : N → R+ ∃b ∈ λ∞ (A) : (Bb1 )o ⊆
                                            D[         E
(1)                                             εk Uko           (by 2.10 ).
                                                                       abs.conv.
                                                           k
 1.24
=  ⇒ 2k ξ k ∈ 2k D1k Uko = εk Uko for k ≤ n0 ⇒
====
                                    0               0
                                  n         n
                                  X
                                         k
                                            X 1 k k D[        E
                          ξ=            ξ =    k
                                                 2 ξ ∈ εk Uko
                                              2                 abs.conv.
                                  k=1              k=1                      k
                                                               4.23
Proof. Suppose λ1 (A) is distinguished =  ⇒ ∀k : Dk := 1, C := 2, n := 0
                                       ====
              ∃D0 : N → R+ ∃n0 ∀i, j : min{2ai,j , sup ai,j /Dk0 } ≤ max0 ai,j
                                                                 (0)            (k)            (k)
                                                                            k            k≤n
a contradiction to ( 2 ) for m := n0 .
Then λ1 (A) is not distinguished, so (λ1 (A))∗ is (DF) but not infra-barrelled.
Proof. In the proof of 3.31 we have shown that each bounded set in a Schwartz
space is precompact.
By definition an lcs E is Schwartz iff
                       ∀U ∃V ∀ε > 0 ∃M finite : V ⊆ M + ε U.
Thus every Schwartz space is quasi-normable. And if every bounded set B is
precompact, then there is a finite set M ⊆ E such that B ⊆ M + ε U , and we have
the converse implication.
4.29 Counter-example.
Note that E := RX is Schwartz and even nuclear for all sets X by 3.73.1 .
However, if X is uncountable then the dual E ∗ = R(X) is not quasi-normable (hence
neither Schwartz nor nuclear).
              P quasi-normable. Recall that the typical seminorms on R
Suppose E ∗ were                                                           (X)
                                                                               are
given by f 7→ x cx |fx | with cx ≥ 0, see [Kri07b, 4.6.1]. Thus for the seminorm
with cx := 1 for all x there exist another seminorm given by some corresponding
cx > 0 such that for all ε > 0 there is some bounded set Bε with
            n X                  o          n              X          o
(1)          f:      cx |fx | ≤ 1 ⊆ Bε + ε · f : kf k`1 :=   |fx | ≤ 1 .
                  x                                               x
                                            1
For some δ > 0 the set I := {x : cx ≤       δ}
                                            has to be (uncountably) infinite. Now
choose ε = 2δ . Then Bε is contained in a finite subsum, so there is some x ∈ I
with prx (Bε ) = {0}. Since δ · ex is an element of the left hand side of 1 , there
has to exist a b ∈ Bε and an f with kf k`1 ≤ 1 with δ · ex = b + ε · f and hence
prx (b) ≥ δ − 2δ > 0, a contradiction.
Proof. This inclusion is continuous (and then an embedding, since EV∗ o  Eβ∗ is
continuous) iff ∀λ > 0 ∃B bounded closed absolutely convex with U o ∩ B o ⊆ λV o .
(⇐) (B + U )o ⊆ U o ∩ B o ⊆ λV o ⇒ V ⊆ λ((B + U )o )o = λ B + U ⊆ λB + 2λU .
(⇒) V ⊆ B + λU ⇒ 2V o ⊇ 2(B + λU )o ⊇ B o ∩ (λU )o ⇒ 2λV o ⊇ U o ∩ λB o .
and such that each finite (resp. bounded) subsets of E is absorbed by (and hence
contained in) An for some n (See [Jar81, 12.3 p.253]).
Proof. It is easy to see, that these absolutely convex hulls form a basis for a locally
convex topology τ A , which is finer than the given one and which coincides with
                                                                               S the
given one on each An : In fact τ A →SE is continuous (Uk := U and use E = k Ak )
and An → τ A is continuous (An ∩ k Ak ∩ Uk ⊇ An ∩ An ∩ Uk ).
Let now τ be another topology with that property and V be an absolutely convex
                          S n there is a 0-nbhd Un with An ∩ Un ⊆ V , hence the
0-nbhd for τ . Thus for each
absolutely convex hull of n An ∩ Un is contained in V , i.e. τ A ≥ τ .
Remains to show T that the two bases are equivalent:
(⊇) Let V := k≥0 (Ak + Uk ). Choose Vk with ((Vk )o )o ⊆ Uk and Uk0 ⊆ i<k Vi .
                                                                             T
                                               0
                                                    T          T           T
Since Am ⊆ Ak for all k ≥ m, we get Am ∩ Um       ⊆ k≥m Ak ∩ k<m Vk ⊆ k Ak +
                                                                           0
          
((Vk )o )o ⊆ V , thus V contains the absolutely convex
                                                               S
                                                  S     hull of m Am ∩ Um     .
(⊆) Let now U be the absolutely convex hull of m≥1 Am ∩ Um . Let kn := 2n + 1.
Then An+1 ⊆ 2−n AkT   n
                        and there exists Vn with Vn ⊆ 2−n Ukn and 2((Vn+1 )o )o ⊆ Vn .
We claim that V := n≥0 (An +Vn+2 ) ⊆ U : Let x ∈ V , i.e. x = yn +vn with yn ∈ An
                                                 Pn
(thus y0 = 0) and vn ∈ Vn+2 . Thus x = vn + i=1 xi , where xi := yi − yi−1 . So
xi ∈ Ai + Ai−1 ⊆ Ai+1 and xi = vi−1 − vi ∈ Vi+1 − Vi+2 ⊆ Vi . Hence xi ∈ Ai+1 ∩ Vi
for all 1 ≤ i ≤ n. By the properties of A := (An )n we have x ∈ An for some n,
hence x − yn = vn ∈ 2An ∩ Vn+2 ⊆ An+2 ∩ Vn+1 , thus
       n
       X                  n
                          X                                      n+1
                                                                 X
 x=          x i + vn ∈         Ai+1 ∩ Vi + An+2 ∩ Vn+1 ⊆              2−i · (Aki ∩ Uki ) ⊆ U.
       i=1                i=1                                    i=1
Proof. (⇒) We use 3.33 , so for every 0-nbhd B o ⊆ E ∗ we have to find a 0-nbhd
                  ∗∗
C o with B oo ⊆ EC oo being compact. Since E is reflexive by 3.22 , this means that
τS is Schwartz by 3.33 , since for every polar Ao of an E-0-sequence x∗n there exist
λn → ∞ such that yn∗ := λn x∗n is still a 0-sequence in EU∗ o and hence (Ao )o (the
σ = β compact closure of the absolutely convex hull of {x∗n : n ∈ N}) is compact in
      o , where B := {yn : n ∈ N}.
  ∗                    ∗
E(B o)
The first statement is sometimes also called Schur’s lemma, see [Jar81, p.218].
There is however no universal (F S)-space, see [Jar81, 10.9 p.218].
Proof. In fact by 4.36 Schwartz spaces have a basis of 0-neighborhoods given by
the polars V := {x∗n : m ∈ N}o of E-0-sequences (x∗n )n∈N in E ∗ . Since pV (x) =
sup{|x∗n (x)| : n ∈ N}, the map T : x 7→ (x∗n (x))n∈N defines a continuous linear
map from E → c0 and factors over ιV : E  Q   EV as T = T̃ ◦ ιV with an isometric
mapping T̃ : EV Q  ,→ c0 . Since (ιV )V : E ,→ V EV is an embedding, we get an
embedding E ,→ V c0 .
It is easy to see that T̃ : (EV )S → (c0 )S is continuous:  Let S ∈ L(E, F ) and yn∗ → 0
         ∗ ∼     ∗               −1
                                     {yn : n ∈ N}o = {S (yn ) : n ∈ N}o and S ∗ (yn∗ ) → 0
                                       ∗               ∗ ∗
                                                     
in (FV ) = FV o . Then S
      ∗            ∗
in ES ∗ (V o ) ⊆ E(S −1 V )o , i.e. S ∈ L(ES , FS ).
It is an embedding, since T̃ ∗ : `1 → (EV )∗ = EV∗ o is a quotient map between
Banach spaces, hence every 0-sequence in the image is the image of a 0-sequence in
the domain.
Remains to show that E embeds into the reduced projective system of the (EV )S :
Obviously E = ES → limV (EV )S is continuous.
                           ←−
Conversely, let V := {x∗n : n ∈ N}o                          "
                                                   M m E
            ∗                 ∗
with xn → 0 in (EV 0 ) for some V              0           _
and A := {x∗n : n ∈ N} ∪ {0} ⊆
(EV 0 )∗ . Since ι∗V 0 (A)o = ι−1  V 0 (Ao ), we
                                                                          &                     ιV 0
                                   −1            lim       E      o id
                                                                        lim     (E       )
have EV 0 ⊇ Ao = ιV 0 (ιV 0 (Ao )) =             ←−V V                  ←−V V S
ιV 0 (ι∗V 0 (A)o ) = ιV 0 (V ), a 0-nbhd in
(EV 0 )S with ιVV (Ao ) = ιVV (ιV 0 (V )) = ιV " "  x
                       0         0
                                                                                                     
                   V 0
ιV (V ), thus ιV : (EV 0 )S → EV is con-              E   V gg
                                                                o         (E   V S   0 )  o o id    o E   V0
tinuous and hence also ιV from E ⊆
limV (EV )S into EV .                                                        ιV
                                                                                0
←−                                                                            V
Thus the identity from (the subspace E of) limV (EV )S → limV EV is continuous.
                                                 ←−                    ←−
     ]
That (c          ∞     ∞ 1
        0 )S = (` , µ(` , ` )) can be found in [Jar81, 10.5.3 p.206].
        DX             E DX                        E X
             xn an , ek =       hx∗ , en i en , ek =        hx∗ , en i hen , ek i = hx∗ , ek i,
          n                   n                          n
Proof. Let (xj )j∈J be a Cauchy net in E and U be a 0-nbhd basis of absolutely
convex sets. We claim the following:
                       ∃n0 ∈ N ∀U ∈ U ∀j ∈ J ∃i  j : xi ∈ U + An0 .
Otherwise, ∀n ∃Un ∃jn ∀j 0  jn : xj 0 ∈
                                                                    T
                                            / Un + An . Put V := n (Un+1 + An ). Let
x ∈ (Um + Am−1 ) ∩ Am−1 and n ≥ m, then x = x + u with x, x0 ∈ Am−1 and
                                                            0
                              0
u ∈ Um , thus    T u = x − x ∈ 2Am−1 ⊆ An , i.e. x = 0 + x ∈ Un+1 + An . Therefore
V ∩ Am = n<m (Un+1 + An ) ∩ Am is a 0-nbhd in Am , hence a 0-nbhd in E by
 1 . Since (xj )j∈J is Cauchy, there exists j ∈ J such that xj 0 − xj 00 ∈ V for all
j 0 , j 00  j. Since (An )n∈N is absorbing there exists an n with xj 00 ∈ An−1 and hence
xj 0 ∈ xj 00 + V ⊆ An−1 + V ⊆ An−1 + (Un + An−1 ) ⊆ Un + An for all j 0  j, a
contradiction.
Now consider the net x̃ : J × U → An0 ⊆ E, which assigns to each (j, U ) an element
x̃j,U := xi − u ∈ An0 with i  j and u ∈ U . This net is Cauchy and hence converges
to some x∞ in E by 2 , since for U ∈ U there exist W ∈ U with 3W ⊆ U and
j ∈ J such that xi0 − xi00 ∈ W for all i0 , i00  j. So
     x̃j 0 ,U 0 − x̃j 00 ,U 00 = (xi0 − u0 ) − (xi00 − u00 ) = (xi0 − xi00 ) − u0 + u00 ∈ 3W ⊆ U
for all (j 0 , U 0 ), (j 00 , U 00 )  (j, W ) and hence u0 ∈ U 0 ⊆ W , u00 ∈ U 00 ⊆ W , i0  j 0  j,
and i00  j 00  j.
It follows that (xj )j∈J converges to x∞ : For any U ∈ U there exist W ∈ U with
4.42 Proposition
(See [Sch12, 2.14 p.38], [Flo71], [Rai59], cf. [Jar81, 12.5.2 p.263]).
An (LB)-space is complete if and only if it is quasi-complete.
Proof. (⇐) Let (E, τ ) = limn (En , τn ) denote an (LB)-space, and let Bn := oEn
                          −→
be the closed unit ball of the Banach space (En , τn ). We will apply 4.41 for
An := 2n Bn .
Since, for each n ∈ N, we may assume Bn to be continuously injected into Bn+1 ,
the sequence (An )n∈N is an absorbing sequence.
To prove 1 , let V ⊆ E be an absolutely convex set such that V ∩ An is a
0-neighborhood of (An , τ |An ) for each n ∈ N and thus also a 0-neighborhood
of (An , τn |An ). Since An is a closed 0-neighborhood of (En , τn ), we see that V ∩ An
and hence V ∩ En ⊇ V ∩ An are also 0-neighborhoods of (En , τn ). This holds
for all n ∈ N, which means that V has to be a 0-neighborhood of the inductive
limit (E, τ ).
Remains to show condition 2 , i.e. that each τ -Cauchy net contained in some An
converges in E. But this is clear by the quasi-completeness of E since the sets An
and hence their Cauchy nets are bounded. .
Let E = limn En be a reduced inductive limit with compact connecting mappings
          −→
Tn : En  En+1 , i.e. which map some absolutely convex 0-nbhd Un ⊆ En to
a relative compact subsets of Tn (Un ) ⊆ En+1  E. Let Bn be the (compact)
closure of the bounded set Tn (Un ) ⊆ E. Thus Tn factors over the normed space EB
generated by B and this space is complete by 4.3 (for τ := pB and τ 0 := E|B ), since
B is compact and hence complete. Thus we can rewrite E as reduced projective
limit of a sequence of Banach spaces with compact connecting homomorphisms.
4.44 Proposition
(See [Jar81, 12.5.9 p.266], [Woz13, 4.30 p.75], [MV92, 25.20 p.57]).
Let F be an lcs. Then
   1. F is the dual of an (FS)-space;
 ⇔ 2. F is a bornological (DF)-space where each bounded set is relative compact
      in FA for some bounded Banach-disk A;
Proof. A typical 0-nbhd in Eβ∗ is B o for some bounded (absolutely convex and
closed) B ⊆ E.
(⇐) By assumption there is some bounded B 0 ⊇ B such that E       fB → E
                                                                       g B 0 is nuclear.
Then its dual mapping (EB 0 ) → (EB ) is nuclear by 4.45 . Now note that (E ∗ )B o
                             ∗          ∗
                                oo
EB → EB 0 (since B = E ∩ B ) is absolutely summing and a composition of 6 such
 f      g
maps is nuclear, see the proof of ( 6 ⇒ 7 ) in 3.70 .
The (DF) condition can be weakend to (df) in this lemma using the same proof,
but with the sharpening mentioned in 4.18 instead of Proposition 4.18 .
The first ≤ holds, since (E, η(E ∗ , E)) := limU EU∗ o and EU∗ o  Eβ∗ is continuous.
                                             −→
The second one holds, since id : E       → ES is continuous, so the injective limit
η(E ∗ , E) has more steps than η(E ∗ , ES ).
The first equality holds since ES is (S): In fact, EU∗ o  (E ∗ , σ(E ∗ , E)) is continuous,
so id : η(E ∗ , E) := limU EU∗ o 
                                 → γ(E ∗ , E) (recall 3.24 ) is continuous. Conversely,
                      −→
let E be Schwartz, i.e. for every 0-nbhd U there exists a 0-nbhd V with U o ⊆ EV∗ o
compact by 3.33 , and hence the induced (compact) topology from EV∗ o on U o
coincides with the restriction of σ(E ∗ , E), and the inclusion from U o with this
topology into EV∗ o is continuous. Thus γ(E ∗ , E)     → η(E ∗ , E) is continuous.
The last equality holds, since ES is a complete Schwarz space, hence semi-Montel
                                  f
by 3.31 , thus the closed bounded subsets coincide with the compact ones.
(β(E ∗ , E) ≥ η(E ∗ , ES )) Since E is (DF), Eβ∗ is (F), by 4.18 . Let x∗n → 0 in Eβ∗ ,
then x∗n is Mackey-convergent by 2.2 , so there exists a sequence λn → ∞ with
λ2n x∗n → 0 in Eβ∗ . Since the (DF)-space E is infra-c0 -barrelled, λ2n x∗n ∈ U o for some
0-nbhd U ⊆ E. Thus λn x∗n is an E-0-sequence and hence W := {λn x∗n : n ∈ N}o
is a 0-nbhd for τS (see 4.36 ). Since x∗n → 0 in EW    ∗
                                                          o and hence in lim       E ∗ =:
      ∗                         ∗          ∗
                                                                            −→W W o
η(E , ES ), the inclusion β(E , E) → η(E , ES ) is (sequentially-)continuous.
The particular case follows, since by the universal property E → ES ,→ E
                                                                       fS factors
                         ∗           ∗           ∗ f
over E ,→ Ẽ. Thus β(E , E) ≤ β(E , Ẽ) ≤ β(E , ES ).
                          (F )
Proof. (nuclear ⇒ co-nuclear) Let pn be an increasing sequence of seminorms
defining the topology of E such that the connectingPmorphisms Tn : Epn+1 → Epn
are nuclear, and hence admit representations Tn = k λn,k x∗n,k ⊗ yn,k with x∗n,k ∈
o(Epn+1 )∗ , yn,k ∈ o(Epn ) and λn := k |λn,k | < ∞. Now let B ⊆ E be a closed
                                          P
bounded
      n disk, σn := sup{p                 o ∈ B}, let ρn := max{σn , λn σn }, and set
                                n+1 (b) : b
C := x ∈ E : qC (x) := n p2nn(x)
                           P
                                  ρn  ≤  1  . For x ∈ B we have pn (x) ≤ pn+1 (x) ≤ σn ,
        P pn (x)      P σn
hence n 2n ρn ≤         n 2n σn = 1, i.e. B ⊆ C. Furthermore C is bounded since
pn (C) ≤ 2n ρn . The connecting morphism EB → EC is absolutely summable, since
for arbitrary finitely many xi ∈ EB ⊆ E we have
X              X                  XX
                                           pn λn,k x∗n,k (xi ) yn,k
                                                                    
    pn (xi ) =    pn (Tn (xi )) ≤
 i                  i                                    i   k
                  X                  X                                              X
             ≤           |λn,k |                 |x∗n,k (xi )|   ≤ λn     sup           |x∗ (xi )|
                                                                             o
                                                                        x∗ ∈Un+1
                   k                     i                                          i
                                         X                                          X                              X
                                                     ∗
             ≤ λn         sup                     |x (xi )| ≤ λn σn sup                 |x∗ (xi )| ≤ ρn sup            |x∗ (xi )|.
                        x∗ ∈σ   n   Bo       i                            x∗ ∈B o   i                    x∗ ∈B o   i
                                          1 pn (xi )
       P                     P                                    P ∗             ∗    o
Thus       i qC (xi )    =           n,i 2n ρn               ≤ sup  i |x (xi )| : x ∈ B                   and hence the
identity EB → EC is absolutely summing by 3.62 . Since                                          S16   ⊆ A1 ⊆ N we may
assume that it is even nuclear, and hence E is co-nuclear.
           (DF )
(nuclear ⇒ co-nuclear) By 4.47 Eβ∗ = Ẽβ∗ , so we may assume that E is a com-
plete nuclear (DF). That Ẽ is (DF) can be seen as follows: By 4.47 we have
β(E ∗ , Ẽ) = β(E ∗ , E) and hence is metrizable and Ẽ has a basis of its bornology
formed by closures of bounded sets in E, since for every bounded B̃ ⊆ Ẽ we find
a bounded set B ⊆ E such that the 0-nbhds B o ⊆ B̃ o and hence B̃ ⊆ ((B̃)o )o ⊆
             Ẽ
(B o )o = B . That Ẽ is quasi-c0 -barrelled is obvious (recall [Kri07b, 4.10.3]).
Let {Bn : n ∈ N} be a basis of the bornology consisting of closed absolutely convex
sets with Bn+1 ⊇ 2Bn . Put En := EBn . Since E is complete (S) hence semi-(M)
and thus semi-reflexive, EB ∼  = ((E ∗ )B o )∗ via x 7→ δ(x)|(E ∗ )Bo :
This mapping is onto, for let λ : (E ∗ )B o → K
be continuous and linear, and x ∈ E be such                                   E∗         EO
                                   ∗∗             o
that δx := δ(x) = λ ◦ ιB ∈ E , i.e. δx (B ) =
                            o                                       ι B o
                                                                                       ι O
                                                                                   B ∗  B
                                                                  xx 
                                                                                 (ι )
{δx (x∗ ) = x∗ (x) : x∗ ∈ B o } is bounded by                                  
                                                            ∗
                                                         (E )B o         / (EB )  ∗
                                                                                        EB
C := kλk and thus x ∈ C (B o )o = C B ⊆ EB .
It is also injective, for let x ∈ EB be such that                                δx
δ(x)|(E ∗ )Bo = 0, hence 0 = δ(x) ◦ (ιB )∗ = δx :
                                                                       λ
                                                                             & 
E ∗ → K, hence x = 0.
                                                                              K
We claim that `1 {E} = cbs- lim `1 {En }, i.e. every bounded S ⊆ `1 {E} is contained
                                −→
and bounded in `1 {En } for some n (recall 3.41 ):
                                                                       P∞     (n)
Suppose indirectly, that for each n we find x(n) ∈ S with πn (x(n) ) := k=0 kxk kn >
2n . So there exists a finite set Fn ⊆ N with k∈Fn kxk kn > 2n . Choose ak ∈ Bno
                                             P         (n)                  (n)
                (n) (n)
with k∈Fn |ak (xk )| > 2n . Then
      P
Thus the sequence (ak )k formed by all these finite subsequences (ak )k∈Fn for n ∈ N
                                                                         (n)
The canonical map `1 [En ] → `1 [E] is continuous and `1 [E] = `1 {E} by 3.70 , so
the image of S := o `1 [En ] is bounded in `1 {E} and, by what we have just shown,
even bounded in `1 {En0 } for some n0 ≥ n, i.e. the connecting mapping En → En0
is absolutely summing, hence E is co-nuclear by 4.46 .
          (F )
(nuclear ⇐ co-nuclear) By assumption and 4.18.1 E ∗ is a nuclear (DF)-space.
Hence by the second part E ∗∗ is nuclear and so is E as a subspace by 3.73.2 .
         (DF )
(nuclear ⇐ co-nuclear) By assumption and 4.18.2 E ∗ is a nuclear Fréchet space.
Hence by the first part E ∗∗ is nuclear. In order to apply 3.73.2 it remains to show
that δ : E → E ∗∗ is an embedding, i.e. E is infra-barrelled: The bounded=pre-
compact (since E ∗ is (S)) sets in E ∗ are contained in the bipolar of some 0-sequence
in E ∗ by 3.6 and, since E is (df) and hence quasi-c0 -barrelled, the 0-sequences
are equicontinuous, hence the topology of E (which is that of uniform convergence
on equicontinuous sets) coincides with that induced from E ∗∗ .
This proof works also for (df) instead of (DF), however the last argument shows,
that (co-)nuclear (df) spaces are infra-barrelled and in particular (DF) spaces.
Proof. Since strict inductive limits are regular this is immediate by 4.46 .
Proof. ( 1 ⇒ 2 ) is trivial.
( 2 ⇒ 1 ) A typical 0-neighborhood in LB (E, F ) is given by NB,V := {T : T (B) ⊆
V } with B ∈ B and V a 0-neighborhood in F . Let ιV : F  FV be the canonical
surjection. Since FV is a normed space ιV ◦ T : E → F  FV ,→ F̃V can be
uniformly approximated on B with respect to pV : FV → K by finite dimensional
operators E → F̃V by 2 . Since FV is dense with respect to pV in F̃V we may
assume that the finite operators belong to L(E, FV ). Taking inverse images of the
vector components, we may even assume that they belong to L(E, F ).
( 1 ⇒ 3 ) and ( 3 ⇒ 4 ) are trivial.
Let E be complete and assume that the equivalent statements of 4.50 are true
for some bornology B. And w.l.o.g. let B ∈ B be absolutely convex. Since the
identity on E can be approximated uniformly on B by finite dimensional operators,
we conclude that the inclusion EB → E can be approximated by finite dimensional
operators EB → E uniformly on the unit ball of EB . Hence it has to have relatively
compact image on the unit ball by the following lemma 4.51 , i.e. B has to be
relatively compact.
4.51 Lemma.
The set K(E, F ) of compact operators from a normed space E into a complete
space F is closed in L(E, F ).
                                                   Q f
Proof. To see this use that F = limV Ff       V ⊆    V FV , hence a subset K of F
                                      ←−
is relatively compact iff ιV (K) is relatively compact in Ff V for all V . Now let
Ti ∈ K(E, F ) converge to T ∈ L(E, F ) = L(E, F ). Then the ιV ◦ Ti ∈ K(E, Ff     V)
converge to ιV ◦ T in L(E, FfV ). Since F
                                        f V is a Banach spaces it can be shown as in
[Kri07b, 6.4.8] that ιV ◦ T ∈ K(E, FV ). Hence ιV (T (oE)) is relatively compact in
                                     f
FfV and thus T (oE) is relatively compact in F .
4.52 Definition.
A complete lcs is said to satisfy the approximation property iff the equivalent
statements in 4.50 are true for the bornology B = cp of all relatively compact
subsets of E. A non-complete space E is said to have the approximation property,
iff its completion Ẽ has it. Note that the finite dimensional operators may be taken
in L(E, E) in this situation.
Note that the topology of F ⊗ε E is by definition 3.44 initial with respect to the
inclusion F ⊗ E ,→ Lequi (F ∗ , E) and has in fact values in L((F ∗ , σ(F ∗ , F )), E) ⊆
L(Fγ∗ , E).
Proof. Note that F ⊗ E is mapped into L(Fγ∗ , E), since for y ∈ F we have
δ(y) ∈ (Fγ∗ )∗ by [Kri14, 5.5.7].
(⇐) Consider the following commuting diagram:
By assumption for F := Eγ∗ the inclined arrow
on the left hand side has dense image. The ar-               Eγ∗ ⊗ E            / Lcp (E, E)
row on the right hand side is an embedding, since                                     pP
(Eγ∗ )∗γ → Ẽ = E is a continuous bijection and the
                                                                          
equi-continuous subsets in (Eγ∗ )∗γ are exactly the
                                                                   Lequi ((Eγ∗ )∗γ , E)
relatively compact subsets of Ẽ = E by 4.54 .
(⇒) Let T ∈ L(Fγ∗ , E) and let a 0-neighborhood NV o ,U in this space be given. Since
T is continuous on the compact space (V o , σ(F ∗ , F )), we have that K := T (V o )
is compact in E. By assumption E ∗ ⊗ E is dense in Lcp (E, E). Hence there
exists a finite dimensional operator S ∈ L(E, E) with (idE −S)(K) ⊆ U . Then
S ◦ T : Fγ∗ → E → E is finite dimensional and since (Fγ∗ )∗ = F̃ by [Kri14, 5.5.7] it
belongs to F̃ ⊗ E and (T − S ◦ T )(V o ) = (id −S)(K) ⊆ U . Thus T − S ◦ T ∈ NV o ,U .
Hence F̃ ⊗ε E is dense in Lequi (Fγ∗ , E) and, since F ⊗ E is dense in F̃ ⊗ε E, it is
also dense in Lequi (Fγ∗ , E).
Proof. We may assume that all Ej and E is complete (since taking completions
commutes with reduced projective limits, see [Jar81, 3.4.6 p.63]). Let K ⊆ E
be compact and U ⊆ E a 0-nbhd, w.l.o.g. of the form ι−1 k (Uk ) for some k ∈ J and
0-nbhd Uk ⊆ Ek . By reducedness Fk := ιk (E) is dense in Ek hence has the approxi-
mation property. So there are ai ∈ Ek∗ and xi ∈ E such that (idFk −S)(ιk (K)) ⊆ Uk
          Pn
for S := i=1 ai ⊗ ιk (xi ). Thus (idE −S̃)(K) ⊆ U for the finite dimensional oper-
           Pn
ator S̃ := i=1 ι∗k (ai ) ⊗ xi .
Let us consider E ∗ ⊗
                    ˆ ε F now. If F is complete and satisfies the approximation
property, then Eγ ⊗ε F ∼
                ∗ˆ
                        = Lequi ((Eγ∗ )∗γ , F ) by 4.55 .
Proof. In the first statement the first isomorphism follows from the definition
 3.44 of E ⊗ε F ⊆ Lequi (Eγ∗ , F ) ,→ L(E ∗ , F ) and the approximation property (that
it hold also if E instead of F satisfies the approximation property follows from
 4.58 ). And the second one follows, since Montel spaces are barrelled by 3.22 and
 3.18 and since Eγ∗ = τc (E ∗ , Ẽ) = β(E ∗ , E) by 3.24 and E being semi-Montel.
Proof. This follows, since the strong dual Eβ∗ of a Montel space E is Montel
by 4.27 . Note that a Montel-space E is reflexive by 3.22 , i.e. (Eβ∗ )∗β = E.
Furthermore Eβ∗ = Eβ0 is complete, provided E is bornological.
Proof. ( 1 ) Recall that we have shown in 4.59 that for complete spaces we have
 ˆ εF ∼
E⊗     = L(Eβ∗ , F ) provided E satisfies the approximation property, is Montel and
 ∗
Eβ is bornological. These conditions are satisfied if E is a nuclear Fréchet space by
4.57 , 3.60 , 3.31 , and 4.39 .
( 2 ) Recall that we have shown in 4.60 that for complete spaces Eβ∗ and F we
         ˆ εF ∼
have Eβ∗ ⊗    = L(E, F ) provided Eβ∗ satisfies the approximation property and E is
Montel and bornological. This is all satisfied if E is a nuclear Fréchet space, since
then Eβ∗ is nuclear by 4.48 .
                                                          ˆ εF ∗ ∼
( 3 ) the same argument as in ( 2 ) applies and hence E ∗ ⊗      = L(E, F ∗ ). In
general we have L(E, F ) = L(E, F ) = L(E, F ; K) = L(E, F ; K) ∼
                        ∗           0 ∼
                                                                   = (E ⊗ˆ π F )∗ ,
since E and F are Fréchet.
Proof. (⇒) is obvious by 3.73.2 , since F and EB∗ can be considered as (comple-
mented) subspaces.
(⇐) First one shows that a 0-neighborhood basis in LB (E, F ) is given by the sets
                                      ∗
N := N{xn },{yn∗ }o := {T : |T (xn )(ym )| ≤ 1 ∀n, m}, where xn is Mackey-convergent
to 0 in E with respect to B and yn is Mackey convergent to 0 in F ∗ with respect
                                      ∗
to the bornology of equicontinuous sets, in fact the polars of these sequences form
bases by 4.36 . Without loss of generality we may replace xn by λn xn and yn∗
by µn yn∗ with λ and µ in c0 . The functionals `j,k : LB (E, F ) → K given by
T 7→ yj∗ (T (xk )) form an equicontinuous family, since N is mapped into {λ ∈ K :
|λ| ≤ 1}. Thus λk µj `j,k are Mackey-convergent to 0 with respect to the bornology of
equicontinuous subsets. Hence its polar (which is a subset of N ) is a neighborhood
in the Schwartzification τS of LB (E, F ).
The proof for nuclearity is analogous using that by 4.38 the nuclearification is
given by the topology of uniform convergence on E-nuclear sequences x∗n ∈ E ∗ .
Proof. This follows from 4.62 since E ⊗ε F ⊆ E ε F ⊆ L(Eγ∗ , F ) and (Eγ∗ )∗γ = Ẽ
is Schwartz.
Dual morphisms
4.65 Remark.
LetQE = limj Ej be a limit. Then E can be identified with the closed subspace
of j∈J Ej formed by all x = (xj )j∈J with F(f )(xj ) = xj 0 for all f : j → j 0 .
We get a short exact sequence 0 → E ,→ j Ej  ( j Ej )/E → 0. We can
                                                 Q           Q
                                                    Q
give an explicite description of the linear space ( j Ej )/E, namely the subspace of
Q                                                         Q          Q
   f :j→j 0 Ej formed by the image of the mapping Q :       j Ej →     f :j→j 0 Ej which
              0                                                                   0
given by prf :j→j 0 ◦Q := F(f ) ◦ prj − pr0j . Even for projective limits of a sequence
it however not clear, whether Q is onto or is a quotient map onto its image.
4.66 Lemma.
Every short exact sequence of (F) spaces is topologically exact.
Now the question arises, whether the dual of a topological short exact sequence is
also topologically exact. Since the topology on the dual space is generated by the
polars of bounded sets and (for infra-barrelled spaces) the bornology is generated
by the polars of 0-nbhds, we need to determine how polars behave under adjoint
mappings:
4.70 Lemma.
If T : E → F is an lcs-embedding, then T : b E → b F is a cbs-embedding.
4.71 Lemma.
If T : E → F is a cbs-quotient mapping, then T : t E → t F is an lcs-quotient
mapping.
The converse is not true, as the example 4.80 (based on 3.36 and 4.79 ) shows:
A Köthe sequence space λp (A) which is (FM), but has `p as quotient, hence the
bounded unit-ball cannot be lifted, since otherwise it would be compact.
4.73 Remark.
Surjectivity of linear operators D, means solvability of inhomogeneous equations
D(u) = s for arbitrary s with respect to u.
For example, by the Malgrange-Ehrenpreis Theorem (see [Kri07b, 8.3.1]) every
linear partial differential operator (PDO) D := P ( 1i ∂) with constant coefficients
C ∞ (Rn ) → C ∞ (Rn ) is onto. This can be shown, by considering the formal adjoint
operator Dt := P t ( 1i ∂) : D → D and its adjoint D̃ := (Dt )∗ on the space of
distributions D∗ (see [Kri07b, 4.9]), proving the existence of a fundamental solution
ε ∈ D∗ (i.e. D̃(ε) = δ) via Fourier transform (see [Kri07b, 8.3.1]), and obtaining
the solution of D(u) = s as u := ε ? s (see [Kri07b,
                                                   P4.7.7]). Here P is a polynomial
z 7→ |k|≤m ak z k and P t is the polynomial z 7→ |k|≤m (−1)|k| ak z k .
     P
Proof.
(⇒) is 4.70 .
has to be bounded and hence has to be contained in some finite subsum KN . Since
T ∗ is injective, it induces a linear isomorphism
                                                                4.68.2 [
                          λ · T (U )o ∼
                       [                 [
    T (U )o lin.sp. =                 =     λ · T ∗ (T (U )o ) ======    λ · U o ∩ img T ∗
                    λ>0                  λ>0                             λ>0
(⇒) Since T (U   o
               ) has to be contained in some               KN , we have that dim   EU∗ o ∩ {x∗k :
k ∈ N} lin.sp. < ∞ for each U .
(⇐) The condition implies that the closed absolutely convex set A := (T ∗ )−1 (U o ) =
T (U )o is contained in a finite dimensional linear subspace KN and contains no R+ ·x∗
for x∗ 6= 0, since otherwise T ∗ (x∗ )|U = 0 and hence T ∗ (x∗ ) = 0, thus x∗ = 0. This
implies that A is bounded, otherwise choose an ∈ A ⊆ KN with 1 ≤ kan k → ∞
and let a∞ ∈ A be an accumulation point of ka1n k an ∈ A. Then λ a∞ ∈ A for all
λ > 0 since A 3 kaλn k an → λ a∞ for kan k ≥ λ.
4.77 Corollary. (F) spaces with KN as quotient (See [MV92, 26.28 p.305]).
Let E be (F) and not Banach. Then KN is a topological quotient of E.
Proof. Let (Un ) be a falling 0-nbhd basis of E. Since E is not Banach, we may
assume that ∃x∗k ∈ EU∗ o \ EU∗ o . Then (x∗k )k is linear independent and the mapping
                        k       k−1
independent (on monomials). For finite sequences ξ the functional j ξj x∗j ∈ E(U
                                                                               ∗
                                                                   P
                                                                                   o
                                                                                 k)
iff ξj = 0 for all j > k (Choose f with small derivatives of order < j but high one
of order j). Thus (x∗k )k∈N is onto by 4.76 .
                (k)                                                 4.74
⇒ kyk`q ≤ 2k a1,k , i.e. (Q∗ )−1 (Uko ) is bounded =  ⇒ Q is a quotient mapping.
                                                   ====
For c0 (A) the proof is analogous.
Proof. The unit ball in `p is not compact and λp (A) is Montel, hence a bounded
lift would be compact.
Proof. Let λp (A) with 1 ≤ p < ∞ be the (FM) space of 3.36 . By 3.22 it is
reflexive and by 4.27 its dual E := λp (A)∗ is Montel (by 4.35 even (S)), hence
reflexive and bornological by 4.16 , and (DF) by 4.18.1 . Let Q : λp (A)  `p be
the quotient mapping as in 4.79 and consider the closed subspace F := img(Q∗ ) =
ker(Q)o in λp (A)∗ , using 4.67 . Let W be the unit ball in `p , then U := Q−1 (W )
is a 0-nbhd with Q(U ) = W . By 4.68.2 we have Q∗ (W o ) = Q∗ (Q(U )o ) = U o ∩
img Q∗ = U o ∩ F , hence Q∗ (W o ) is absolutely convex and closed in F . It is a
bornivorous barrel, since each bounded set B in F has bounded inverse (Q∗ )−1 (B)
in `q by 4.74 and hence is absorbed by the unit-ball W o . Infra-barrelledness of F
would imply that Q∗ (W o ) is a 0-nbhd in F and is bounded as image of the unit-ball.
4.83 Theorem on closed image (See [MV92, 26.3 p.290], [Kri14, 9.11]).
Let T : E → F be continuous linear between (F) spaces. Then
   1.   img(T ) is closed;
 ⇔ 2.   img(T ) = ker(T ∗ )o ;
 ⇔ 3.   U o ∩ img(T ∗ ) is a Banach-disk for each 0-nbhd U ;
 ⇔ 4.   U o ∩ img(T ∗ ) is (σ(E ∗ , E) or) β(E ∗ , E) closed for each U ;
 ⇔ 5.   img(T ∗ ) is closed;
 ⇔ 6.   img(T ∗ ) = ker(T )o ;
 ⇔ 7.   T : E/ ker(T ) → img(T ) is a homeomorphism.
      ker(S) = img(S ∗ )o = (E ∗ )o = 0
                           [Kri07b, 7.4.3]
      img(S) = img(S) ============ ker(S ∗ )o = img(Q∗ )o = ker(Q)
                                             [Kri07b, 7.4.3]
     img(Q) = img(Q) = (img(Q)o )o ============ ker(Q∗ )o = {0}o = G.
Proof.
                                         By [Kri14, 5.4.4] the vertical arrows in this di-
           
  (F/E)∗       / F∗
                  ;
                              / / E∗
                                   OO    agram are continuous bijections. The left one
                                        is an iso, since (F/E)∗ → F ∗ is assumed to
     
        -                    "" O        be an embedding and the right one is an iso,
    Eo                       F ∗ /E o    since F ∗ → E ∗ is assumed to be a quotient map-
                                         ping.
                                                                     Q
Proof. Let E = limn En with (F) spaces En , i.e. E = {x ∈ k Ek : xk =
                       ←−
fkk+1 (xk+1 ) for all k} which is the kernel of the mapping π : k Ek → k Ek
                                                                   Q          Q
                                                                        k
                                      X                                
                                           yk∗        ∗           k
                                                                             (xk )k∈N
                                                                                         
                                 =               −   yk−1    ◦   fk−1                        ⇒
                                      k
               π ∗ : (yk∗ )k∈N 7→ (yk∗ − yk−1
                                          ∗      k
                                              ◦ fk−1               ∗
                                                     )k∈N , where y−1 := 0.
Since ι is an embedding, ι∗ is onto by Hahn-Banach. The adjoint π ∗ is injective,
since yk∗ − yk−1
             ∗    k
                 fk−1 = 0 for all k recursively gives yk∗ = 0 for all k.
              = 0∗ = (π ◦ ι)∗ = ι∗ ◦ π ∗ . So let x∗ ∈ k Ek∗ with ι∗ (x∗ ) = 0. Remains
                                                      `
Obviously 0 `
to find y ∗ ∈ k Ek∗ with x∗ = π ∗ (y ∗ ), i.e.
                        x∗k = yk∗ − yk−1
                                     ∗      k
                                         ◦ fk−1  for all k ∈ N.
                     ∗
Recursively we get y−1 := 0 and yk := j≤k x∗j ◦ fjk :
                                    ∗
                                         P
                                        X                        X
      yk∗ = x∗k + yk−1
                   ∗      k
                       ◦ fk−1 = x∗k +         x∗j ◦ fjk−1 ◦ fk−1
                                                              k
                                                                 =   x∗j ◦ fjk .
                                                        j≤k−1                                    j≤k
          ∗        ∗
                                                        N with     x∗j
              `
Since x ∈       k Ek     there exists an n ∈                                = 0 for all j ≥ n. For m ≥ n we
thus have
                    X                 X                                     X              
  0 = ι∗ (x∗ ) =         x∗k ◦ ιk =         x∗k ◦ (fkm ◦ ιm ) =                    x∗k ◦ fkm ◦ ιm = ym
                                                                                                     ∗
                                                                                                       ◦ ιm .
                     k                k<n                                    k≤m
                                              ∗
                                                     = 0 for all m ≥ n, i.e. y ∗ ∈                     Ek∗ .
                                                                                                 `
Since ιm has dense image we get              ym                                                    k
Thus the dual sequence is exact and by 4.84 the canonical resolution itself is
exact.
4.90 Sequences of bounded subsets in (F) spaces (See [MV92, 26.6 p.292]).
Let E be metrizable. Then
                                S
    1. Bn ⊆ E bounded ⇒ ∃δn > 0: n∈N δn Bn bounded.
       2. ∀ B bounded ∃ C ⊇ B bounded: EC ⊇ B → E is an embedding.
                ⇒ B ∩ (x + Uk ) = x + (B − x) ∩ Uk ⊆ x + 2B ∩ Uk ⊆ x + ε C
                                            ⇒ B ∩ (x + Uk ) ⊆ B ∩ (x + ε C).
i.e. B0 ⊆ Q(M ).
4.95 Theorem. Dual of sequences in (F) with (M) quotient (See [MV92,
26.22 p.303]).
Let 0 → E → F → G be a sequence in (F) and let G be (M). Then
0 → E → F → G → 0 is exact ⇔ 0 ← E ∗ ← F ∗ ← G∗ ← 0 is topologically exact.
Proof. G (M) ⇒ bounded sets are relatively compact, hence have bounded lifts
along Q by 4.94 . Thus Q is a cbs-quotient mapping, hence the dual sequence is
topologically exact by 4.93 . The converse follows from 4.84 .
                                              4.88.4
                                       ⇒ ∃nk , Uk := Wnk , Vk := Uk ∩ E:
Proof. (Wn )n 0-nbhd-basis of F . ======
(1)                     ∀k ∀ε > 0 ∃ε0 > 0 : Vk ⊆ ε0 Vk+1 + ε Vk−1 .
                     Q open
Let B ⊆ G bd. =  =====⇒ ∀k ∃Ck : B ⊆ Ck Q(Uk ). Put Ck0 := Ck + Ck+1 . We use
recursion to construct
                                        1
(2)    ∀k ≥ 2 ∃εk > 0 : Vk ⊆ εk Vk+1 + k   Vk−1 with Dk := Ck0 + Dk−1 εk−1 .
                                      2 Dk
In fact, put D1 := 0, and in the induction step let ε := 1/(2k Dk ) and take εk := ε0
as in 1 .
For k kk := pUk let M := {x ∈ F : ∀k ≥ 2 : kxkk ≤ Ck + Dk εk + Ck0 + 1}. Then M
is bounded.
Claim: Q(M ) ⊇ B.
Let ξ ∈ B ⊆ Ck Q(Uk ) = Q(Ck Uk ) ⇒ ∃xk ∈ Ck Uk , Q(xk ) = ξ. Put yk :=
xk − xk+1 . ⇒ Q(yk ) = Q(xk − xk+1 ) = ξ − ξ = 0, i.e. yk ∈ E = ker(Q) and
              yk = xk − xk+1 ∈ Ck Uk + Ck+1 Uk+1 ⊆ (Ck + Ck+1 ) Uk = Ck0 Uk
⇒ ∀k : yk ∈ Ck0 Uk ∩ E = Ck0 Vk .
We use induction to construct vk ∈ Dk εk Vk+1 and uk ∈ 2−k Vk−1 such that
                                         yk + vk−1 = vk + uk .
Let v0 := 0 and vk−1 already be given. Then yk + vk−1 ∈ (Ck0 + εk−1 Dk−1 )Vk =
          2
       ⇒ ∃vk ∈ Dk εk Vk+1 , uk ∈ 2−k Vk−1 : yk + vk−1 = vk + uk .
Dk Vk ==
                                                                                      1
                 P                    P               P              P
⇒ ∃bk := vk−1 − j≥k uj ∈ E, since j>k kuj kk ≤ j>k kuj kj−1 ≤ j>k                     2j   =
 1
2k
   .
                           X                            1
     kbk kk = vk−1 − uk −      uj k ≤ kvk kk + kyk kk + k ≤ Dk εk + Ck0 + 1.
              | {z }                                   2
                                   j>k
                     =vk −yk
Proof. ( 1 ⇒ 2 ) by 4.96 .
( 2 ⇒ 3 ) by 4.93 .
( 3 ⇒ 1 ) 4.89 for the canonical resolution of 4.87 .
           4.97
Proof. = ====⇒ dual of canonical resolution of 4.87 is topological exact. ⇒ E ∗ is
quotient of countable coproduct of Banach spaces, hence bornological by 2.5 .
4.99 Dual sequences for Schwartz spaces (See [MV92, 26.24 p.303]).
Let 0 → E → F → G → 0 be exact in (F) and let one of these 3 spaces be in (S).
Then the dual sequence is topologically exact.
Proof. Closed subspaces and quotients of (FS) are (FS) by 3.73 , hence (M) and
quasi-normable. Thus 4.97 and 4.95 yield the result.
Proof. In 4.25 we have shown that λ1 (A) is not distinguished and hence (λ1 (A))∗
is (DF) but not infra-barrelled. By 4.98 λ1 (A) is not quasi-normable.
Splitting sequences
In this section we describe situations, where short exact sequences split and refor-
mulate this in terms of the derived functor Ext of the Hom-functor. For sequences
with a power series space as kernel the characterizing property on the quotient is
(DN). And for sequences with a power series space of infinite (resp. finite) type
as quotient the characterizing property on the kernel is (Ω) (resp. (Ω)). We show
that the spaces in (DN) are exactly the subspaces of generalized power series spaces
λ∞
 ∞ (a) of infinite type. And the spaces in (Ω) are exactly the quotients of gener-
alized power series spaces λ1∞ (a) of infinity type. We give some applications to
extensions of non-linear mappings and introduce universal linearizer for that.
Whereas, by [Tid79, Beispiel 2 p.301] the set A := {(x, y) : x ≥ 0, |y| ≤ ϕ(x)} does
not have it, when ϕ ∈ C ∞ (R, R) is ∞-flat at 0.
4.107 Proposition.
The functor L( , F ) : lcsop → lcs is left exact,
i.e. if 0 ← E + ←Q− E ←S− E − is topologically exact, then
                                        ∗                ∗
                    0 → L(E + , F ) −Q → L(E, F ) −S → L(E − , F )
is also exact, i.e. L( , F ) is a left exact functor.
The functor L( , F ) is not exact (i.e. maps short exact sequences to such sequences)
in general, since exactness at L(E − , F ) would mean that for closed embeddings
S : E − ,→ E the adjoint S ∗ : L(E, F ) → L(E − , F ) is onto, i.e. every morphism
ϕ : E − → F must have an extension to E.
4.109 Proposition.
The functor L(E, ) : lcs → lcs is also left exact.
The functor L(E, ) is not exact in general, since exactness at L(E, F + ) would mean
that for quotient mappins p : F  F + the adjoint p∗ : L(E, F ) → L(E, F + ) is onto,
i.e. every morphism ϕ : E → F + can be lifted along p : F  F + to a morphism
ϕ̃ : E → F .
                                                                             H`o o
                                                                                         Q
An (F) space E is called projective Fréchet space iff for every                             GO
quotient mapping Q : G  H of (F) spaces every T ∈ L(E, H)
                                                                                              T̃
has a lift T̃ ∈ L(E, G), i.e. Q ◦ T̃ = T .                                           T
                                                                                             E
Obviously every finite dimensional space is projective, since all linear mappings
on it are continuous. It was shown in [Gej78] that there are no other projective
Fréchet spaces.
    0      /F       /G         //H                 /0
                        O           O                       The bottom row is again an exact
                              pr1            T
                                                            sequence, hence splits by 4 , and
    0      /F    inj1
                      / G ×H E      pr2
                                           //E       /0     thus gives a lifting
                           O
                           ∼
                           = Φ                                       T̃ := pr1 ◦Φ ◦ inj2 ,
    0      /F    inj1
                          / F ⊕E    pr2
                                           //E       /0
                                h                           where Φ is the isomorphism.
                                    inj2
Proof.
( 1 ) By 4.113 there is an injective resolution I of F :
                              0 → F → I0 → I1 → I2 → · · ·
0 / E− /E / E+ /0
                                                             
                          0          / I−        /I         / I+               /0
Since Ik0 is injective, the sequences 0 → Ik− → Ik → Ik+ → 0 split and hence also
0 → L(E, Ik0 ) → L(E, Ik ) → L(E, Ik00 ) → 0 splits and, in particular, is exact. By
[KriSS, 7.30] we get a long exact sequence in (co)homology:
                                 ∼
                                 =                            ∼
                                                              =
                                                           
                                               ∼
        ...        /0      / ExtkR (E, F + )   =   / Extk+1    −
                                                        R (E, F )
                                                                             /0       / ...
4.115 Proposition.
The statement Ext1 (E, F ) = 0 is equivalent to the equivalent conditions of 4.111 .
For an additive description of (DN) and later of (Ω) similar to 4.88.2 for quasi-
normed spaces, we need the following:
4.119. Lemma.                                                                 b     b     a
                                                   1                   a+b a a+b
Let a, b > 0 and α, β ≥ 0. Then inf{ra α +         rb
                                                      β   : r > 0} =    a (b)    α a+b β a+b
Proof. Let f (r) := ra α + r1b β = ra+b α + β r−b . Then f 0 (r) = a α ra−1 − b rb+1
                                                                                  1
                                                
                                                                                     β
and hence f 0 (r) = 0 ⇔ ra+b α = a β. Thus
                                    b
                               − a+b  b
                                                                        b
                    b            bβ              b        b      b  a − a+b
         f (r) ≥       β+β                  = α a+b β 1− a+b 1 +
                    a            aα                              a   b
                                          b
                          a a + b    a a+b
                     b
                                      
               = α a+b β a+b                .
                               a     b
Note that f (r) → +∞ for r & 0 if α > 0 and for r % +∞ if β > 0, hence the
infimum is attained if α, β > 0. Otherwise f (r) → 0 for r → 0 or r → +∞, hence
the statement is valid in this case as well.
                   0                  o        o  C o
 ⇔ 3. ∃q      ∀p ∃p ∃C > 0 ∀r > 0 : Up ⊆ r Uq + r Up0 ;
 ⇔ 4. ∃q      ∀ 0 < δ < 1 ∀p ∃p0 ∃C > 0 : k kp ≤ Ck k1−δ
                                                      q   ·k        kδp0 ;
                           0            1+d        d
 ⇔ 5. ∃q      ∃d > 0 ∀p ∃p ∃C > 0 : k kp ≤ Ck kq · k kp0 ;
Note that in all these conditions we may assume w.l.o.g. that q < p < p0 , since for
p00 ≥ p0 we have k kp00 ≥ k kp0 and for p ≤ q we may take p0 = p and C = 1.
Note furthermore, that for k kp0 ≥ k kq only δ near 0 (and hence d = 1−δ     δ near ∞)
are relevant, since for δ < δ 0 (and kykq 6= 0) we get
                                  kyk 0 δ         kyk 0 δ0          0      0
                                      p                 p
          kyk1−δ
              q   kykδp0 = kykq             ≤ kykq             = kyk1−δ
                                                                    q     kykδp0
                                   kykq              kykq
                                  | {z }
                                               ≥1
Proof.
( 1 ⇔ 2 ) the minimum of r 7→ rkxkq + Cr kxkp0 is 2 Ckxkq kxkp0 by 4.119 .
                                                         p
Hence the inequality in 6 for all r > 0 is equivalent to k k2p ≤ 4Ck kq k kp0 .
                                               C
( 2 ⇔ 3 ) From k kp ≤ rk kq +                  rk   kp0 we get
                                                                                           
                  1
                  r Uq   ∩   r
                             C Up
                                  0    ⊆ 2 Up and hence Upo ⊆ 2 r Uqo +             C o
                                                                                    r Up0       .
Conversely,
                                              Upo ⊆ r Uqo +     C o
                                                                r Up0
implies that any u ∈ Upo can be written as u = r v + Cr u0 with v ∈ Uqo , u0 ∈ Upo0 , i.e.
                                                    C 0                      C
                         |u(x)| ≤ r|v(x)| +         r |u (x)|   ≤ rkxkq +    r kxkp
                                                                                    0
( 1 ⇒ 6 ) Define a new basis of semi norms ||| |||k recursively by: ||| |||0 := k kq ;
                     2
∃p00 ∃C0 ≥ 1: ||| |||0 ≤ ||| |||0 · C0 k kp00 = ||| |||0 · ||| |||1 , where ||| |||1 := C0 k kp00 ;
                     2
∃p0k ∃Ck ≥ 1: ||| |||k ≤ ||| |||0 · Ck k kp0k ≤ ||| |||k−1 · ||| |||k+1 , with ||| |||k+1 := Ck k kp0k .
                             2
( 6 ⇒ 1 ) From ||| |||k ≤ ||| |||k−1 ||| |||k+1 we obtain that all ||| |||k are norms and
using |||x|||k /|||x|||k−1 ≤ |||x|||k+1 /|||x|||k for all x 6= 0, we get for all k ∈ N the
inequality
                  k                2k
      |||x|||k   Y      |||x|||j   Y     |||x|||j   |||x|||2k               2
               =                 ≤                =           , i.e. |||x|||k ≤ |||x|||0 |||x|||2k .
      |||x|||0   j=1
                     |||x||| j−1      |||x||| j−1    |||x|||k
                                        j=k+1
                                                              1−δ
( 4 ⇒ 5 ) This follows directly with d :=                      δ .
t∈J
These Fréchet spaces λp∞ (a) are usally denoted Λp (J, a) for p ∈ {1, ∞} and for
p = 1 the index is often dropped.
taking the `∞ -Norm of (kxi kk )i∈I and corresponds to the seminorm k kk of λ∞ ∞ (a).
Replacing the supremum by the 1-Norm, gives the second isomorphism.
4.127 Proposition (See [Vog82, 1.1 p.540], [Vog85, Lemma 1.3 p.258], [Vog87,
4.3 p.185], and [Vog77a, Satz 1.5 p.111]).
Let F and G be Fréchet spaces and assume that G has property (DN ).
Then any exact sequence
                                                       Q
                                 0 → λ∞
                                      ∞ (a) → F −→ G → 0
with a ∈ RJ≥1 splits.
In fact, it can be shown that the condition (DN) yields even a characterization:
                                                  ∞
                                               = C[−2,−1] (R) ⊕ C[1,2]
                                                                 ∞
                                                                       (R) ∼
                                                                           =s⊕s∼
                                                                               =s:
In fact s ∼
          = s × s via (xk )k∈N 7→ ((x2k )k∈N , (x2k+1 )k∈N ): This mapping is obviously
linear and injective. It is continuous, since |(k + 1)q x2k | ≤ |(2k + 1)q x2k | and
|(k + 1)q x2k+1 | ≤ |(2k + 2)q x2k+1 |. It is onto s × s, since given y, z ∈ s the inverse
image is given by x2k := yk and x2k+1 := zk with
                           (
                   q        |(2k + 1)q yk | ≤ |2q (k + 1)q yk | for n = 2k,
           |(n + 1) xn | =
                            |(2k + 2)q zk | ≤ |2q (k + 1)q zk | for n = 2k + 1.
Thus we have a short exact sequence s ,→ s  s, which splits by 4.127 since s is
a power series space of infinite type by 1.15.4 and hence has property (DN) by
 3.14.3 .
Using translation it suffices to consider the restriction map C ∞ (R) → C ∞ (R≥−1 ).
We choose a function ϕ ∈ C ∞ (R, [0, 1]) with ϕ(t) = 0 for all t ≥ 0 and ϕ(t) = 1 for
all t ≤ − 21 and decompose f ∈ C ∞ (R≥−1 ) as f = (1 − ϕ) · f + ϕ · f . Since (1 − ϕ) · f
is 0 on [−1, − 12 ] we can extend it by 0 to f˜0 ∈ C ∞ (R). By what we have shown
before the restriction of ϕ · f to [−1, 1] has an extension f˜1 ∈ C[−2,2]
                                                                       ∞
                                                                            (R) ⊆ C ∞ (R).
Then f˜2 : t 7→ ϕ(t − 1 ) · f˜1 (t) is an extension of ϕ · f restricted to [−1, +∞), since
                      2
ϕ(t − 21 ) = 1 for all t with ϕ(t) 6= 0 and ϕ(t − 21 ) = 0 = ϕ(t) for all t ≥ 12 . Thus
f˜ := f˜0 + f˜2 is the desired extension of f , and it depends continuously and linearly
on f , since all intermediate steps do so.
More generally, it is shown in [Tid79, Folgerung 2.4 p.296] for compact K ⊆ Rn :
C ∞ (Rn )  E(K) has a continuous linear right inverse ⇔ E(K) (DN) ⇔ E(K) ∼ = s.
Here E(K) ∼ = C ∞ (Rn )/{f ∈ C ∞ (Rn ) : f |K = 0} denotes the Fréchet space of
Whitney jets on K.
Another application is:
4.131 Corollary (See [Vog83, 6.1. Satz p.197], [Vog85, 2.6 p.260] ).
A Fréchet space F is (DN ) ⇔ ∃J ∃a ∈ RJ≥1 : F ,→ λ∞
                                                   ∞ (a).
Proof.
(⇒) Let J := k Bk for some basis of equicontinuous sets Bk ⊆ F ∗ . Then F can
             S
                         N
be embedded into (`∞ (J)) in a natural way.
                            ∞
By Borel’s theorem 0 → C[−1,0]  (R) × C[0,1]
                                        ∞
                                             (R) ,→ C[−1,1]
                                                       ∞
                                                              RN → 0 is exact and
 ∞
C[a,b] (R) ∼
           = s by 1.16.3 . Moreover s ∼
                                      = s × s via (xk )k∈N 7→ ((x2k )k∈N , (x2k+1 )k∈N )
by what we have shown in 4.129 . By tensoring this exact sequence of nuclear (F)
spaces with `∞ (J) (i.e. applying L(( )∗ , `∞ (J)) with the injective (F) space `∞ (J))
we get the (using 4.99 ) exact sequence of (F) spaces:
                   0 → s⊗`         ˆ ∞ (J) → RN ⊗`
                        ˆ ∞ (J) → s⊗`           ˆ ∞ (J) → 0.
                      0 → λ∞        ∞          ∞          Q                  N
                           ∞ (a) → λ∞ (a) −→ (` (J)) → 0.
                                   N
Since F embeds into (`∞ (J)) we may consider the pullback(=preimage) Q−1 (F )
of F under Q, and get the short exact sequence
                          0 → λ∞
                               ∞ (a) → Q
                                        −1
                                           (F ) → F → 0.
By 4.127 the sequence splits if F has property (DN ). We therefore get the
embedding F ,→ Q−1 (F ) ⊆ λ∞
                           ∞ (a).
                                 
                       ˆ ∞ (J) 
                    / s⊗`                    / s⊗`                  //   RN ⊗`            /0
                                                              Q
            0                                   ˆ ∞ (J)                     ˆ ∞ (J)
                     / λ∞                   / λ∞            Q
                                                                         / / (`∞ (J))N    /0
            0           ∞ (a)                    ∞O (a)                           O
                                                                                    (1)
                                                 ?       s
                                                              (3)
                                                                               ?
                                       (2)
            0        / λ∞
                        ∞ (a)
                                             / Q−1 (F )                      //F          /0
(⇐) Since λ∞
           ∞ (a) has property (DN ) by 4.125 , the converse follows from 3.14.2 .
A Fréchet space E is said to be (Ω) iff these equivalent conditions are satisfied.
Note that we may assume that p0 > p and it suffices that k > p0 and d ∈ N, since
q ≥ p0 ⇔ k kq ≥ k k0p ⇔ Up0 ⊇ Uq ⇔ k k−p0 ≥ k k−q thus 1 holds for each p00 > p0
as well and 2 holds for each d0 > d as well.
Proof.
                   d
(1⇔2)δ=           d+1 .
                                                          0               k0 +1
                                                                             p
( 2 ⇔ 3 ) since the infimum of r 7→ αrk + β 1r is Ck0                             αβ k0 by 4.119 .
                                            0
( 3 ⇒ 4 ) Let k k−p0 ≤ Crk k k−k + 1r k k−p . Then
                                                  1      o   r o
                                                      0 Uk ∩  U ⊆ Upo0
                                                2Cr k        2 p
and by taking polars
                                                     1           r o       0    2
                Up0 ⊆ ((Up0 )o )o ⊆                     k 0 U o
                                                              k ∩   Up ⊆ 3Crk Uk + Up .
                                                    2Cr           2    o          r
                                        0
( 3 ⇐ 4 ) Let Up0 ⊆ C rk Uk + 1r Up . Then every x ∈ Up0 can be written as as
x = Crn a + 1r b with a ∈ Uk and b ∈ Up . Thus for x∗ ∈ E ∗ we get
                                                 0         1
                                   |x∗ (x)| ≤ Crk kx∗ k−k + kx∗ k−p
                                                           r
and taking the sup over x ∈ Up0 gives 3 .
4.134 Theorem
(See [Vog77b, Theorem 2.3], [Vog85, Lemma 1.3 p.258], and [Vog87, 4.1 p.183]).
Let E and F be Fréchet spaces and assume that E has property (Ω).
Then any exact sequence
                                                            Q
                                       0 → E → F −→ λ1∞ (a) → 0
with a ∈ RJ≥1 splits.
we have kej kk = a(j)k . By the open mapping theorem, Q(Wk ) ⊆ λ1∞ (a) is open.
Hence, for every k there exists an nk ∈ N and a Ck ≥ 1 with
                        ej      
                           n
                               ∈ x : kxknk ≤ 1 ⊆ Ck Q(Wk ).
                      a(j)   k
Thus there are dkj ∈ Ck a(j)nk Wk ∩ Q−1 (ek ) ⊆ F . We may assume that
           nk+1 ≥ (1 + νk−1 )nk ≥ nk and Ck+1 ≥ 2kνk−1 (3Ck )1+νk−1 ≥ Ck
for all k ∈ N. Thus
                                                        
    dkj − dk−1
           j   ∈  C k a(j) nk
                              W k + C k−1 a(j)nk−1
                                                   W k−1   ∩ ker Q ⊆ 2Ck a(j)nk Uk−1
                                       k
Multiplying 1 for r := ρ 2 with ρ gives the existence of
           akj ∈ ρ rνk−1 Uk = (3Ck a(j)nk )1+νk−1 2k νk−1 Uk ⊆ Ck+1 a(j)nk+1 Uk
with
                                                                       ρ
                Rjk − Rjk−1 = (dkj − dk−1
                                      j   + ak−1
                                             j   ) − akj ∈               Uk−2 = 2−k Uk−2
                                                                       r
Thus
                                       X                                     X
  ∃ Rj := lim Rjl = Rjk +               (Rjl − Rjl−1 ) ∈ 2Ck+1 a(j)nk+1 Wk +   2−l Ul−2 ⊆
             l→∞
                                       l>k                                            l>k
                                                
                            nk+1            −k                                nk+1
                                                                        
          ⊆ 2Ck+1 a(j)             +2                Wk−1 ⊆ 1 + 2Ck+1 a(j)           Wk−1 ⊆ F.
So we can define
                              X
                    R(x) :=                xj Rj ∈ F for all x = (xj )j∈N ∈ λ1∞ (a),
                                   j
since
                       X                                Rj
             R(x) =         a(j)nk+1 xj                       ∈ kxknk+1 (1 + 2Ck+1 ) Wk−1 .
                        j
                                                     a(j)nk+1
Thus, for each k > 0,
                             pWk−1 (R(x)) ≤ (1 + 2Ck+1 ) kxknk+1 ,
we get Q ◦ R = id.
In fact, it has be shown that the condition (Ω) gives even a characterization:
      1. F is (Ω);
 ⇔ 2. Ext1 (λ1∞ (α), F ) = 0, i.e. any ses 0 → F → G → λ1∞ (α) → 0 splits.
 ⇔ 3. If Q : G  H is a quotient mapping with kernel F
      then Q∗ : L(λ1∞ (α), G) → L(λ1∞ (α), H) is onto;
 ⇔ 4. If S : H ,→ G is a closed embedding with quotient λ1∞ (α)
      then S ∗ : L(G, F ) → L(H, F ) is onto.
Proof. ( 1 ⇒ 2 ) is 4.134 .
Proof.
(⇒) We have the canonical resolution
                                  Y       Y
                       0→E→          Ek →   Ek → 0.
                                        k           k
                           Q                   P
Let F := {x = (xk )k ∈ k Ek : kxk :=           kxk kk <       ∞}, a Banach space which
contains each Ek as direct summand (and let Fk be a           complement of Ek in F ).
Let {xi : i ∈ I} be a (w.l.o.g. infi-
nite) dense subset in F and 0 →                  0O                   0O        0O
K ,→ `1 (I)  F → 0 be the
resulting exact sequence. Taking
the tensor product with the ses 0            / F ⊗s
                                                  ˆ             / F ⊗s
                                                                    ˆ         / FN      /0
                                                    O               O            O
                   N
0 → s → s → K → 0 gives by
 4.112 a diagram with exact rows
                                         / `1 (I)⊗s     ˆ     / `1 (I)⊗s
                                                                       ˆ    / `1 (I)N   /0
and columns (since all factors are 0                  O              O            O
Fréchet and always one of them is
nuclear). This gives a right exact 0        / K ⊗sˆ             / K ⊗s
                                                                    ˆ         / KN      /0
diagonal sequence                                 O                 O            O
(`1 (I)⊗s)⊕(K
       ˆ       ˆ → `1 (I)⊗s−
               ⊗s)          ˆ Q→ F N → 0
                                                 0                    0         0
and let N denote the kernel of Q.
For power series spaces λ10 (α) of finite type one needs the stronger condition (Ω):
4.137 Proposition. [Vog87, 4.2 p.184].
Let limn→∞ ααn+1
              n
                 = 1 and E be a Fréchet space. Then
      1. E has (Ω), i.e. ∀p ∃p0 ∀k ∀d > 0 ∃C > 0 : k k1+d                d
                                                      −p0 ≤ C k k−k · k k−p (cf.
         4.132.2 );
 ⇔ 2. Ext1 (λ10 (α), E) = 0, i.e. any ses 0 → E → G → λ10 (α) → 0 splits.
If all involved Fréchet spaces have a basis of Hilbert seminorms then 4.127 and
 4.134 can be generalized to
4.138 Splitting theorem (See [MV92, 30.1 p.357], [Vog87, 5.1 p.186]).
Let 0 → E → G → F → 0 be a short exact sequence of (F) spaces having a basis of
Hilbert seminorms.
If E is (Ω) and F is (DN), then the sequence splits.
Proposition.
Let F(U, E) be function spaces with the following properties:
      f ∈ F, T ∈ L ⇒ T ◦ f ∈ F.
      1.
      If ι : G ,→ E is a closed embedding, then f ∈ F(U, G) ⇔ ι ◦ f ∈ F(U, E).
      2.
      F(U, E 0 ) ∼
      3.         = L(E, F(U )) by switch of variables.
      F(U ) carries a reflexive lc-topology.
      4.
Let λ(U ) := F(U )∗ then δ ∗ : L(λ(U ), E) ∼
                                           = F(U, E) is a linear bijection for each
complete lcs E with complete dual E ∗ .
Examples.
( 1 ) T ∈ L, f ∈ F ⇒ T ◦ f ∈ F:
For `∞ , C ∞ (See [KM97, 2.11 p.24]), H ([KN85, 2.6 p.283]), and C ω ([KM90,
1.9 p.10]) this is easily checked.
( 2 ) ι ◦ f ∈ F ⇒ f ∈ F:
For `∞ , C ∞ , H, and C ω this is obvious since these mappings can be tested by the
continuous linear functionals.
( 3 ) F(U, E 0 ) ∼
                 = L(E, F(U )):
For C ∞ see [FK88, 4.4.5], for H see [KN85, 2.14 p.288], for C ω see [KM90, 6.3.3
p.37], and for `∞ see [Kri07a, 4.7.4].
( 4 ) F(U ) reflexive:
                                                   = sN ).
C ∞ (U ) is nuclear (F) and has (Ω), but not (DN) (∼
                                                                                =C
H(U ) is nuclear (F) and a power series space, it has always (Ω) and only for U ∼
(DN).
C ω (U ) is complete ultrabornological (N) and its dual is complete nuclear (LF).
`∞ : For bornological spaces X one has `∞ (X) = (`1 (X))∗ by [FK88, 5.1.25] and
`1 (X) = (c0 (X))∗ by [FK88, 5.1.19], where
  `1 (X) := {f ∈ RX : carr(f ) is bounded and kf k1 < ∞} and
 c0 (X) := {f ∈ RX : carr(f ) countable and ∀B ∀ε > 0 : {x : |f (x)| > ε} finite.}.
However, λ(X) = `1 (X) for F := `∞ by [Kri07a, 4.7.4].
In many situations one can show better density conditions for the image of δ (like
Mackey-denseness) and hence gets the universal property for spaces E being less
complete (like Mackey-complete).
For U ⊆ Rn is open, it has been show in [FK88, 5.1.8] that λ(U ) = C ∞ (U, R)∗ is
universal for C ∞ -mappings into Mackey-complete spaces. For open U ⊆ Cn , it has
been shown in [Sie95] that λ(U ) = H(U )∗ is universal for H-mappings into Mackey-
complete spaces. The free convenient vector space for real-analytic mappings has
been considered in [KM90] and for sequentially complete spaces in [BD01]. In
[FK88, 5.1.24] it is shown that λ(X) = `1 (X) is universal for `∞ -mappings into
Mackey-complete spaces.
[BD98, Corollary 39 p.34] If D : C ω (R)  C ω (R) onto then one can find solutions
depending holomorphically on a parameter in C. By [BD01, Proposition 9 p.501]
for every elliptic surjective linear PDO D := P (∂) : C ω (U ) → C ω (U ) with constant
coefficients and open U ⊆ Rn the extension D ⊗ E : C ω (U, E) → C ω (U, E) is
surjective if E is (F) or the strong dual of a (F)-space with (DN).
In contrast, by [BD01, Theorem 8 p.501] for every elliptic surjective linear PDO
D := P (∂) : C ω (R2 ) → C ω (R2 ) with constant coefficients the extension D ⊗ H(D̄) :
C ω (R2 , H(D̄)) → C ω (R2 , H(D̄)) is not surjective.
[BD01, Theorem 6 p.499] and [BD98, Theorem 38 p.33]: For open sets Ui ⊆ Rni
let T : C ω (U1 ) → C ω (U2 ) be a continuous linear surjective mapping. Then T ⊗ E :
C ω (U1 , E) → C ω (U2 , E) is onto provided E is (F)+(DN) or (E is complete+(LB)
and E ∗ is (Ω)) or E is a (F)-quojection, i.e. every quotient with a continuous norm
is a Banach space.
For (sequentially) complete lcs E and open U ⊆ Rn one has a linear bijection
C ω (U, E) ∼
           = C ω (U )εE = L(C ω (U )∗β , E) ∼
                                            = L((E ∗ , τc ), C ω (U )) by [BD01, Theorem
2 p.496]
In this section, we describe situations where continuous linear mappings are even
locally bounded. If the domain space is a power series space of finite type, then
the characterizing property for the range space is (DN). And if the range space is
such a power series space, then the characterizing property of the domain space
is (Ω). For power series spaces of infinite type, the characterizing properties for
the other involved space are (LB∞ ) and (LB ∞ ). We give applications to vector
valued real-analytic mappings and mention applications to holomorphic functions
on Fréchet spaces.
                                ?                           ?
                          LB(C ω (U )∗ , E) /        / / Ctω (U, E)
Sketch of proof. It is easy to see that f ∈ Ctω (R, E) is locally C ω into some EB
and by 4.90.1 even globally, hence corresponds to an element in L(C ω (R)∗ , EB ) =
LB(C ω (R)∗ , EB ) ⊆ LB(C ω (R)∗ , E).
4.143 Definition.
Let E and F be (F) with increasing bases of seminorms (k kk )k∈N and (k kn )n∈N .
For linear T : E → F consider
                        kT kk,n := sup kT xkn ∈ [0, +∞].
                                    kxkk ≤1
Proof.
( 1 ) T ∈ L(E, F ) ⇔ ∀n ∈ N ∃kn ∈ N ∃C > 0 : kT (x)kn ≤ Ckxkkn .
( 2 ) T ∈ LB(E, F ) ⇔ ∃k 0 ∈ N ∀n ∈ N ∃C > 0 ∀x : kxkk0 ≤ 1 ⇒ kT (x)kn ≤ C.
Hence
                               kT kk0 ,n < ∞ for all n,
i.e. T ∈ Hk0 ⊆ LB(E, F ).
W.l.o.g. k % ∞.
Proof.
( 1 ⇒ 2 ) follows from 4.144 for T := prj ⊗y with y ∈ F and prj (x) := xj for
x ∈ λ1 (B) =: E, since
                                        n                              o kyk
                                                                             n
   kT kk,n = sup kT xkn : kxkk ≤ 1 = sup |xj | kykn : kx · b(k) k`1 ≤ 1 = (k) .
                
                                                                          bj
              kykn e−ρk0 βj+1 ≤ Ckykk0 < kykn e−ρk0 βj (& 0 for j → ∞),
then
       kykn ≤ eρk0 βj C max kykk0 , kykn0 e−ρk0 +1 βj = C kykn0 e(ρk0 −ρk0 +1 )βj
                              
                                 ρ 0    −ρ 0          kyk 0 d
                        −ρ 0 βj+1 k b+1   k
                                                           k
            ≤ Ckykn0 e k              ρ 0
                                       k     ≤ Ckykn0 C         ,
                                                        kykn
              ρ   0     −ρ   0
where d := k b+1ρ 0 k , i.e. kyk1+d
                                n     ≤ C 1+d kykn0 kykdk0 .
                 k
                             −ρk0 β0
If no such j exists, then e          kykn ≤ Ckykk0 and we get
                         kyk1+d
                            n   ≤ kykn0 kykdn ≤ kykn0 (C eρk0 β0 )d kykdk0 .
Hence in both cases
                  kyk1+d
                     n   ≤ C 0 kykn0 kykdk0 with C 0 := C d max{C, eρk0 β0 d },
which is equivalent to (DN) by 4.123.5 with q := k 0 , p := n, and p0 := n0 .
by 2 . This implies
                               kT kk0 ,n ≤ C max0 kT kkm ,m ,
                                               m≤n
                  ∞
i.e. T ∈ LB(E, λ (A)) by 4.144 .
Minkowski-functional of Uko ⊆ E ∗ .
( 1 ⇐ 2 ) Let k ∈ NN be given. For p := k0 choose p0 according to (Ω), i.e.
                      ∀n ∀d > 0 ∃C ≥ 1 : k k1+d                d
                                            −p0 ≤ C k k−n · k k−p
For every n ∈ N let n0 ≥ p with ρn0 > ρn and d > 0 with d(ρn − ρ0 ) ≤ ρn0 − ρn .
Thus there exists a C ≥ 1 such that
                              k k1+d                  d
                                 −p0 ≤ C k k−kn0 · k k−k0 .
For x∗ ∈ E ∗ and j ∈ N either eρn αj kx∗ k−p0 < eρ0 αj kx∗ k−k0 or
          kx∗ k1+d       ∗          ∗ d         ∗
               −p0 ≤ C kx k−kn0 · kx k−k0 ≤ C kx k−kn0 e
                                                         d(ρn −ρ0 )αj
                                                                      kx∗ kd−p0 ,
i.e. kx∗ k−p0 ≤ C e(ρn0 −ρn )αj kx∗ k−kn0 .
In both cases we have 4.147.2
  eρn αj kx∗ k−p0 ≤ max eρ0 αj kx∗ k−k0 , C eρn0 αj kx∗ k−kn0       ≤ C max0 eρm αj kx∗ k−km ,
                        
                                                                          m≤n
since                                 (
                                          eρn−1 αj kx∗ k−k0           for m < n,
                  eρm αj kx∗ k−km ≤
                                          eαj kx∗ k−kn                for n ≤ m ≤ n0 .
Let x∗ ∈ E ∗ . If there exists a j ∈ N such that
     e(ρn −ρn−1 )αj−1 kx∗ k−p0 ≤ Cn kx∗ k−p < e(ρn −ρn−1 )αj kx∗ k−p0 (% 0 for j → ∞),
                             (1−ρn )αj
then, since dn := supj    (ρn −ρn−1 )αj−1 ,
                             n                              o
 kx∗ k−p0 ≤ e−ρn αj Cn max eρn−1 αj kx∗ k−p , eαj kx∗ k−kn = Cn e(1−ρn )αj kx∗ k−kn
                                                      kx∗ k−p dn ∗
          ≤ Cn e(ρn −ρn−1 )αj−1 dn kx∗ k−kn ≤ Cn Cn ∗             kx k−kn ,
                                                      kx k−p0
          1+dn
i.e. kx∗ k−p0  ≤ Cn1+dn kx∗ k−kn kx∗ kd−p
                                        n
                                          .
If no such j exists, then Cn kx k−p < e(ρn −ρn−1 )α0 kx∗ k−p0 . Hence
                                ∗
                             n                             o
  kx∗ k−p0 ≤ e−ρn αj Cn max eρn−1 αj kx∗ k−p , eαj kx∗ k−kn = Cn e(1−ρn )α0 kx∗ k−kn
and thus we obtain in both cases
        kx∗ k1+d    0   ∗         ∗ dn          0            dn (1−ρn )α0
             −p0 ≤ Cn kx k−kn · kx k−p , where Cn := Cn max{Cn , e
                n
                                                                          }.
Since kn → +∞ and dn → 0, condition
(Ω)               ∀p ∃p0 ∀k ∀d > 0 ∃C ≥ 1 : k k1+d                d
                                               −p0 ≤ C k k−k · k k−p
follows.
For power series spaces λp∞ (α) of infinite type one needs new (smaller) classes:
Proof.
( 1 ⇐ 2 ) Let k ∈ NN with k % +∞ be arbitrary. By 2 we have for ρ := k
           ∃k 0 ∀n ∃n0 ≥ k 0 ∃C > 0 ∀y ∃m ∈ [n, n0 ] : kyk1+k
                                                          n
                                                             m
                                                               ≤ C kykkkm
                                                                        0 kykm .
and hence
                      00                                         00
           kykn e|−k βj
                   {z } ≤ kykn ≤ C kykm e
                                          (kk0 −k                     )βj km
                                                                               = C kykm e−km βj .
                    ≤1
Proof.
( 1 ⇐ 2 ) We will verify condition 4.147.2 :
( 1 ⇒ 2 ) Let ρ % ∞ and p ∈ N.
                                   (n)
By 4.147.2 for k : m 7→ p + m and aj := eρn αj we get:
         ∃p0 > p ∀n ∃n0 ∃C ≥ 1 ∀x∗ ∀j : eρn αj kx∗ k−p0 ≤ C max0 eρm αj kx∗ k−km .
                                                                        m≤n
holds. Then
                                  n                                            o
           eρn αj kx∗ k−p0 ≤ C max eρm αj kx∗ k−km : m ∈ [0, p0 − p] ∪ [n, n0 ] ,
                            kx∗ k1+d
                                 −p0 ≤ C
                                         1+d
                                             kx∗ k−km kx∗ kd−p ,
For given n we may now choose the n from above such that n > max{p0 − p, n} and
    b
ρn −ρ 0  ≤ 1 and thus d ≤ ρn −ρb 0    ρm ≤ ρm . Hence, in both cases we have for
      p −p                             p −p
                              L(E, λ∞               ∞
                                    ∞ (a)) = LB(E, λ∞ (a)).
Thus, by 4.141 ,
                                    L(E, F ) = LB(E, F )
as well.
Proof. For sake of simplicity we consider only the case U = R treated in [BD98,
Thm. 18 p.23]. By 4.142 : 2 ⇔ L(C ω (R)∗ , F ) = LB(C ω (R)∗ , F ).
(⇐) By [BD98, Proposition 5 p.17] there exists a quotient map q : C ω (R)∗  H(D)
         ω
(since Cper (R) ∼
                = H(D)) thus L(H(D), F ) = LB(H(D), F ).
                                                               = H(D) by [BD98,
(⇒) Let T ∈ L(C ω (R)∗ , F ). Since C ω (R)∗ = limn En with En ∼
                                               −→
Proposition 3 p.16] and H(D) = λ0 (id) by 1.15.6 , there exists for every n ∈ N a
     S Un ⊆ En with T (Un ) bounded by 4.146 . By 4.90.1 there are δn > 0 such
0-nbhd
that n δn T (Un ) is bounded. SThus T is bounded on the absolutely convex hull U∞
(which is a 0-nbhd in lim) of n δn Un .
                       −→
Similarly, the following can be shown:
This property has been used in [DMV84, Theorem 9 p.54] to characterize (NF)
spaces in which not every bounded set is uniformly polar. One has the implications:
(Ω) ⇒ (Ω̃) ⇒ (LB∞ ) ⇒ (Ω).
Note: Quotient and subspaces of s via (N) and Ext1 = 0 ([Vog84, 2.4 p.362] and
[Vog84, 2.3 p.361]) [Vog84, 2.5 p.363] Quotient and subspaces of s [MV92, 31
p.369],
nuclear-(DN) are the subspaces of s [MV92, 31.5 p.372],
nuclear-(Ω) are the quotients of s [MV92, 31.6 p.373],
nuclear-(DN ∩ Ω) are the direct summand of s [MV92, 31.7 p.375]
Supplied with the norms pk (x) := sup{(1 + n)k p(xn ) : n ∈             N}    for k ∈       N   and
seminorms p of F it is an lcs and Fréchet if F is Fréchet.
                   n                     o
              ≤ sup p(T (x∗ )) : x∗ ∈ Uko ,
since for the standard seminorms (given by kxkk := supn (1 + n)k |xn |, see 1.15.4 )
on s = c0 (A) the polar of the corresponding 0-nbhd Uk is by 1.24
It is bijective, since for x = (xn )n∈N ∈ s(N, F ) the only possible inverse image
T ∈ L(s∗ , F ) is given by
                                X             X                    
           x∗ : x 7→ x∗ (x) = x∗     prn (x) en =     x∗ (en ) prn (x) 7→
                                    n                      n
                        X                   X                         X
              ∗               ∗                        ∗
        7→ T (x ) = T        x (en ) prn :=        x (en ) T (prn ) =       x∗ (en ) xn .
                         n                     n                        n
                                                       ∗       ∗
This
P definition     for T makes sense, since any x ∈ S is contained in some Uko , i.e.
      ∗
  n |x (en )|(1 + n)
                     −k
                         ≤ 1 and {(1 + n)k xn : n ∈ N} is bounded.
Moreover, the so defined T is continuous, since
                             X               X |x∗ (e )|
                                                           n
           (p ◦ T )(x∗ ) = p     x∗ (en ) xn ≤                k
                                                                (1 + n)k p(xn )
                              n                    n
                                                     (1  + n)
                           X |x∗ (en )|
                         ≤            k
                                          · sup(1 + n)k p(xn ) ≤ kx∗ kUko pk (x).
                            n
                              (1 +  n)       n
This shows at the same time, that the inverse s(N, F ) → L(s∗ , F ), (xn )n∈N 7→ T ,
is continuous as well.
4.158 s(N, s) ∼
              = s (See [MV92, 31.1 p.369]).
So the seminorms of s(N, s) and s can be dominated by each other under this
bijection.
Proof. By 4.78 (see the proof of 4.131 ) we have the short exact sequence
0 → s ,→ s  KN → 0. By 4.99 the dual sequence 0 → K(N) → s∗ → s∗ → 0 is
             Q                                                                  Q∗
                 0          ˆ /
                         / s⊗s             / s⊗s
                                              ˆ
                                                      Q⊗s
                                                               //   KN ⊗s
                                                                       ˆ             /0
                               4.61                                  4.61
                 0     / L(s∗ , s) /    / L(s∗ , s)    Q∗∗
                                                              / L(K(N) , s)
                              4.157
                 0     / s(N, s) /      / s(N, s)             / L(K, s)N
                              4.158
                 0         /s /             /s                      / / sN           /0
In order to see that these isomorphic sequences are short exact we use that any
z ∈ KN ⊗ ˆ π s can be represented by 3.40 as z = n λn xn ⊗ yn with λ ∈ `1 , {xn :
                                                   P
Proof. (⇒) By 1.15.4 s ∼ = λ∞ (α) with α(n) := ln(n + 1), by 3.78.1 and 4.125
λ∞ (α) is (N) and (DN), and by 3.73.2 and 3.14 E is (N) and (DN).
(⇐) By 4.159 there is an exact sequence 0 → s → s → sN → 0 and by 3.81
there is an embedding E ,→ sN . So the pullback gives another short exact sequence
(where α(n) := ln(n + 1))
                                          /s         / / sN
                               s             O              O
                           1.15.4
                                                       ? u            3 S ?
                                λ∞
                                 ∞ (α)
                                       /         / s ×sN E             //E
Proof.
(⇒) s ∼
      = λ∞ (α) has (N) and (Ω) by 3.78.1 and 4.133.4 . Thus E has (N) and
(Ω) by 3.73.4 and 4.133.2 .
(⇐) By 3.81 there is an embedding E ,→ sN .
Then Q := sN /E is (NF), and thus there exists a                                 0O          0O
short exact sequence 0 → s −j2→ Q̃ −p2→ Q → 0
                                                              0    /E      / sN       //Q       /0
as in the proof of 4.160 with Q̃ ,→ s and hence                                 O         O
Q̃ has (DN) by 3.14.2 .                                       0    /E        /H          / Q̃     /0
                                                                               O            O
Let H := {(x, y) ∈ sN × Q̃ : p1 (x) = p2 (x)} be the
pullback. Then the diagram on the right side has                                 sO          sO
exact rows and columns and by 4.138 H ∼     = E × Q̃
since E is (Ω).                                                                  0           0
Proof.
   ∃E0 : E ∼
           = E0 × s and ∃E1 : s ∼
                                = E × E1 ⇒
     ⇒ s = E × E1 = E0 × s × E1 ∼
          ∼         ∼             = E0 × E2 with E2 := s × E1
                = s(N, s) ∼
                          = s(N, E0 ) × s(N, E2 )
            4.158
      =====
      =    ⇒s∼
   s(N, E0 ) =
             ∼ E0 × s(N, E0 ) ⇒
          = s(N, E0 ) × s(N, E2 ) ∼
        ⇒s∼                       = E0 × s(N, E0 ) × s(N, E2 ) ∼
                                                               = E0 × s ∼
                                                                        =E
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Ao . . . polar of A, 11                                c0 (A), 4
E⊗  ˆ π F . . . completed projective tensor prod-      p-approximable operators, 40
        uct, 34                                        p-nuclear operators, 40
E ⊗ F . . . algebraic tensor product, 33               p-summing operators, 40
E ⊗π F . . . projective tensor product, 33             s. . . space of fast falling sequences, 6
E ⊗ε F . . . injective tensor product, 37              b E. . . cbs given by von Neumann bornology,
Mackey convergent, 18
Mackey-complete, 23
Minkowski-functional, 1, 2
Montel space, 28
nuclear operator, 47
nuclear space, 38
nuclearification, 73
polar set, 11
power series space, 6
precompact, 23
probability measure µ, 45