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FTex8 Solutions

This document contains solutions to functional analysis exercises. The key points summarized are: 1) A sequence xn in a Banach space X is bounded if the linear maps Ln defined by each xn are uniformly bounded. 2) A sequence xn converges weakly to x in a Hilbert space H if and only if kxn - xk → 0. 3) For operators Pn projecting onto spans of the first n basis vectors, Pn does not converge in norm but does converge weakly to the identity operator. 4) Strong convergence of operators An does not guarantee AnBn converges strongly to AB if Bn only converges weakly to B.

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Manuel Chaves
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100 views3 pages

FTex8 Solutions

This document contains solutions to functional analysis exercises. The key points summarized are: 1) A sequence xn in a Banach space X is bounded if the linear maps Ln defined by each xn are uniformly bounded. 2) A sequence xn converges weakly to x in a Hilbert space H if and only if kxn - xk → 0. 3) For operators Pn projecting onto spans of the first n basis vectors, Pn does not converge in norm but does converge weakly to the identity operator. 4) Strong convergence of operators An does not guarantee AnBn converges strongly to AB if Bn only converges weakly to B.

Uploaded by

Manuel Chaves
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Functional Analysis Exercise sheet 8 — solutions

1. Each element xn defines a linear map Ln : X ∗ → C by Ln (f ) = f (xn ).


Since the sequence f (xn ) is Cauchy, it is bounded. Hence, there exists
c = c(f ) > 0 such that |Ln (f )| ≤ c for all n. We note that X ∗ is also
a Banach space. By the Uniform Boundedness Theorem, supn kLn k <
∞. Finally,
kLn k = sup{|Ln (f )| : kf k = 1} = sup{|f (xn )| : kf k = 1} = kxn k.
This proves that the sequence xn is bounded.
2. (a) Suppose that kxn − xk → 0. Then for every y ∈ H,
| hxn , yi − hx, yi | ≤ kxn − xkkyk → 0.
Hence, xn → x weakly. Also by triangle inequality,
|kxn k − kxk| ≤ kxn − xk → 0.
To prove the converse, we observe that
kx − xn k2 = kxk2 − hxn , xi − hxn , xi + kxn k2 .
By weak convergence, hxn , xi → kxk2 . Hence, kx − xn k2 → 0 as
required.
(b) We have
| hxn , yn i − hx, yi | ≤ | hxn , yn i − hx, yn i | + | hx, yn i − hx, yi |
= | hxn − x, yn i | + | hx, yn − yi |.
Since yn → y weakly, the second term converges to zero. By
Cauchy-Schwarz inequality,
| hxn − x, yn i | ≤ kxn − xkkyn k.
Since yn is weakly convergent, it is bounded. Hence, it follows
that | hxn − x, yn i | → 0.
(c) This is not true in general. For example, consider H = `2 and
xn = yn = en . Then en → 0 weakly, but hen , en i = 1 does not
converge to zero.
3. (a) Let L(f ) = f (0). We claim that Ln → L weak∗ . Since f is
continuous, for every  > 0 and n ≥ n0 (), we have |f (t)−f (0)| <
 for all t ∈ [0, 1/n]. Then
Z 1/n
|Ln (f ) − L(f )| ≤ n |f (t) − f (0)| dt < .
0

Hence, Ln (f ) → L(f ) for all f ∈ C([0, 1]).

1
(b) It follows form (a) that if Ln → S in norm, then S = L. Consider
a function fn such that 0 ≤ fn ≤ 1, fn (0) = 0, and fn = 1 on
1 2
[ 3n , 3n ]. Then kfn k = 1, and
Z
1/n 1
|Ln (fn ) − L(fn )| = n fn (t) dt ≥ .

0 3

Hence, kLn − Lk ≥ 31 , and the sequence Ln does not converge in


norm.

4. (a) For n < m,


2
m
X m
X
kPn x − Pm xk2 = hx, ek i ek = | hx, ek i |2 .


k=n+1 k=n+1

Hence, by the Bessel inequality, kPn x − Pm xk2 ≤ kxk2 . This


proves that kPn − Pm k ≤ 1. For x = em , kPn x − Pm xk = kem k.
Hence, kPn − Pm k = 1.
If we suppose Pn → P for some P in norm, then kPn − Pm k ≤
kPn − P k + kP − Pm k → 0 as n, m → ∞, which is impossible.
(b) We recall that for every x ∈ H, x = ∞
P P∞ 2
k=1 hx, ek i ek and k=1 | hx, ek i | <
∞. Hence,
∞

X XN
kPn x − Ixk = hx, ek i ek = lim hx, ek i ek

N →∞
k=n+1 k=n+1
v v
u ∞
u N u X
u X
= lim t | hx, ek i |2 = t | hx, ek i |2 ,
N →∞
k=n+1 k=n+1

and kPn x − Ixk → 0 as n → ∞.

5. (a) We first note that since An strongly converge, the sequence of


norms kAn k is bounded. We fix c > 0 such that kAn k ≤ c for all
n. For every x ∈ X,

kAn Bn x − ABxk ≤ kAn Bn x − An Bxk + kAn Bx − ABxk


≤ kAn kkBn x − Bxk + kAn (Bx) − A(Bx)k
≤ ckBn x − Bxk + kAn (Bx) − A(Bx)k
s s
Since An → A and Bn → B, we have kBn x − Bxk → 0 and
kAn (Bx) − A(Bx)k → 0. Hence,

kAn Bn x − ABxk → 0.

2
(b) Consider the operators An , Bn : `2 → `2 defined by Bn x =
hx, e1 i en and An x = hx, en i e1 . For every x, y ∈ `2 , hBn x, yi =
s
hx, e1 i hen , yi → 0, and kAn xk → 0. Hence, An → A and
w w
Bn → B. We also have An Bn x = hx, e1 i e1 . Hence, An Bn → 6 AB.

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