(November 1, 2020)

Examples of operators and spectra

Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/egarrett/
[This document is http://www.math.umn.edu/˜garrett/m/fun/notes 2012-13/06b examples spectra.pdf]
1. Generalities on spectra
2. Positive examples
3. Simplest Rellich compactness lemma
4. Cautionary examples: non-normal operators
The usefulness of the notion of spectrum of an operator on a Hilbert space is the analogy to eigenvalues of
operators on finite-dimensional spaces. Naturally, things become more complicated in infinite-dimensional
vector spaces.

1. Generalities on spectra
It is convenient to know that spectra of continuous operators are non-empty, compact subsets of C.
Knowing this, every non-empty compact subset of C is easily made to appear as the spectrum of a continuous
operator, even normal ones, as below.

[1.0.1] Proposition: The spectrum σ(T ) of a continuous linear operator T : V → V on a Hilbert space V
is bounded by the operator norm |T |op .

Proof: For |λ| > |T |op , an obvious heuristic suggests an expression for the resolvent Rλ = (T − λ)−1 :
T
−1  T T
2 
(T − λ)−1 = −λ−1 · 1 − = −λ−1 · 1 + + + ...
λ λ λ
The infinite series converges in operator norm for |T /λ|op < 1, that is, for |λ| > |T |op . Then
 T T
2 
(T − λ) · (−λ−1 ) · 1 + + + ... = 1
λ λ
giving a continuous inverse (T − λ)−1 , so λ 6∈ σ(T ). ///

[1.0.2] Remark: The same argument applied to T n shows that σ(T n ) is inside the closed ball of radius
|T n |op . By the elementary identity

T n − λn = (T − λ) · (T n−1 + T n−2 λ + . . . + T λn−2 + λn−1 )

1/n
(T − λ)−1 exists for |λn | > |T n |op , that is, for |λ| > |T n |op . That is, σ(T ) is inside the closed ball of radius
1/n
inf n≥1 |T n |op . The latter expression is the spectral radius of T . This notion is relevant to non-normal
operators, such as the Volterra operator, whose spectral radius is 0, while its operator norm is much larger.

[1.0.3] Proposition: The spectrum σ(T ) of a continuous linear operator T : V → V on a Hilbert space V
is compact.

Proof: That λ 6∈ σ(T ) is that there is a continuous linear operator (T −λ)−1 . We claim that for µ sufficiently
close to λ, (T − µ)−1 exists. Indeed, a heuristic suggests an expression for (T − µ)−1 in terms of (T − λ)−1 .
The algebra is helpfully simplified by replacing T by T + λ, so that λ = 0. With µ near 0 and granting
existence of T −1 , the heuristic is
 
(T − µ)−1 = (1 − µT −1 )−1 · T −1 = 1 + µT −1 + (µT −1 )2 + . . . · T −1

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The geometric series converges in operator norm for |µT −1 |op < 1, that is, for |µ| < |T −1 |−1
op . Having found
the obvious candidate for an inverse,
 
(1 − µT −1 ) · 1 + µT −1 + (µT −1 )2 + . . . = 1

and  
(T − µ) · 1 + µT −1 + (µT −1 )2 + . . . · T −1 = 1

so there is a continuous linear operator (T − µ)−1 , and µ 6∈ σ(T ). Having already proven that σ(T ) is
bounded, it is compact. ///

[1.0.4] Proposition: The spectrum σ(T ) of a continuous linear operator on a Hilbert space V 6= {0} is
non-empty.

Proof: The argument reduces the issue to Liouville’s theorem from complex analysis, that a bounded entire
(holomorphic) function is constant. Further, an entire function that goes to 0 at ∞ is identically 0.

Suppose the resolvent Rλ = (T − λ)−1 is a continuous linear operator for all λ ∈ C. The operator norm is
readily estimated for large λ:

T T
2
|Rλ |op = |λ|−1 · 1 + + + ...
λ λ op

 T T 2
 1 1
≤ |λ|−1 · 1 + + + ... = ·
λ op λ op |λ| |T |op
1−
|λ|
This goes to 0 as |λ| → ∞. Hilbert’s identity asserts the complex differentiability as operator-valued function:

Rλ − Rµ (T − µ) − (T − λ)
= Rλ · · Rµ = Rλ · Rµ −→ Rλ2 (as µ → λ)
λ−µ λ−µ

since µ → Rµ is continuous for large µ, by the same identity:

|Rλ − Rµ |op ≤ |λ − µ| · |Rµ · Rλ |op

Thus, the scalar-valued functions λ → hRλ v, wi for v, w ∈ V are complex-differentiable, and satisfy

1 1
|hRλ v, wi| ≤ |Rλ v| · |w| ≤ |Rλ |op · |v| · |w| ≤ · · |v| · |w|
|λ| |T |op
1−
|λ|

By Liouville, hRλ v, wi = 0 for all v, w ∈ V , which is impossible. Thus, the spectrum is not empty. ///

[1.0.5] Proposition: The entire spectrum, both point-spectrum and continuous-spectrum, of a self-adjoint
operator is a non-empty, compact subset of R. The entire spectrum of a unitary operator is a non-empty,
compact subset of the unit circle.

Proof: For self-adjoint T , we claim that the imaginary part of h(T − µ)v, vi is at least hv, vi times the
imaginary part of µ. Indeed, hT v, vi is real, since

hT v, vi = hv, T ∗ vi = hv, T vi = hT v, vi

so
h(T − µ)v, vi = hT v, vi − µ · hv, vi

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and
|Im h(T − µ)v, vi| = |Im µ| · hv, vi
and by Cauchy-Schwarz-Bunyakowsky

|(T − µ)v| · |v| ≥ |h(T − µ)v, vi| ≥ |Im µ| · hv, vi = |Im µ| · |v|2

Dividing by |v|,
|(T − µ)v| ≥ |Im µ| · |v|
This inequality shows more than the injectivity of T − µ. Namely, the inequality gives a bound on the
operator norm of the inverse (T − µ)−1 defined on the image of T − µ. The image is dense since µ is not an
eigenvalue and there is no residual spectrum for normal operators T . Thus, the inverse extends by continuity
to a continuous linear map defined on the whole Hilbert space. Thus, T − µ has a continuous linear inverse,
and µ is not in the spectrum of T .
For T unitary, |T v| = |v| for all v implies T |op = 1. Thus, σ(T ) is contained in the unit disk, by the general
bound on spectra in terms of operator norms. From (T − λ)∗ = T ∗ − λ, the spectrum of T ∗ is obtained by
complex-conjugating the spectrum of T . Thus, for unitary T , the spectrum of T −1 = T ∗ is also contained
in the unit disk. At the same time, the natural

T − λ = −T · (T −1 − λ−1 ) · λ

gives
(T − λ)−1 = −λ−1 · (T −1 − λ−1 )−1 · T −1
so λ−1 ∈ σ(T −1 ) exactly when λ ∈ σ(T ). Thus, the spectra of both T and T ∗ = T −1 are inside the unit
circle. ///

2. Positive examples
This section gives non-pathological examples. Let `2 be the usual space of square-summable sequences
(a1 , a2 , . . .), with standard orthonormal basis

ej = (0, . . . , 0, 1, 0, . . .)
| {z }
1 at jth position

[2.1] Multiplication operators with specified eigenvalues Given a countable, bounded list of complex
numbers λj , the operator T : `2 → `2 by

T : (a1 , a2 , . . .) −→ (λ1 · a1 , λ2 · a2 , . . .)

has λj -eigenvector the standard basis element ej . Clearly

T ∗ : (a1 , a2 , a3 , . . .) −→ (λ1 · a1 , λ2 · a2 , λ3 · a3 , . . .)

so T is normal, in the sense that T T ∗ = T ∗ T . To see that there are no other eigenvalues, suppose T v = µ · v
with µ not among the λj . Then

µ · hv, ej i = hT v, ej i = hv, T ∗ ej i = hv, λj ej i = λj · hv, ej i

Thus, (µ − λj ) · hv, ej i = 0, and hv, ej i = 0 for all j. Since ej form an orthonormal basis, v = 0. ///

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[2.2] Every compact subset of C is the spectrum of an operator Grant for the moment a countable
dense subset {λj } of a non-empty compact subset [1] C of C, and as above let

T : (a1 , a2 , a3 , . . .) −→ (λ1 · a1 , λ2 · a2 , λ3 · a3 , . . .)

We saw that there are no further eigenvalues. Since spectra are closed, the closure C of {λj } is contained in
σ(T ).
It remains to show that the continuous spectrum is no larger than the closure C of the eigenvalues, in this
example. That is, for µ 6∈ C, exhibit a continuous linear (T − µ)−1 .
For µ 6∈ C, there is a uniform lower bound 0 < δ ≤ |µ − λj |. That is, supj |µ − λj |−1 ≤ δ −1 . Thus, the
naturally suggested map
 
(a1 , a2 , . . .) −→ (λ1 − µ)−1 · a1 , (λ2 − µ)−1 · a2 , . . .

is a bounded linear map, and gives (T − µ)−1 .

[2.3] Two-sided shift has no eigenvalues Let V be the Hilbert space of two-sided sequences
(. . . , a−1 , a0 , a1 , . . .) with natural inner product

h(. . . , a−1 , a0 , a1 , . . .), (. . . , b−1 , b0 , b1 , . . .)i = . . . + a−1 b−1 + a0 b0 + a1 b1 + . . .

The right and left two-sided shift operators are

(R · a)n = an−1 (L · a)n = an+1

These operators are mutual adjoints, mutual inverses, so are unitary. Being unitary, their operator norms
are 1, so their spectra are non-empty compact subsets of the unit circle.

They have no eigenvalues: indeed, for Rv = λ · v, if there is any index n with vn 6= 0, then the relation
Rv = λ · v gives vn+k+1 = λ · vn+k for k = 0, 1, 2, . . .. Since |λ| = 1, such a vector is not in `2 .
Nevertheless, we claim that λ ∈ σ(L) for every λ with |λ| = 1, and similarly for R. Indeed, for λ not in the
spectrum, there is a continuous linear operator (L − λ)−1 , so |(L − λ)v| ≥ δ · |v| for some δ > 0. It is easy
to make approximate eigenvectors for L for any |λ| = 1: let

v (`) = (. . . , 0, . . . , 0, 1, λ, λ2 , λ3 , . . . , λ` , 0, 0, . . .)

Obviously it doesn’t matter where the non-zero entries begin. From

(L − λ)v (`) = (. . . , 0, . . . , 0, 1, 0, . . . , 0, λ`+1 , 0, 0, . . .)

√ √
|(L − λ)v (`) | = 1 + 1, while |v (`) | = ` + 1. Thus, |(L − λ)v (`) |/|v (`) | −→ 0, and there can be no (L − λ)−1 .
Thus, every λ on the unit circle is in σ(R).

[2.4] Compact multiplication operators on `2 For a sequence of complex numbers λn → 0, we claim

that the multiplication operator

T : (a1 , a2 , . . .) −→ (λ1 · a1 , λ2 · a2 , . . .)

[1] To make a countable dense subset of C, for n = 1, 2, . . . cover C by finitely-many disks of radius 1/n, each meeting
C, and in each choose a point of C. The union over n = 1, 2, . . . of these finite sets is countable and dense in C.

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is compact. We already showed that it has eigenvalues exactly λ1 , λ2 , . . ., and spectrum the closure of {λj }.
Thus, the spectrum includes 0, but 0 is an eigenvalue only when it appears among the λj , which may range
from 0 times to infinitely-many times.
To prove that the operator is compact, we prove that the image of the unit ball is pre-compact, by showing
that it is totally bounded. Given ε > 0, take k such that |λi | < ε for i > k. The ball in Ck of radius
max{|λj | : j ≤ k} is precompact, so has a finite cover by ε-balls, centered at points v 1 , . . . , v N . For
v = (v1 , v2 , . . .) with |v| ≤ 1,

T v = (λ1 v1 , λ2 v2 , . . . , λk vk , 0, 0, . . .) + (0, . . . , 0, λk+1 vk+1 , λk+2 vk+2 , . . .)

With v j the closest of the v 1 , . . . , v N to (λ1 v1 , λ2 v2 , . . . , λk vk , 0, 0, . . .),

|T v−v j | < ε+|(0, . . . , 0, λk+1 vk+1 , λk+2 vk+2 , . . .)| < ε + ε·|(0, . . . , 0, vk+1 , vk+2 , . . .)| ≤ ε+ε·|v| ≤ 2ε

Thus, the image of the unit ball under T is covered by finitely-many 2ε-balls. ///

[2.5] Multiplication operators on L2 [a, b] For ϕ ∈ C o [a, b], we claim that the multiplication operator
Mϕ : L2 [a, b] −→ L2 [a, b]

by
Mϕ f (x) = ϕ(x) · f (x)
is normal, and has spectrum the image ϕ[a, b] of ϕ. The eigenvalues are λ such that ϕ(x) = λ on a subset of
[a, b] of positive measure. The normality is clear, so, beyond eigenvalues, we need only examine continuous
spectrum, not residual.
On one hand, if ϕ(x) = λ on a set of positive measure, there is an infinite-dimensional sub-space of L2 [0, 1]
of functions supported there, and all these are eigenvectors. On the other hand, if f 6= 0 in L2 [0, 1] and
ϕ(x) · f (x) = λ · f (x), even if f is altered on a set of measure 0, it must be that ϕ(x) = λ on a set of positive
measure.
To understand the continuous spectrum, for ϕ(xo ) = λ make approximate eigenvectors by taking L2 functions
f supported on [xo − δ, xo + δ], where δ > 0 is small enough so that |ϕ(x) − ϕ(xo )| < ε for |x − xo | < δ. Then
Z
2
|(Mϕ − λ)f |L2 = |ϕ(x) − λ|2 · |f (x)|2 dx ≤ ε2 · |f |2L2

Thus, inf f 6=0 |(Mϕ − λ)f |L2 /|f |L2 = 0, so Mϕ − λ is not invertible. If λ is not an eigenvalue, it is in the
continuous spectrum. On the other hand, if ϕ(x) 6= λ, then there is some δ > 0 such that |ϕ(x) − λ| ≥ δ for
all x ∈ [0, 1], by the compactness of [0, 1]. Then
Z 1 Z 1
|(Mϕ − λ)f |2L2 = |ϕ(x) − λ|2 · |f (x)|2 dx ≥ δ 2 · |f (x)|2 dx = δ 2 · |f |2L2
0 0

Thus, there is a continuous inverse (Mϕ − λ), and λ is not in the spectrum.

3. Simplest Rellich compactness lemma

One characterization of the sth Levi-Sobolev space of functions H s (A) on a product A = (S 1 )×n of circles
S 1 = R/2πZ is as the closure of the function space of finite Fourier series with respect to the Levi-Sobolev
norm (squared)
X 2 X
cξ eiξ·x = |cξ |2 · (1 + |ξ|2 )s (s ∈ R, on finite Fourier series)
Hs
ξ∈Zn ξ∈Zn

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The standard orthonormal basis for H s (A) is

1 eiξ·x
· (with ξ ∈ Zn )
(2π)n/2 (1 + |ξ|2 )s/2

By the Plancherel theorem, the map from L2 (Zn ) (with counting measure) to L2 (A) by

1 X eiξ·x
{cξ : ξ ∈ Zn } −→ n/2
cξ
(2π) n
(1 + |ξ|2 )s/2
ξ∈Z

is an isometric isomorphism.
For s > t, there is a continuous inclusion H s (A) → H t (A). In terms of these orthonormal bases, there is a
commutative diagram
L2 (A)
T / L2 (A)

≈ ≈
 
H s (A) / H t (A)
inc

given by

/ (1 + |ξ|2 ) t−s
n o
T
{cξ } 2 · cξ

≈ ≈
 
1 X eiξ·x / 1 X
2 t−s eiξ·x
cξ (1 + |ξ| ) 2 · c ξ
(2π)n/2 ξ (1 + |ξ|2 )s/2 inc (2π)n/2 ξ (1 + |ξ|2 )t/2

t−s
Since s > t, the number λξ = (1 + |ξ|2 ) 2 are bounded by 1, and have unique limit point 0. In particular,
T : L2 (A) → L2 (A) is compact.
Thus, we have the simplest instance of Rellich’s compactness lemma: the inclusion H s (A) → H t (A) is
compact for s > t.

4. Cautionary examples: non-normal operators

[4.1] Shift operators on one-sided `2 We claim the following: The right-shift

R : (a1 , a2 , . . .) −→ (0, a1 , a2 , . . .)

and the left-shift

L : (a1 , a2 , a3 , . . .) −→ (a2 , . . .)
are mutual adjoints. These operators are not normal, since L ◦ R = 1`2 but

R ◦ L : (a1 , a2 , . . .) −→ (0, a2 , . . .)

The eigenvalues of the left-shift L are all complex numbers in the open unit disk in C. In particular, there
is a continuum of eigenvalues and eigenvectors, so they cannot be mutually orthogonal. The spectrum σ(L)
is the closed unit disk.

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The right-shift R has no eigenvalues, has continuous spectrum the unit circle, and residual spectrum the
open unit disk with 0 removed.
Indeed, suppose
(0, a1 , a2 , . . .) = R(a1 , a2 , . . .) = λ · (a1 , a2 , . . .)
With n the lowest index such that an 6= 0, the nth component in the eigenvector relation gives 0 = an−1 =
λ · an , so λ = 0. Then, the (n + 1)th component gives an = λ · an+1 = 0, contradiction. This proves that R
has no eigenvalues.
Oppositely, for |λ| < 1,
L(1, λ, λ2 , . . .) = (λ, λ2 , . . .) = λ · (1, λ, λ2 , . . .)
so every such λ is an eigenvector for L. On the other hand, for |λ| = 1, in an eigenvector relation

(a2 , . . .) = L(a1 , a2 , . . .) = λ · (a1 , a2 , . . .)

let n be the smallest index n with an 6= 0. Then an+1 = λ · an , an+2 = λ · an+1 , . . ., so

(a1 , a2 , . . .) = (0, . . . , 0, an , λan , λ2 an , . . .)

But this is not in `2 for |λ| = 1 and an 6= 0, so λ on the unit circle is not an eigenvalue.
For |λ| = 1, we can make approximate λ-eigenvectors for L by

v [N ] = (1, λ, λ2 , . . . , λN , 0, 0, . . .)

since

(L − λ) v [N ] = (λ, λ2 , . . . , λN , 0, 0, 0, . . .) − λ · (1, λ, λ2 , . . . , λN , 0, 0, . . .) = (0, 0, . . . , 0, 0, λN +1 , 0, 0, . . .)

Since
|(L − λ)v [N ] | |λ|N +1 1
[N ]
= 2 2N 1/2
= √ −→ 0
|v (1 + |λ| + . . . + |λ| ) N +1
there can be no continuous (L − λ)−1 . Thus, λ on the unit circle is in the spectrum, but not in the point
spectrum.
That the unit circle is in the spectrum also follows from the observation above that all λ with |λ| < 1 are
eigenvalues, and the fact that the spectrum is closed.
The spectrum of L is bounded by the operator norm |L|op , and |L|op is visibly 1, so is nothing else in the
spectrum.
To see that the unit circle is the continuous spectrum of L, as opposed to residual, we show that L − λ has
dense image for |λ| = 1. Indeed, for w such that, for all v ∈ `2 ,

0 = h(L − λ)v, wi = hv, (L∗ − λ)wi = hv, (R − λ)wi

we would have (R − λ)w = 0. However, we have seen that R has no eigenvalues. Thus, L − λ always has
dense image, and the unit circle is continuous spectrum for L.
Reversing that discussion, every λ with |λ| < 1 is in the residual spectrum of R, because such λ is not an
eigenvalue, and R − λ does not have dense image: for w a λ-eigenvector for L,

h(R − λ)v, wi = hv, (R∗ − λ)wi = hv, (L − λ)wi = hv, 0i = 0

That is, the image (R − λ)`2 is in the orthogonal complement to the eigenvector w. The same computation
shows that the unit circle is continuous spectrum for R, because it is not eigenvalues for L.

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Rx
[4.2] Volterra operator We will show that the Volterra operator V f (x) = 0
f (t) dt on L2 [0, 1] is compact,
but not self-adjoint, that its spectrum is {0}, and that it has no eigenvalues.
A relation T f = λ · f for f ∈ L2 and λ 6= 0 implies f is continuous:
Z x+h
1
|λ| · |f (x + h) − f (x)| = |T f (x + h) − T f (x)| ≤ 1 · |f (t)| dt = ≤ |h| 2 · |f |L2
x

The fundamental theorem of calculus would imply f is continuously differentiable and λ · f 0 = (T f )0 = f .

Thus, f would be a constant multiple of ex/λ , by the mean value theorem. However, by Cauchy-Schwarz-
Bunyakowsky, for a λ-eigenfunction
1
|λ| · |f (x)| ≤ |x| 2 · |f |L2
No non-zero multiple of the exponential satisfies this. Thus, there are no eigenvectors for non-zero
eigenvalues.

For f ∈ L2 [0, 1] and T f = 0 ∈ L2 [0, 1], T f is almost everywhere 0. Since x → T g(x) is unavoidably
continuous, T f (x) is 0 for all x. Thus, for all x, y in the interval,
Z y
0 = 0 − 0 = T f (y) − T f (x) = f (t) dt
x

That is, x → T f (x) is orthogonal in L2 [0, 1] to all characteristic functions of intervals. Finite linear
combinations of these are dense in C o [0, 1] in the L2 topology, and C o [0, 1] is dense in L2 [0, 1]. Thus
f = 0, and there are no eigenvectors for the Volterra operator.
To see that T is compact, rewrite it as being given by an integral kernel K(x, y):
Z x Z 1 
0 (for 0 ≤ y < x)
T f (x) = f (y) dy = K(x, y) f (y) dy (with K(x, y) = )
0 0 1 (for x < y ≤ 1)

Thus, T is Hilbert-Schmidt, and compact. The adjoint T ∗ is given by the integral kernel K ∗ (x, y) = K(y, x),
visibly different from K(x, y), so T is not self-adjoint.
To see that the spectrum is at most {0}, show that the spectral radius is 0:
Z x Z xn−1 Z x2 Z x1 Z x Z xZ xn−1 Z x2 
T n f (x) = ... f (t) dt dx1 . . . dxn−1 = f (t) ... dx1 . . . dxn−1 dt
0 0 0 0 0 t t t

x
(x − t)n−1
Z
= f (t) · dt
0 (n − 1)!
n 1
From this, |T |op ≤ n! , and
 1 1/2n 1 1 X
log lim = − lim · log(2n)! = − lim log k
2n (2n)! 2n 2n 2n 2n
1≤k≤2n

1 X 1 X
= − lim (log k + log(2n − k + 1)) ≤ − lim (log k + log(2n − k + 1))
2n 2n 2n 2n n
1≤k≤n 1≤k≤ k

1 X log 2n
≤ − lim log 2n = − lim = −∞
2n 2n n
2n 2
1≤k≤ k

1/n
since k(2n − k) ≥ 2n for 1 ≤ k ≤ n, noting the sign. That is, limn |T n |op = 0, so the spectral radius is 0.
Since the spectrum is non-empty, it must be exactly {0}.