On the definition of pseudospectra
Eugene Shargorodsky
Department of Mathematics
King’s College London
The ε–pseudospectrum:
σε (A) = {λ ∈ C : k(A − λI)−1 k > 1/ε}
or
Σε (A) = {λ ∈ C : k(A − λI)−1 k ≥ 1/ε},
where ε > 0 and k(A − λI)−1 k is assumed to be infinite if λ
belongs to the spectrum of A.
The ε–pseudospectrum:
σε (A) = {λ ∈ C : k(A − λI)−1 k > 1/ε}
↑
or
↓
−1
Σε (A) = {λ ∈ C : k(A − λI)
k ≥ 1/ε},
where ε > 0 and k(A − λI)−1 k is assumed to be infinite if λ
belongs to the spectrum of A.
The ε–pseudospectrum:
σε (A) = {λ ∈ C : k(A − λI)−1 k > 1/ε}
↑
or
↓
−1
Σε (A) = {λ ∈ C : k(A − λI)
k ≥ 1/ε},
where ε > 0 and k(A − λI)−1 k is assumed to be infinite if λ
belongs to the spectrum of A.
Questions:
Is Σε (A) the closure of σε (A)?
Does the ε–pseudospectrum depend continuously on ε?
Can the level set
{λ ∈ C : k(A − λI)−1 k = 1/ε}
have an open subset?
Can the level set
{λ ∈ C : k(A − λI)−1 k = 1/ε}
have an open subset?
m
Main question:
Can k(A − λI)−1 k take a finite constant value on an open set?
Background
Let Ω ⊂ C be open and connected, f : Ω → C be analytic. Then
|f | cannot be constant on an open subset of Ω unless f is
constant in Ω (by the maximum modulus principle).
Background
Let Ω ⊂ C be open and connected, f : Ω → C be analytic. Then
|f | cannot be constant on an open subset of Ω unless f is
constant in Ω (by the maximum modulus principle).
This does not generalise to analytic operator-valued functions:
λ 0
A(λ) =
: C2 → C2 ,
0 1
where C2 is equipped with the `p –norm, 1 ≤ p ≤ ∞. It is clear
that kA(λ)k = 1 if |λ| ≤ 1.
Background
Let Ω ⊂ C be open and connected, f : Ω → C be analytic. Then
|f | cannot be constant on an open subset of Ω unless f is
constant in Ω (by the maximum modulus principle).
This does not generalise to analytic operator-valued functions:
λ 0
A(λ) =
: C2 → C2 ,
0 1
where C2 is equipped with the `p –norm, 1 ≤ p ≤ ∞. It is clear
that kA(λ)k = 1 if |λ| ≤1.
1 0
0
The derivative A (λ) =
is not invertible.
0 0
Background
Let Ω ⊂ C be open and connected, f : Ω → C be analytic. Then
|f | cannot be constant on an open subset of Ω unless f is
constant in Ω (by the maximum modulus principle).
This does not generalise to analytic operator-valued functions:
λ 0
A(λ) =
: C2 → C2 ,
0 1
where C2 is equipped with the `p –norm, 1 ≤ p ≤ ∞. It is clear
that kA(λ)k = 1 if |λ| ≤1.
1 0
0
The derivative A (λ) =
is not invertible.
0 0
On the other hand,
d
dλ (A
− λI)−1 = (A − λI)−2 is invertible.
Cases where the resolvent norm of a bounded linear
operator cannot be constant on an open set
1976
J. Globevnik
1994
Hilbert space (A. Daniluk)
1997
Lp − spaces, 1 < p < ∞ (A. Böttcher,
S. Grudsky and B. Silbermann)
1998
finite-dimensional case (A. Harrabi)
Theorem. (J. Globevnik 1976)
Let X be a Banach space, A ∈ B(X ) and let Ω be an open
subset of the unbounded component of C \ Spec(A). If
k(A − λI)−1 k ≤ M, ∀λ ∈ Ω,
then
k(A − λI)−1 k < M, ∀λ ∈ Ω.
In particular, if C \ Spec(A) is connected, then k(A − λI)−1 k
cannot take a finite constant value on an open set.
Theorem. (J. Globevnik 1976)
Let X be a Banach space, A ∈ B(X ) and let Ω be an open
subset of the unbounded component of C \ Spec(A). If
k(A − λI)−1 k ≤ M, ∀λ ∈ Ω,
then
k(A − λI)−1 k < M, ∀λ ∈ Ω.
In particular, if C \ Spec(A) is connected, then k(A − λI)−1 k
cannot take a finite constant value on an open set.
This theorem applies, e.g., if A is compact and in particular if X
is finite dimensional.
Definition.
A Banach space X is called
complex strictly convex (strictly convex) if
x, y ∈ X , kxk = 1 and
kx + ζy k ≤ 1, ∀ζ ∈ C (∀ζ ∈ R) with |ζ| ≤ 1
⇓
y = 0;
Definition.
A Banach space X is called
complex strictly convex (strictly convex) if
x, y ∈ X , kxk = 1 and
kx + ζy k ≤ 1, ∀ζ ∈ C (∀ζ ∈ R) with |ζ| ≤ 1
⇓
y = 0;
complex uniformly convex (uniformly convex) if for every
ε > 0 there exists δ > 0 such that
x, y ∈ X , ky k ≥ ε and
kx + ζy k ≤ 1, ∀ζ ∈ C (∀ζ ∈ R) with |ζ| ≤ 1
⇓
kxk ≤ 1 − δ.
Uniform convexity
Strict convexity
Complex uniform convexity
Complex strict convexity
Hilbert spaces and the Lp –spaces with 1 < p < ∞ are
uniformly convex (J. Clarkson, 1936).
Hilbert spaces and the Lp –spaces with 1 < p < ∞ are
uniformly convex (J. Clarkson, 1936).
L1 is complex uniformly convex (J. Globevnik, 1975) but not
uniformly convex, not even strictly convex.
Hilbert spaces and the Lp –spaces with 1 < p < ∞ are
uniformly convex (J. Clarkson, 1936).
L1 is complex uniformly convex (J. Globevnik, 1975) but not
uniformly convex, not even strictly convex.
L∞ does not have any of the above properties, but (L∞ )∗ is
complex uniformly convex.
Theorem. (Almost completely from J. Globevnik 1976)
Let Ω be an open subset of C, X be a Banach space such that
X or X ∗ is complex uniformly convex, and A ∈ B(X ). If
k(A − λI)−1 k ≤ M < +∞, ∀λ ∈ Ω,
then
k(A − λI)−1 k < M, ∀λ ∈ Ω.
Theorem. (Almost completely from J. Globevnik 1976)
Let Ω be an open subset of C, X be a Banach space such that
X or X ∗ is complex uniformly convex, and A ∈ B(X ). If
k(A − λI)−1 k ≤ M < +∞, ∀λ ∈ Ω,
then
k(A − λI)−1 k < M, ∀λ ∈ Ω.
Examples of X : Hilbert spaces and Lp (S, Σ, µ) with 1 ≤ p ≤ ∞,
where (S, Σ, µ) is an arbitrary measure space.
Examples where the resolvent norm is constant on an
open set
Example 1. (ES, 2008) The resolvent norm of a bounded linear
operator on `∞ cannot be constant on an open set, but ...
Examples where the resolvent norm is constant on an
open set
Example 1. (ES, 2008) The resolvent norm of a bounded linear
operator on `∞ cannot be constant on an open set, but ...
... modify the standard norm on `∞ (Z):
kxk∗ = sup |xk | + |x0 |,
x = (xk )k ∈Z .
k 6=0
kxk∞ ≤ kxk∗ ≤ 2kxk∞ ,
∀x ∈ `∞ (Z).
Examples where the resolvent norm is constant on an
open set
Example 1. (ES, 2008) The resolvent norm of a bounded linear
operator on `∞ cannot be constant on an open set, but ...
... modify the standard norm on `∞ (Z):
kxk∗ = sup |xk | + |x0 |,
x = (xk )k ∈Z .
k 6=0
kxk∞ ≤ kxk∗ ≤ 2kxk∞ , ∀x ∈ `∞ (Z).
Let X := `∞ (Z), k · k∗ and let A : X → X be the following
weighted shift operator

 xk +1 , k ∈ Z \ {0},
(Ax)k =
 1
k = 0.
3 x1 ,
Then
k(A − λI)−1 k = 3,
|λ| <
1
4
.
The space X in Example 1 is neither reflexive nor complex
strictly convex, and it is natural to ask whether a similar
example can be constructed in a “nicer” space.
The space X in Example 1 is neither reflexive nor complex
strictly convex, and it is natural to ask whether a similar
example can be constructed in a “nicer” space.
Theorem. (ES & S. Shkarin, 2009)
There exist a separable, reflexive and strictly convex Banach
space X and an operator T ∈ B(X ) such that k(T − λI)−1 k is
constant in a neighborhood of λ = 0.
Example 2. (ES, 2008) Let B : `2 (N)
→ `2 (N),
0 αn
B = diag(B1 , B2 , B3 , . . . ) with Bn =
,
βn 0
where 2 ≤ αn → ∞ as n → ∞ and βn := 1 + 1/αn .
Then B is a closed operator with the dense domain
n
Dom(B) = x = (x1 , x2 , . . . ) ∈ `2 (N)
∞
X
o
αn2 |x2n |2 < ∞
n=1
and
k(B − λI)−1 k = 1,
|λ| <
1
2
.
Question (J. R. Partington):
Let H be a Hilbert space and suppose a closed densely defined
operator A : H → H is the generator of a semigroup. Can
k(A − λI)−1 k be constant on an open set?
Question (J. R. Partington):
Let H be a Hilbert space and suppose a closed densely defined
operator A : H → H is the generator of a semigroup. Can
k(A − λI)−1 k be constant on an open set?
Theorem. (ES, 2010)
Let Ω be an open subset of C and X be a complex uniformly
convex Banach space. Suppose A is the infinitesimal generator
of a C0 semigroup on X and A − λI is invertible for all λ ∈ Ω. If
k(A − λI)−1 k ≤ M for all λ ∈ Ω, then k(A − λI)−1 k < M for all
λ ∈ Ω.
Question (J. R. Partington):
Let H be a Hilbert space and suppose a closed densely defined
operator A : H → H is the generator of a semigroup. Can
k(A − λI)−1 k be constant on an open set?
Theorem. (ES, 2010)
Let Ω be an open subset of C and X be a complex uniformly
convex Banach space. Suppose A is the infinitesimal generator
of a C0 semigroup on X and A − λI is invertible for all λ ∈ Ω. If
k(A − λI)−1 k ≤ M for all λ ∈ Ω, then k(A − λI)−1 k < M for all
λ ∈ Ω.
E.B. Davies & ES, 2014: By duality, this result extends to the
case when X ∗ is complex uniformly convex. (Not entirely trivial
if X is not reflexive.)
Theorem. (S. Bögli & P. Siegl, 2014)
Let T be a closed operator in a complex uniformly convex
Banach space X . If there exist an open subset U of the
resolvent set of T and a constant M > 0 such that
k(T − λI)−1 k = M, ∀λ ∈ U, then k(T − λI)−1 k ≥ M for all λ in
the resolvent set of T .
Theorem. (S. Bögli & P. Siegl, 2014)
Let T be a closed operator in a complex uniformly convex
Banach space X . If there exist an open subset U of the
resolvent set of T and a constant M > 0 such that
k(T − λI)−1 k = M, ∀λ ∈ U, then k(T − λI)−1 k ≥ M for all λ in
the resolvent set of T .
Theorem.
Let Ω0 be a connected open subset of C and Z be a Banach
space. Suppose F : Ω0 → Z is an analytic vector valued
function, kF (λ)k ≤ M for all λ in an open subset Ω ⊂ Ω0 , and
kF (µ)k < M for some µ ∈ Ω0 . Then kF (λ)k < M for all λ ∈ Ω.
Theorem. (E.B. Davies & ES, 2014)
Suppose a Banach space X or its dual X ∗ is complex strictly
convex, and B : X → X is a closed densely defined operator
with a compact resolvent R(λ) := (B − λI)−1 . Let Ω be an open
subset of the resolvent set of B. If kR(λ)k ≤ M for all λ ∈ Ω,
then kR(λ)k < M for all λ ∈ Ω.
Let X denote the space `2 (Z) equipped with the following
equivalent norm:
kxk∗ = max kx 0 k2 , |x1 | + |x0 |,
x = (xk )k ∈Z , x 0 = (xk )k ∈Z\{0,1} .
Theorem. (E.B. Davies & ES, 2014)
There exists a closed densely defined operator A : X → X with
a compact resolvent such that k(A − λI)−1 k is constant in a
neighbourhood of λ = 0.
Let X denote the space `2 (Z) equipped with the following
equivalent norm:
kxk∗ = max kx 0 k2 , |x1 | + |x0 |,
x = (xk )k ∈Z , x 0 = (xk )k ∈Z\{0,1} .
Theorem. (E.B. Davies & ES, 2014)
There exists a closed densely defined operator A : X → X with
a compact resolvent such that k(A − λI)−1 k is constant in a
neighbourhood of λ = 0.
Theorem. (E.B. Davies & ES, 2014)
There exists an invertible bounded linear operator B : X → X
such that k(B − λI)−1 k is constant in a neighbourhood of λ = 0.
A norm k · kψ on C2 is called absolute if
k(z, w)kψ = k(|z|, |w|)kψ ,
∀(z, w) ∈ C2
and normalized if
k(1, 0)kψ = k(0, 1)kψ = 1.
A norm k · kψ on C2 is called absolute if
k(z, w)kψ = k(|z|, |w|)kψ ,
∀(z, w) ∈ C2
and normalized if
k(1, 0)kψ = k(0, 1)kψ = 1.
Let X and Y be Banach spaces and let k · kψ be absolute and
normalised. Then X ⊕ψ Y denotes the space X ⊕ Y equipped
with the norm
k(x, y )k = k(kxkX , ky kY )kψ .
Theorem. (E.B. Davies & ES, 2014)
Let Ω be an open subset of C, let Y be a complex strictly
convex Banach space with a complex strictly convex dual Y ∗ ,
and let Y := Y ⊕ψ C. Suppose A : Y → Y is a closed densely
defined operator with a compact resolvent (A − λI)−1 defined
for all λ ∈ Ω. If k(A − λI)−1 k ≤ M for all λ ∈ Ω, then
k(A − λI)−1 k < M for all λ ∈ Ω.
Theorem. (E.B. Davies & ES, 2014)
Let Ω be an open subset of C, let Y be a complex strictly
convex Banach space with a complex strictly convex dual Y ∗ ,
and let Y := Y ⊕ψ C. Suppose A : Y → Y is a closed densely
defined operator with a compact resolvent (A − λI)−1 defined
for all λ ∈ Ω. If k(A − λI)−1 k ≤ M for all λ ∈ Ω, then
k(A − λI)−1 k < M for all λ ∈ Ω.
Theorem. (E.B. Davies & ES, 2014)
Let Ω be an open subset of C, let Y be a complex uniformly
convex Banach space with a complex uniformly convex dual
Y ∗ , and let Y := Y ⊕ψ C. Suppose B is the infinitesimal
generator of a C0 semigroup on Y and B − λI is invertible for all
λ ∈ Ω. If k(B − λI)−1 k ≤ M for all λ ∈ Ω, then k(B − λI)−1 k < M
for all λ ∈ Ω.