where X is a complex Banch space. We eliminate from
consideration the stupid case that X is only the 0 vector. To begin with, here is a fundamental lemma
which will be used whenever convenient. It is about taking a continuous linear transformation through the
integral sign.
Lemma 18.1.1Let f : γ^{∗}→ℒ
(X, X)
be continuous where γ :
[a,b]
→ ℂ has finite total variation and Xis a Banach space. Let A ∈ℒ
(X,X )
. Then
∫ ∫
A f (z)dz = Af (z)dz
γ γ
(∫ ) ∫
f (z)dz A = f (z)Adz
γ γ
When we write AB for A,B ∈ℒ
(X,X )
, it means A ∘ B. That is, A ∘ B
(x)
= A
(B (x))
Proof: This follows from the definition of the integral. Let P denote a sequence of partitions such that
∥P∥
→ 0.
∫ ∑
f (z)dz ≡ lim f (γ(τi))(γ (ti)− γ(ti−1))
γ ∥P∥→0 P
Now multiplication of an element of ℒ
(X, X)
by A ∈ℒ
(X,X )
is continuous because
∥AB1 − AB2∥ ≤ ∥A∥∥B1 − B2∥
∥B1A − B2A∥ ≤ ∥B1 − B2 ∥∥A∥
Therefore,
∫
A f (z)dz ≡ A lim ∑ f (γ(τ ))(γ (t )− γ(t ))
γ ∥P∥→0 P i i i−1
∑
= ∥liPm∥→0 A f (γ(τi))(γ (ti)− γ(ti−1))
∑ P
= lim Af (γ(τi))(γ(ti) − γ (ti−1))
∥∫P∥→0 P
≡ γAf (z)dz
There are no issues regarding existence of the various quantities because the functions are continuous and
the curve is of bounded variation. The other claim is completely similar. ■
Corresponding to A there are two sets defined next.
Definition 18.1.2The resolvent set, denoted as ρ
(A )
is defined as
{ − 1 }
λ ∈ ℂ : (λI − A ) ∈ ℒ(X, X)
The spectrum of A, denoted as σ
(A)
is ℂ ∖ρ
(A )
.When λ ∈ ρ
(A)
, we call
(λI − A )
^{−1}the resolvent. Thus,in particular, when λ is in ρ
(A)
,λI − A is one to one and onto.
There is a fundamental identity involving the resolvent which should be noted first. In order to
remember this identity, simply write
(λI − A)−1 ≈ --1--
λ− A
and proceed formally.
1 1 μ− λ
-----− ----- = --------------
λ− A μ − A (λ − A)(μ− A )
This suggests that
(λI − A )−1 − (μI − A)−1 = (μ − λ)(λI − A)−1(μI − A)−1
−1 −1
= (μ − λ)(μI − A) (λI − A) (18.1)
It is easy to verify that this is in fact the case. Multiply on the left by
(λI − A)
and on the right by
(μI − A)
in the top line. This yields
μI − A− (λI − A)
on the left and it yields
(μ − λ)
I on the right which is seen to be the same thing. Thus we have two
quantities in 18.1, x on the left and y on the right and we have
(λI − A )x(μI − A) = (λI − A )y(μI − A )
Since both μI −A and λI −A are one to one and onto, it follows that x = y. You can do a similar thing to
the expression on the second row. Multiply on the left by
The notion of spectrum is just a generalization of the concept of eigenvalues in linear algebra. In linear
algebra, everything is finite dimensional and the spectrum is just the set of eigenvalues. You can get them
by looking for solutions to the characteristic equation which involves a determinant and they exist because
of the fundamental theorem of algebra. No such thing is available in a general Banach space. However,
many of the same results still hold.
Proposition 18.1.6For A ∈ℒ
(X,X )
,σ
(A)
≠∅ and σ
(A)
is a compact set. If λ ∈ σ
(A)
, then forall n ∈ ℕ, it follows that λ^{n}∈ σ
n
(A )
.
Proof:Suppose first that σ
(A)
= ∅. Then you would have λ →
(λI − A)
^{−1} is analytic and in fact
from estimate 4 above,
∥ ∥
lim ∥∥(λI − A )−1∥∥ = 0 (18.3)
|λ|→∞
(18.3)
Thus there exists r such that if
|λ|
< r, then
∥ ∥
∥∥(λI − A)−1∥∥
< 1. However,
∥ ∥
∥∥(λI − A )− 1∥∥
is bounded for
|λ|
≤ r and so λ →
(λI − A)
^{−1} is analytic on all of ℂ (entire) and is bounded. Therefore, it
is constant thanks to Liouville’s theorem, Theorem 15.6.2. But the constant can only be 0
thanks to 18.3. Therefore,
(λI − A )
^{−1} = 0 which is nonsense since we are not considering
X =
{0}
.
σ
(A )
is the complement of the open set ρ
(A )
and so it is closed. It is bounded thanks to 4 of
Proposition 18.1.5. Therefore, it is compact by the Heine Borel theorem.
It remains to verify the last claim.
n∑−1 n∑−1 n∑−1
(λI − A ) λkA(n−1)− k = λk+1A (n−1)−k − λkAn −k
k=0 k=0 k=0
∑n n∑−1
= λkAn− k − λkAn− k = λnI − An (18.4)
k=1 k=0
n∑−1
= λkA (n−1)−k(λI − A )
k=0
If λ ∈ σ
(A)
, then this means
(λI − A)
^{−1} does not exist. Thanks to the open mapping theorem, this is
equivalent to either λI −A not being one to one or not onto because by this theorem, if it is both, then it is
continuous and has continuous inverse because it maps open sets to open sets. Now consider the identity
18.4. If λI −A is not one to one, then the bottom line and the middle line shows that λ^{n}I −A is not one to
one. If
(λI − A)
is not onto, then the top line and the middle line shows that λ^{n}I −A^{n} is not onto. Thus
λ^{n}∈ σ
(An)
. ■
Definition 18.1.7σ
(A)
is closed and bounded. Let r
(A )
denote the spectral radiusdefinedas
max {|λ| : λ ∈ σ (A)}
It is very important to have a description of this spectral radius. The following famous result is due to
Gelfand.