where the K_{k} are disjoint compact sets. Let δ be small enough that no point of any K_{k} is within δ
of any other K_{j} as in the proof of Theorem 16.5.2. Let the open set containing K_{j} be given
by U_{j}≡ K_{j} + B
(0,δ∕2)
, defined as all numbers in ℂ of the form k + d where
|d|
<
δ2
and
k ∈ K_{j}. Then by the construction of Theorem 16.5.2, K_{j}⊆ U_{j} and Γ_{j}^{∗}⊆ U_{j} also. In addition
to this, the open sets just described are disjoint. Now let f_{k}
(z)
= 1 on U_{k} and let it be 0
everywhere else. Thus f_{k} is not analytic on ℂ but it is analytic on ∪_{j}U_{j}≡ Ω and Ω contains
σ
(A )
.
Now f_{k}
(λ)
(λI − A )
^{−1} is not analytic on Ω because Ω contains σ
(A)
. Recall Theorem 16.5.2 about the
existence of cycles in this context.
Theorem 18.2.1Let K_{1},K_{2},
⋅⋅⋅
,K_{m}be disjoint compact subsets of an open set Ω in ℂ. Then there existcontinuous, closed, bounded cycles
{Γ j}
_{j=1}^{m}for which Γ_{j}^{∗}∩K_{k} = ∅ for each k,j,Γ_{j}^{∗}∩ Γ_{k}^{∗} = ∅, Γ_{j}^{∗}⊆ Ω.Also, if p ∈ K_{k}and j≠k,
n(Γ k,p) = 1, n (Γ j,p) = 0
so if p is in some K_{k},
∑m
n(Γ j,p) = 1
j=1
each Γ_{j}being the union of oriented simple closed curves, while for all z
∈∕
Ω
∑m
n (Γ k,z) = 0.
k=1
Also, if p ∈ Γ_{j}^{∗}, then for i≠j,n
(Γ i,p)
= 0.
Also recall the corollary to this theorem, Corollary 16.5.3.
Corollary 18.2.2In the context of Theorem 16.5.2, there exist continuous, closed, bounded cycles
{ }
ˆΓ j
_{j=1}^{m}for which
ˆΓ
_{j}^{∗}∩ K_{k} = ∅ for each k≠j,
ˆΓ
_{j}^{∗}∩
ˆΓ
_{k}^{∗} = ∅,
ˆΓ
_{j}^{∗}⊆ Ω. Also, if p ∈ K_{k}andj≠k,
( ) ( )
n ˆΓ k,p = 1, n ˆΓ j,p = 0
so if p is in some K_{k},
∑m (ˆ )
n Γ j,p = 1
j=1
each
ˆ
Γ
_{j}being the union of oriented simple closed curves, while for all z
∈∕
Ω
m∑ ( )
n Γˆk,z = 0.
k=1
Also, if p ∈
ˆΓ
_{j}^{∗}, then for i≠j,n
( )
ˆΓ i,p
= 0. Additionally, n
( )
ˆΓ k,p
= 0 if p ∈ Γ_{k}^{∗}.
There is a general notion of finding functions of linear operators.
Definition 18.2.3Let A ∈ ℒ
(X, X )
and let Ω be an open set which contains σ
(A )
. Let Γ be acycle which has Γ^{∗}⊆ Ω ∩ σ
(A)
^{C}, and suppose that
n
(Γ ,z)
= 1 if z ∈ σ
(A )
.
n
(Γ ,z)
is an integer if z ∈ Ω.
n
(Γ ,z)
= 0 if z
∈∕
Ω.
Then if f is analytic on Ω,f
(A )
≡
21πi
∫_{Γ}f
(λ)
(λI − A )
^{−1}dλ.
First of all, why does this make sense for things which have another meaning? In particular, why does it
make sense if λ^{n} = f
(λ)
? Is A^{n} correctly given by this formula? If not, then this isn’t a very good way to
define f
(A)
. This involves the following theorem which says that if you look at f
(A)
g
(A )
defined above, it
gives the same thing as the above applied to f
(λ)
g
(λ)
.
Theorem 18.2.4Suppose Ω ⊇ σ
(A)
where Ω is an open set. Let Γ be an oriented cycle, the union oforiented simple closed curves such that n
(Γ ,z)
= 1 if z ∈ σ
(A)
,n
(Γ ,z)
is an integer if z ∈ Ω, andn
(Γ ,z)
= 0 if z
∕∈
Ω. Then for f,g analytic on Ω,
∫
f (A ) ≡ -1- f (λ)(λI − A )−1dλ
2πi∫Γ
-1- − 1
g (A ) ≡ 2πi Γ g(λ)(λI − A ) dλ
∫ ∫
−1 --1--
= Γ f (λ)(λI − A) ˆΓ g (μ)μ − λ dμdλ
∫ − 1∫ 1
+ ˆg(μ)(μI − A ) f (λ )λ−-μ-dλdμ
Γ Γ
The first term is 0 from Cauchy’s integral formula, Theorem 15.6.1 applied to the individual simple closed
curves whose union is Γ because the winding number n
( )
ˆΓ ,λ
= 0. Thus the inside integral vanishes. From
the Cauchy integral formula, the second term is
∫
2πi g(μ)(μI − A)−1 f (μ)dμ
ˆΓ
and so
1 ∫ −1
f (A)g (A ) = 2πi ˆf (μ)g (μ) (μI − A) dμ
∫ Γ
= -1- f (λ)g (λ)(λI − A)−1dλ ■
2πi Γ
Now consider the case that f
(λ)
= λ. Is f
(A )
, defined in terms of integrals as above, equal to A? If so,
then from the theorem just shown, λ^{n} used in the integral formula does lead to A^{n}. Thus one
considers
1 ∫ −1
2πi λ (λI − A) dλ
Γ
where Γ is a cycle such that n
(Γ ,z)
= 1 if z ∈ σ
(A )
, n
(Γ ,z)
is an integer if z ∈ Ω, and n
(Γ ,z)
= 0 if
z
∕∈
Ω,Γ^{∗}∩ σ
(A)
= ∅. Recall the following corollary of the Cauchy integral formula.
Corollary 18.2.5Let Ω be an open set (note that Ω might not be simply connected) and letγ_{k} :
[ak,bk]
→ Ω, k = 1,
⋅⋅⋅
,m, be closed, continuous and of bounded variation. Suppose alsothat
m∑
n(γk,z) = 0
k=1
for all z
∕∈
Ω and∑_{k=1}^{m}n
(γk,z)
is an integer for z ∈ Ω. Then if f : Ω → X is analytic,
∑m ∫
f (w)dw = 0.
k=1 γk
Thus if Γ is the union of these oriented curves,
∫
f (w)dw = 0.
Γ
Now let
ˆΩ
≡ σ
(A )
^{C} and let
ˆγ
_{R} be a large circle of radius R >
∥A∥
oriented clockwise, which includes
σ
(A )
∪ Γ^{∗}on its inside.
Consider the following picture in which σ
(A)
is the union of the two compact sets, K_{1},K_{2} which are
contained in the closed curves shown and Γ is the union of the oriented cycles Γ_{i}. A similar picture would
apply if there were more than two K_{i}. All that is of interest here is that there is a cycle Γ oriented such
that for all z ∈ σ
(A )
,n
(Γ ,z)
= 1, n
(Γ ,z)
is an integer if z ∈
ˆΩ
, and
ˆγ
_{R} is a large circle oriented clockwise
as shown.
∫ ∫
-1- f (λ)(λI − A)−1 dλ+-1- f (λ)(λI − A )−1dλ = 0
2πi Γ 2πi ˆγR
and so,
∫ ∫
f (A) ≡-1- f (λ)(λI − A )− 1dλ = 1 f (λ )(λI − A)−1dλ
2πi Γ 2πi γR
The integrand is
∑∞ -Ak-
λ λk+1
k=0
for λ ∈ γ_{R}^{∗} and convergence is uniform on γ_{R}^{∗}. Then all terms vanish except the one when k = 1 because
all the other terms have primitives. The uniform convergence implies that the integral of the sum is the
sum of the integrals and there is only one which survives. Therefore,