Lemma 15.6.5Let L be a compact linear map. Then the eigenspace of L is finitedimensional for each eigenvalue λ≠0.
Proof:Consider
(L − λI)
^{−1}
(0)
∩S where S is the unit sphere. The eigenspace is
just
(L − λI)
^{−1}
(0)
. Let Y be this inverse image. If Y is infinite dimensional, then the
above Lemma 15.6.4 applies. There exists
{xn}
⊆
(L − λI)
^{−1}
(0)
∩ S where
∥xn − xm ∥
≥ 1∕2 for all n≠m. Then there is a subsequence, still denoted with
subscript n such that
{Lxn}
is a Cauchy sequence. Thus Lx_{n} = λx_{n} and so,
since λ≠0, it follows that
{xn}
is also a Cauchy sequence and converges to
some x. But this is impossible because of the construction of the
{xn}
which
prevents there being any Cauchy sequence. Thus Y must be finite dimensional.
■
This lemma is useful in proving the following major spectral theorem about the
eigenvalues of a compact operator. I found this theorem in Deimling [?].
Theorem 15.6.6Let L ∈ℒ
(X, X)
with L compact. Let Λ be the eigenvalues ofL. That is λ ∈ Λ means there exists x≠0 such that Lx = λx. It is assumed the fieldof scalars is ℝ or ℂ. Let R_{λ}≡ L − λI. Then the following hold.
If μ ∈ Λ then
|μ |
≤
∥L∥
,Λ is at most countable and has no limit points otherthan possibly 0.
R_{λ}is a homeomorphism onto X whenever λ
∕∈
Λ ∪
{0}
.
For all λ ∈ Λ ∖
{0}
, there exists a smallest k = k
(λ)
,
R_{λ}^{k}X⊕N
(Rk )
λ
= X where N
(Rk)
λ
is the vectors x such that R_{λ}^{k}x = 0.R_{λ}^{k}X is closed, dim
( ( k))
N Rλ
< ∞.
R_{λ}^{k}X and N
( )
Rkλ
are invariant under L and R_{λ}|_{RkkX}is ahomeomorphism onto R_{λ}^{k}X.
N
(Rk )
μ
⊆ R_{λ}^{k}X for all λ,μ ∈ Λ ∖
{0}
where λ≠μ.
Proof:Consider λ≠0. The N
( k)
R λ
are increasing in k and R_{λ}
( ( k+1))
N R λ
⊆ N
( k)
Rλ
.
This follows from the definition. (It isn’t necessary to assume in most of this that λ ∈ Λ,
just a nonzero number will do.) Now
( 1 )
Rλ = − λ I − λ-L
If these things are strictly increasing for infinitely many k, then by Lemma 15.6.3, there
is an infinite sequence x_{k},x_{k}∈ N
(Rk+1 )
λ
∖ N
(Rk )
λ
, dist
(Lxk,LN (Rk))
λ
≥ 1∕2. Hence
∥Lxk − Lxk− 1∥
≥ 1∕2 and this can’t happen because L is compact so
{Lxk }
has a
Cauchy subsequence. Therefore there exists a smallest k such that
( k) m
N R λ = N (Rλ ),m ≥ k
On the other hand,
{ }
RkλX
are decreasing in k. By similar reasoning using Lemma
15.6.3 and the observation that R_{λ}
( )
RkλX
⊇ R_{λ}^{k+1}X (in fact they are equal) it follows
that the
{ }
RkλX
are also eventually constant, say for m ≥ l.
Now if you have y ∈ N
( )
Rkλ
∩ R_{λ}^{k}X, then y = R_{λ}^{k}w and also R_{λ}^{k}y = 0. Hence
R_{λ}^{2k}w = 0 and so, w ∈ N
(R2kλ )
= N
(Rkλ)
which implies R_{λ}^{k}w = 0 and so y = 0. It
follows N
(Rk )
λ
∩ R_{λ}^{k}X =
{0}
.
Now suppose l > k. Then there exists y ∈ R_{λ}^{l−1}X ∖ R_{λ}^{l}X and so
R_{λ}y ∈ R_{λ}^{l}X = R_{λ}^{l+1}X = R_{λ}R_{λ}^{l}X. So R_{λ}y = R_{λ}z for some z ∈ R_{λ}^{l}X. Thus y − z≠0
because y
∕∈
R_{λ}^{l}X but z is. However, R_{λ}
(y− z)
= 0 and so
(y − z) ∈ N (R λ)∩ RkλX ⊆ N (Rkλ)∩ RkλX
which cannot happen from the above which showed that N
( k)
Rλ
∩ R_{λ}^{k}X =
{0}
. Thus
l ≤ k.
Next suppose l < k. Then you would have R_{λ}^{l}X = R_{λ}^{k}X and N
( )
Rkλ
⊋ N
( )
Rlλ
.
Thus there exists y ∈ N
( )
Rkλ
but not in N
( )
Rlλ
. Hence R_{λ}^{k}y = 0 but R_{λ}^{l}y≠0. However,
R_{λ}^{l}y is in R_{λ}^{k}X from the definition of l and so there is u such that R_{λ}^{l}y = R_{λ}^{k}u.
Thus
k k−l+l k−l l k−l k 2k−l
0 = Rλy = Rλ y = Rλ Rλy = Rλ Rλu = R λ u
Now it follows that u ∈ N
(R2kλ−l)
= N
(Rkλ)
. This is a contradiction because it says that
R_{λ}^{k}u = 0 but right above the displayed equation, we had R_{λ}^{l}y = R_{λ}^{k}u and R_{λ}^{l}y≠0.
Thus, with the above paragraph, k = l.
What about the claim that R_{λ} restricted to R_{λ}^{k}X is a homeomorphism? It maps
R_{λ}^{k}X to R_{λ}^{k+1}X = R_{λ}^{k}X. Also, if R_{λ}
(y)
= 0 for y ∈ R_{λ}^{k}X, then R_{λ}^{k}y = 0 also and so
y ∈ R_{λ}^{k}X ∩ N
(Rk)
λ
. It was shown above that this implies y = 0. Thus R_{λ}
appears to be one to one. By assumption, it is continuous. Also from Lemma
15.6.2,
RkλX is closed.
This follows from the observation that
k ( ) k ( )
Rk = (L − λI)k = ∑ k Lj(− λI)k−j = (− λ)kI +∑ k Lj(− λI)k−j
λ j=0 j j=1 j
(15.6.17)
(15.6.17)
which is a multiple of I − C where C is a compact map. Then by the open mapping
theorem, it follows that R_{λ} is a homeomorphism onto R_{λ}^{k+1}X = R_{λ}^{k}X.
What about R_{λ}^{k}X ⊕N
(Rk)
λ
= X? It only remains to verify that R_{λ}^{k}X + N
(Rk )
λ
= X
because the only vector in the intersection was shown to be 0. Thus if you have x + y = 0
where x is in one of these and y in the other, then x = −y so each is in both and
hence both are 0. Pick x ∈ X. Then R_{λ}^{k}x ∈ R_{λ}^{k}
(RkX )
λ
= R_{λ}^{k}X. Therefore,
R_{λ}^{k}x = R_{λ}^{k}
(Rky)
λ
for some y and so R_{λ}^{k}
(x− Rk y)
λ
= 0. Hence
( )
x − Rkλy ∈ N Rkλ
showing that x ∈ R_{λ}^{k}X + N
( )
Rkλ
.
It is obvious that R_{λ}^{k}X and N
(Rkλ)
are invariant under L. If λ_{0}
∕∈
Λ ∖
{0}
,
then L − λ_{0}I is one to one and so the compactness of L and Lemma 15.6.2
implies that
(L − λ0I)
X is closed. Hence the open mapping theorem implies
L − λ_{0}I is a homeomorphism onto
(L − λ0I)
X. Is this last all of X? There is
nothing in the above argument which involved an essential assumption that λ ∈ Λ.
Hence, repeating this argument, you see that
(L− λ0I)
X ⊕ N
(L − λ0I)
= X
but N
(L− λ0I)
= 0. Hence
(L− λ0I)
X = X and so indeed
(L − λ0I)
is a
homeomorphism.
For μ ∈ Λ,Lx = μx and so
|μ|
∥x∥
≤
∥L∥
∥x∥
so
|μ|
≤
∥L∥
. Why is Λ at most
countable and has only one possible limit point at 0? It was shown that R_{λ} is a
homeomorphism when restricted to R_{λ}^{k}X. It follows that for x ∈ R_{λ}^{k}X,