is continuous and linear. By the Rieszrepresentation theorem, there exists a unique element of H, denoted by A^{∗}y suchthat
(Ax,y) = (x,A∗y).
It is clear y → A^{∗}y is linear and continuous. A^{∗}is called the adjoint of A. A is a self adjointoperator if A = A^{∗}. Thus for a self adjoint operator,
(Ax, y)
=
(x,Ay )
for all x,y ∈ H. A isa compact operator if whenever
{x }
k
is a bounded sequence, there exists a convergentsubsequence of
{Ax }
k
. Equivalently, A maps bounded sets to sets whose closures arecompact.
The big result is called the Hilbert Schmidt theorem. It is a generalization to arbitrary
Hilbert spaces of standard finite dimensional results having to do with diagonalizing a
symmetric matrix.
Theorem 24.5.2Let A be a compact self adjoint operator defined on a Hilbert space,H. Then there exists a countable set of eigenvalues,
{λi}
and an orthonormal set ofeigenvectors, u_{i}satisfying
λi is real, |λn| ≥ |λn+1|, Aui = λiui, (24.16)
(24.16)
and either
lnim→∞ λn = 0, (24.17)
(24.17)
or for some n,
span (u1,⋅⋅⋅,un) = H. (24.18)
(24.18)
In any case,
span({ui}∞i=1) is dense in A (H ). (24.19)
(24.19)
and for all x ∈ H,
∞∑
Ax = λk(x,uk)uk. (24.20)
k=1
(24.20)
This sequence of eigenvectors and eigenvalues also satisfies
|λn| = ||An||, (24.21)
(24.21)
and
An : Hn → Hn. (24.22)
(24.22)
where H ≡ H_{1}and H_{n}≡
{u1,⋅⋅⋅,un−1}
^{⊥}and A_{n}is the restriction of A to H_{n}.
Proof: If A = 0 then pick u ∈ H with
||u||
= 1 and let λ_{1} = 0. Since A
(H)
= 0 it follows
the span of u is dense in A
(H )
and this proves the theorem in this uninteresting
case.
Assume from now on A≠0. Let λ_{1} be real and λ_{1}^{2}≡
||A ||
^{2}. From the definition of
||A ||
there exists x_{n},
||xn||
= 1, and
||Axn||
→
||A||
=
|λ1|
. Now it is clear that A^{2} is also a
compact self adjoint operator. Consider
(( 2 2) ) 2 2
λ1 − A xn,xn = λ1 − ||Axn|| → 0.
Since A is compact, there exists a subsequence of
{xn}
still denoted by
{xn }
such that Ax_{n}
converges to some element of H. Thus since λ_{1}^{2}− A^{2} satisfies
(( ) )
λ21 − A2 y,y ≥ 0
in addition to being self adjoint, it follows x,y →
((λ2− A2)x,y)
1
satisfies all the axioms for
an inner product except for the one which says that
(z,z)
= 0 only if z = 0. Therefore, the
Cauchy Schwarz inequality may be used to write
where e_{n}→ 0 as n →∞. Therefore, taking the sup over all
||y||
≤ 1,
||||( 2 2) ||||
nli→m∞ λ1 − A xn = 0.
Since A^{2}x_{n} converges, it follows since λ_{1}≠0 that
{xn}
is a Cauchy sequence converging to x
with
||x||
= 1. Therefore, A^{2}x_{n}→ A^{2}x and so
||( ) ||
|| λ21 − A2 x|| = 0.
Now
(λ1I − A)(λ1I + A) x = (λ1I + A )(λ1I − A)x = 0.
If
(λ1I − A )
x = 0, let u_{1}≡ x. If
(λ1I − A )
x = y≠0, let u_{1}≡
-y-
||y||
.
Suppose
{u1,⋅⋅⋅,un}
is such that Au_{k} = λ_{k}u_{k} and
|λk|
≥
|λk+1|
,
|λk|
=
||Ak ||
and
A_{k} : H_{k}→ H_{k} for k ≤ n. If
span(u1,⋅⋅⋅,un) = H
this yields the conclusion of the theorem in the situation of 24.18. Therefore, assume
the span of these vectors is always a proper subspace of H. It is shown next that
A_{n+1} : H_{n+1}→ H_{n+1}. Let
y ∈ Hn+1 ≡ {u1,⋅⋅⋅,un}⊥
Then for k ≤ n
(Ay, u ) = (y,Au ) = λ (y,u ) = 0,
k k k k
showing A_{n+1} : H_{n+1}→ H_{n+1} as claimed. There are two cases. Either λ_{n} = 0 or it is not. In
the case where λ_{n} = 0 it follows A_{n} = 0. Every element of H is the sum of one in
span
(u1,⋅⋅⋅,un)
and one in span
(u1,⋅⋅⋅,un)
^{⊥}. (note span
(u1,⋅⋅⋅,un)
is a closed
subspace.) Thus, if x ∈ H, x = y + z where y ∈span
(u1,⋅⋅⋅,un)
and z ∈span
(u1,⋅⋅⋅,un)
^{⊥}
and Az = 0. Say y = ∑_{j=1}^{n}c_{j}u_{j}. Then
The conclusion of the theorem holds in this case because the above equation holds if with
c_{i} =
(x,ui)
.
Now consider the case where λ_{n}≠0. In this case repeat the above argument used to find
u_{n+1} and λ_{n+1} for the operator, A_{n+1}. This yields u_{n+1}∈ H_{n+1}≡
|( ( ) ) |
|| ∑n ||
|| A − λkuk ⊗ uk x,y ||
|( k=1 )|
|| ∑n ||
= || Ax − λk(x,uk)uk,y ||
|( ∞ k=1 )|
= || ∑ λ (x,u )u ,y ||
|| k=n k k k ||
|| ∞ ||
= ||∑ λk (x,uk) (uk,y)||
|k=n |
( ∞ )1∕2( ∞ )1 ∕2
≤ |λ | ∑ |(x,u )|2 ∑ |(y,u )|2
n k=n k k=n k
≤ |λ |||x||||y||
n
It follows
||||( ∑n ) ||||
|||| A − λkuk ⊗ uk (x)||||≤ |λn|||x|| ■
|| k=1 ||
Corollary 24.5.4Let A be a compact self adjoint operator defined on a separable Hilbertspace, H. Then there exists a countable set of eigenvalues,
{λi}
and an orthonormal set ofeigenvectors, v_{i}satisfying
Avi = λivi,||vi|| = 1, (24.23)
(24.23)
∞
span({vi}i=1) is dense in H. (24.24)
(24.24)
Furthermore, if λ_{i}≠0, the space, V_{λi}≡
{x ∈ H : Ax = λix}
is finite dimensional.
Proof: In the proof of the above theorem, let W ≡span
({u })
i
^{⊥}. By Theorem 24.4.2,
there is an orthonormal set of vectors,
{w }
i
_{i=1}^{∞} whose span is dense in W. As shown in the
proof of the above theorem, Aw = 0 for all w ∈ W. Let
{v}
i
_{i=1}^{∞} =
{u}
i
_{i=1}^{∞}∪
{w }
i
_{i=1}^{∞}.
It remains to verify the space, V_{λi}, is finite dimensional. First observe that
A : V_{λi}→ V_{λi}. Since A is continuous, it follows that A :V_{λi}→V_{λi}. Thus A is a compact self
adjoint operator on V_{λi} and by Theorem 24.5.2, 24.18 holds because the only eigenvalue is λ_{i}.
■
Note the last claim of this corollary holds independent of the separability of H. This
proves the corollary.
Suppose λ
∕∈
{λ }
n
and λ≠0. Then the above formula for A, 24.20, yields an interesting
formula for
(A − λI)
^{−1}. Note first that since lim_{n→∞}λ_{n} = 0, it follows that λ_{n}^{2}∕
(λ − λ)
n
^{2}
must be bounded, say by a positive constant, M.
Corollary 24.5.5Let A be a compact self adjoint operator and let λ
∕∈
{λn}
_{n=1}^{∞}andλ≠0 where the λ_{n}are the eigenvalues of A. Then
∞
(A − λI)−1x = − 1-x+ 1-∑ --λk--(x,u )u . (24.25)
λ λ k=1λk − λ k k
and so for m large enough, the first term in 24.26 is smaller than ε. This shows the infinite
series in 24.25 converges. It is now routine to verify that the formula in 24.25 is the inverse.
■