Topics In Analysis

17.7.1 Compact Operators In Hilbert Space

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. There is another statement and proof of this theorem around Page 2056.

Proof: If A = 0 then pick u ∈ H with

= 1 and let λ₁ = 0. Since A = 0 it follows the span of u is dense in A and this proves the theorem in this uninteresting case.

Assume from now on A≠0. Let λ₁ be real and λ₁² ≡

². From the definition of there exists x_n, = 1, and → = . Now it is clear that A² 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

still denoted by such that Ax_n converges to some element of H. Thus since λ₁² − A² satisfies

((λ21 − A2)y,y) ≥ 0

in addition to being self adjoint, it follows x,y →

satisfies all the axioms for an inner product except for the one which says that = 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 ≤ 1,

||( ) || lim || λ21 − A2 xn|| = 0. n→ ∞

Since A²x_n converges, it follows since λ₁≠0 that

is a Cauchy sequence converging to x with = 1. Therefore, A²x_n → A²x and so

||||(λ2 − A2 )x|||| = 0. 1

Now

(λ1I − A )(λ1I + A)x = (λ1I + A)(λ1I − A )x = 0.

If

x = 0, let u₁ ≡ x. If x = y≠0, let u₁ ≡.

Suppose

is such that Au_k = λ_ku_k and ≥, =  and A_k : H_k → H_k for k ≤ n. If

span(u ,⋅⋅⋅,u ) = H 1 n

this yields the conclusion of the theorem in the situation of 17.7.29. 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

and one in span^⊥. (note span is a closed subspace.) Thus, if x ∈ H, x = y + z where y ∈ span and z ∈ span^⊥ and Az = 0. Say y = ∑ _j=1ⁿc_ju_j. Then

The conclusion of the theorem holds in this case because the above equation holds if with c_i = .

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 ≡

^⊥ such that

||u || = 1,||Au || = |λ | = ||A || ≤ ||A || = |λ | n+1 n+1 n+1 n+1 n n

and if it is ever the case that λ_n = 0, it follows from the above argument that the conclusion of the theorem is obtained.

I claim lim_n→∞λ_n = 0. If this were not so, then for some ε > 0, 0 < ε = lim_n→∞

but then

and so there would not exist a convergent subsequence of _k=1^∞ contrary to the assumption that A is compact. This verifies the claim that lim_n→∞λ_n = 0.

It remains to verify that span

is dense in A. If w ∈ span^⊥ then w ∈ H_n for all n and so for all n,

||Aw || ≤ ||An||||w || ≤ |λn|||w||.

Therefore, Aw = 0. Now every vector from H can be written as a sum of one from

span ({u })⊥ = span-({u-})⊥ i i

and one from span

. Therefore, if x ∈ H, x = y + w where y ∈span

({ui})

and w ∈span

({ui})

^⊥. It follows Aw = 0. Also, since y ∈span

({ui})

, there exist constants, c_k and n such that

|| || |||| ∑n |||| ||||y − ckuk|||| < ε. k=1

Therefore, from Corollary 17.5.3,

|| || || || |||| ∑n |||| |||| ∑n |||| ||||y− (y,uk)uk|||| = ||||y− (x,uk)uk|||| < ε. k=1 k=1

Therefore,

Since ε is arbitrary, this shows span is dense in A and also implies 17.7.31. This proves the theorem.

Define v ⊗ u ∈ℒ

by

v⊗ u (x ) = (x,u )v,

then 17.7.31 is of the form

∑∞ A = λkuk ⊗ uk k=1

This is the content of the following corollary.

Proof: Using 17.7.31 

It follows

||||( ∑n ) |||| |||| A − λkuk ⊗ uk (x)||||≤ |λn|||x|| || k=1 ||

and this proves the corollary.

Proof: In the proof of the above theorem, let W ≡span

^⊥. By Theorem 17.5.2, there is an orthonormal set of vectors, _i=1^∞ whose span is dense in W. As shown in the proof of the above theorem, Aw = 0 for all w ∈ W. Let _i=1^∞ = _i=1^∞∪_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 17.7.2, 17.7.29 holds because the only eigenvalue is λ_i. This proves the corollary.

Note the last claim of this corollary holds independent of the separability of H. This proves the corollary.

Suppose λ

and λ≠0. Then the above formula for A, 17.7.31, yields an interesting formula for ⁻¹. Note first that since lim_n→∞λ_n = 0, it follows that λ_n²∕² must be bounded, say by a positive constant, M.

Proof: Let m < n. Then since the

form an orthonormal set,

But from Bessel’s inequality,

∞∑ 2 2 |(x,uk)|≤ ||x|| k=1

and so for m large enough, the first term in 17.7.37 is smaller than ε. This shows the infinite series in 17.7.36 converges. It is now routine to verify that the formula in 17.7.36 is the inverse.

+ class=”left” align=”middle”(U)17.8.  STURM LIOUVILLE PROBLEMS