Analysis

24.5.3 Hilbert Schmidt Operators

Definition 24.5.11 Let H and G be two separable Hilbert spaces and let T map H to G be linear. Then T is called a Hilbert Schmidt operator if there exists some orthonormal basis for H,

such that

\sum ||Tej||2 < \infty. j

The collection of all such linear maps will be denoted by ℒ₂

Theorem 24.5.12 ℒ₂

⊆ℒ

and ℒ₂

is a separable Hilbert space with norm given by

( \sum )1∕2 ||T||ℒ \equiv ||Tek||2 2 k

where

is some orthonormal basis for H. Also ℒ₂

⊆ℒ

and

||T|| \leq ||T||ℒ2 . (24.27)

(24.27)

All Hilbert Schmidt opearators are compact. Also for X ∈ H and Y ∈ G,X ⊗ Y ∈ℒ₂

and

||X \otimes Y||ℒ2 = ||X||H ||Y ||G (24.28)

(24.28)

Proof: First I want to show ℒ₂

⊆ℒ

and

≤

_ℒ₂. Pick an orthonormal basis for H,

and an orthonormal basis for G,

. Then letting

\sumn x = xkek, k=1

( n ) n Tx = T \sum x e = \sum x T (e ) k=1 k k k=1 k k

where x_k ≡

. Therefore using Minkowski’s inequality,

(\sum\infty )1∕2 ||Tx|| = |(T x,fk)|2 k=1 ( ||( ) ||2) 1∕2 |\sum\infty ||( n\sum ) ||| = ( || xjTej,fk ||) k=1 j=1 ( || ||2)1∕2 |\sum\infty ||\sumn || | = ( || (xjTej,ek)|| ) k=1 j=1 n\sum (\sum\infty )1∕2 \leq |(xjT ej,ek)|2 j=1 k=1 n ( \infty )1∕2 \leq \sum |x | \sum |(Te ,e)|2 j=1 j k=1 j k ( ) 1∕2 n\sum \sumn 2 = |xj|||Tej|| \leq ( |xj|) ||T||ℒ2 j=1 j=1 = ||x||||T||ℒ2

Therefore, since finite sums of the form ∑ _k=1ⁿx_ke_k are dense in H, it follows T ∈ℒ

and

≤

_ℒ₂

Next consider the norm. I need to verify the norm does not depend on the choice of orthonormal basis. Let

be an orthonormal basis for G. Then for

an orthonormal basis for H,

\sum 2 \sum \sum 2 \sum \sum * 2 ||Tek|| = |(T ek,fj)| = |(ek,T fj)| k \sumk \sumj k\sum j = |(ek,T*fj)|2 = ||T*fj||2. j k j

The above computation makes sense because it was just shown that T is continuous. The same result would be obtained for any other orthonormal basis

and this shows the norm is at least well defined. It is clear this does indeed satisfy the axioms of a norm.and this proves the above claims.

It only remains to verify ℒ₂

is a separable Hilbert space. It is clear it is an inner product space because you only have to pick an orthonormal basis,

and define the inner product as

(S,T) \equiv \sum (Sek,Tek). k

The only remaining issue is the completeness. Suppose then that

is a Cauchy sequence in ℒ₂

. Then from 24.27

is a Cauchy sequence in ℒ

and so there exists a unique T such that lim_n→∞

= 0. Then it only remains to verify T ∈ℒ₂

. But by Fatou’s lemma,

\sum ||Tek||2 \leq lim inf \sum ||Tnek||2 k n\to\infty k 2 = lim nin\tof\infty ||Tn||ℒ2 < \infty.

All that remains is to verify ℒ₂

is separable and these Hilbert Schmidt operators are compact. I will show an orthonormal basis for ℒ₂

where

is an orthonormal basis for G and

is an orthonormal basis for H. Here, for f ∈ G and e ∈ H,

f \otimes e (x) \equiv (x,e)f.

I need to show f_j ⊗ e_k ∈ℒ₂

and that it is an orthonormal basis for ℒ₂

as claimed.

\sum \sum ||fj \otimes ei(ek)||2 = ||fjδik||2 = ||fj||2 = 1 < \infty k k

so each of these operators is in ℒ₂

. Next I show they are orthonormal.

\sum (fj \otimesek,fs \otimes er) = (fj \otimes ek(ep) ,fs \otimeser(ep)) \sump \sum = δrpδkp(fj,fs) = δrpδkpδjs p p

If j = s and k = r this reduces to 1. Otherwise, this gives 0. Thus these operators are orthonormal. Now let T ∈ℒ₂

. Consider

n n T \equiv \sum \sum (Te ,f )f \otimes e n i=1 j=1 i j j i

Then

n n T e = \sum \sum (Te ,f)(e ,e)f n k i=1j=1 i j k i j n = \sum (Tek,fj)fj j=1

It follows

||Tnek|| \leq ||T ek||

and

lnim\to\infty Tnek = T ek.

Therefore, from the dominated convergence theorem,

\sum lim ||T - Tn||2ℒ2 \equiv lim ||(T - Tn)ek||2 = 0. n\to\infty n\to \infty k

Therefore, the linear combinations of the f_j ⊗ e_i are dense in ℒ₂

and this proves completeness of the orthonomal basis.

This also shows ℒ₂

is separable. From 24.27 it also shows that every T ∈ℒ₂

is the limit in the operator norm of a sequence of compact operators. This follows because each of the f_j ⊗ e_i is easily seen to be a compact operator because if x_m → x weakly, then

fj \otimes ei(xm ) = (xm,ei)fj \to (x,ei)fj = fj \otimes ei(x)

and since if

is any bounded sequence, there exists a subsequence,

which converges weakly and by the above, f_j ⊗ e_i

→ f_j ⊗ e_i

showing bounded sets are mapped to precompact sets. Therefore, each T ∈ℒ₂

must also be a compact operator. Here is why.

Let B be a bounded set in which

< M for all x ∈ B and consider TB. I need to show TB is totally bounded. Let ε > 0 be given. Then let

where T_m is a compact operator like those described above and let

_j=1^N be an ε∕3 net for T_m

. Then

||Tx - T x || < ε j m j 3

and so letting x ∈ B, pick x_j such that

< ε∕3. Then

||Tx - Txj|| \leq ||Tx- Tmx ||+||Tmx - Tmxj ||+ ||Tmxj - T xj|| < ε+ -ε+ ε = ε 3 3 3

showing

_j=1^N is an ε net for TB.

Finally, consider 24.28. Let

be an orthonormal basis for H and consider the following computation which establishes this equation.

2 \sum\infty 2 \sum\infty 2 ||Y \otimes X ||ℒ2 \equiv ||Y \otimes X (ek)|| = ||(ek,X)Y || k=1 \infty k=1 = ||Y ||2\sum |(e ,X )|2 = ||Y||2||X ||2 < \infty. ■ (24.29) Gk=1 k G H

This proves the theorem.

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