592 CHAPTER 22. HILBERT SPACES

This is just like it is for a matrix. Recall the trace of a matrix is the sum of the eigen-values.

It is also easy to see that in any Hilbert space, there exist nuclear operators. Let∑

∞k=1 |λ k|< ∞. Then let {ek} be an orthonormal set of vectors. Let A≡ ∑

∞k=1 λ kek⊗ ek,λ k

real. It is not too hard to verify this works.Much more can be said about nuclear operators.

22.5.2 Hilbert Schmidt OperatorsIn all of this, H,G will be separable Hilbert spaces.

Definition 22.5.14 Let T be a continuous linear mapping from H to G and when-ever {ek} is an orthonormal basis for H, then ∑k ∥Tek∥2 < ∞. Such an operator is called aHilbert Schmidt operator.We write T ∈L2 (H,G) . Picking an orthonormal basis (completeorthonormal set), define ∥T∥2

L2≡ ∑k ∥Tek∥2

G .

It is necessary to show that this is well defined and does not depend on the orthonormalbasis. For now let the orthonormal basis be fixed.

Lemma 22.5.15 If T is Hilbert Schmidt, then ∥T∥L (H,G) ≤ ∥T∥L2(H,G). Also T ∗ isHilbert Schmidt and T ∗T is Hilbert Schmidt. In fact, if T ∈L2 (H,G) and S ∈L (G,H)then ST ∈L2 (H,H).

Proof: Pick an orthonormal basis for H,{ek} ∑k ∥Tek∥2 < ∞ and an orthonormal basisfor G,{ fk}. Then let x = ∑

nk=1 (x,ek)ek ≡ ∑

nk=1 xkek. Then

∥T x∥=

(∞

∑k=1|(T x, fk)|2

)1/2

=

 ∞

∑k=1

∣∣∣∣∣(

n

∑j=1

x jTe j, fk

)∣∣∣∣∣21/2

=

 ∞

∑k=1

∣∣∣∣∣ n

∑j=1

(x jTe j, fk)

∣∣∣∣∣21/2

≤n

∑j=1

(∞

∑k=1

∣∣(x jTe j, fk)∣∣2)1/2

≤n

∑j=1

∣∣x j∣∣( ∞

∑k=1

∣∣(Te j, fk)∣∣2)1/2

=n

∑j=1

∣∣x j∣∣∥∥Te j

∥∥≤( n

∑j=1

∣∣x j∣∣2)1/2( n

∑j=1

∥∥Te j∥∥2

)1/2

= ∥x∥∥T∥L2

Therefore, since finite sums of the form ∑nk=1 xkek are dense in H, it follows ∥T∥ ≤ ∥T∥L2

.

Letting { fi} be orthonormal in G,∥Tek∥2 = ∑ j∣∣(Tek, f j)

∣∣2 and so

∑k∥Tek∥2 = ∑

k∑

j

∣∣(Tek, f j)∣∣2 = ∑

j∑k

∣∣(ek,T ∗ f j)∣∣2 = ∑

j

∥∥T ∗ f j∥∥2

so T ∗ is also Hilbert Schmidt.