22.5. COMPACT OPERATORS IN HILBERT SPACE 593
Let {ei} be an orthonormal basis for H so ∑i ∥Tei∥2 < ∞. Then
∑i∥T ∗Tei∥2 ≤ ∥T ∗∥2
∑i∥Tei∥2 < ∞
For the last claim, let {ek} be an orthonormal basis. Then
∑k∥ST (ek)∥2 ≤∑
k∥S∥2 ∥Tek∥2 < ∞. ■
Definition 22.5.16 Define (S,T ) ≡ ∑k (Sek,Tek) where {ek} is a given orthonor-mal basis. This is well defined because the sum converges absolutely. Indeed,
∑k|(Sek,Tek)| ≤∑
k|Sek| |Tek| ≤
(∑k|Sek|2
)1/2(∑k|Tek|2
)1/2
< ∞
Definition 22.5.17 For X ∈ G and Y ∈ H,X ⊗Y (h) ≡ X (h,Y ) . This is a contin-uous linear map from H to G because ∥X⊗Y (h)∥ = ∥X (h,Y )∥ ≤ ∥h∥∥Y∥∥X∥ . Next weshow X⊗Y ∈L2 (H,G) among other things.
Theorem 22.5.18 L2 (H,G) is a separable Hilbert space with norm given by
∥T∥L2≡
(∑k∥Tek∥2
)1/2
where {ek} is some fixed orthonormal basis for H. Also L2 (H,G)⊆L (H,G) and
∥T∥ ≤ ∥T∥L2. (22.20)
All Hilbert Schmidt opearators are compact. Also for X ∈G and Y ∈H,X⊗Y ∈L2 (H,G)and
∥X⊗Y∥L2= ∥X∥G ∥Y∥H (22.21)
If T is Hilbert Schmidt, then so is T ∗T and T ∗and ST for any S ∈L (G,H). If T = T ∗ andG = H, then the choice of orthonormal basis in computing ∥T∥L2
is not important.
Proof: Is ∥T∥L2really a norm? This obviously is so except for the triangle inequality.
But this follows from the triangle inequality.
∥T +S∥L2≡
(∑k∥Tek +Sek∥2
)1/2
≤
(∑k∥Tek∥2
)1/2
+
(∑k∥Sek∥2
)1/2
= ∥T∥L2+∥S∥L2
Next is the claim that L2 is a Hilbert space. So pick an orthonormal basis {ek}. It isclear that L2 is an inner product space with respect to the inner product described above inDefinition 22.5.16.
Consider completeness. Suppose that {Tn} is a Cauchy sequence in L2 (H,G) . Thenfrom 22.20 {Tn} is a Cauchy sequence in L (H,G) and so there exists a unique T such