Definition 24.5.11Let H and G be two separable Hilbert spaces and let T map Hto G be linear. Then T is called a Hilbert Schmidt operatorif there exists some orthonormalbasis for H,
{ej}
such that
∑
||Tej||2 < ∞.
j
The collection of all such linear maps will be denoted by ℒ2
(H,G )
.
Theorem 24.5.12ℒ2
(H, G)
⊆ℒ
(H,G )
and ℒ2
(H, G)
is a separable Hilbert spacewith norm given by
( ∑ )1∕2
||T||ℒ ≡ ||Tek||2
2 k
where
{ek}
is some orthonormal basis for H. Also ℒ2
(H, G)
⊆ℒ
(H,G )
and
||T|| ≤ ||T||ℒ2 . (24.27)
(24.27)
All Hilbert Schmidt opearators are compact. Also for X ∈ H and Y ∈ G,X ⊗ Y ∈ℒ2
Therefore, the linear combinations of the fj⊗ ei are dense in ℒ2
(H, G)
and this proves
completeness of the orthonomal basis.
This also shows ℒ2
(H,G )
is separable. From 24.27 it also shows that every T ∈ℒ2
(H,G )
is the limit in the operator norm of a sequence of compact operators. This follows because
each of the fj⊗ ei is easily seen to be a compact operator because if xm→ x weakly,
then
fj ⊗ ei(xm ) = (xm,ei)fj → (x,ei)fj = fj ⊗ ei(x)
and since if
{xm}
is any bounded sequence, there exists a subsequence,
{xn }
k
which
converges weakly and by the above, fj⊗ ei
(xn )
k
→ fj⊗ ei
(x)
showing bounded sets are
mapped to precompact sets. Therefore, each T ∈ℒ2
(H,G )
must also be a compact operator.
Here is why.
Let B be a bounded set in which
||x||
< M for all x ∈ B and consider TB. I need to
show TB is totally bounded. Let ε > 0 be given. Then let
||Tm − T ||
<
-ε-
3M
where Tm is a
compact operator like those described above and let