65.4. SOME HILBERT SPACE THEORY 2239

Then denote by L (U1,H)0 the space of restrictions of elements of L (U1,H) to the Hilbertspace JQ1/2U ⊆U1.

Here is a diagram to keep this straight.

U↓ Q1/2

U1 ⊇ JQ1/2U J←1−1

Q1/2U

Φn ↘ ↓ Φ

H

Lemma 65.4.5 In the context of the above definition, L (U1,H)0 is dense in

L2

(JQ1/2U,H

),

the Hilbert Schmidt operators from JQ1/2U to H. That is, if f ∈L2(JQ1/2U,H

), there

existsg ∈L (U1,H)0 ,∥g− f∥L2(JQ1/2U,H) < ε.

Proof: The operator JJ∗ ≡ Q1 : U1 →U1 is self adjoint and nonnegative. It is alsocompact because J is Hilbert Schmidt. Therefore, by Theorem 21.3.9 on Page 663,

Q1 =L

∑k=1

λ kek⊗ ek

where the λ k are decreasing and positive, the {ek} are an orthonormal basis for U1, and λ Lis the last positive λ j. (This is a lot like the singular value matrix in linear algebra.) Thusalso

Q1ek = λ kek

If the λ k are all positive, then L≡ ∞. Then for k ≤ L if L < ∞, k < ∞ otherwise,(J∗ek√

λ k,

J∗e j√λ j

)Q1/2(U)

=

(JJ∗ek√

λ k,

e j√λ j

)U1

=

(λ kek√

λ k,

e j√λ j

)U1

=

√λ k√λ j

δ k j = δ jk

Now in case L < ∞, J(Q1/2 (U)

)⊆ span(e1, · · · ,eL) . Here is why. First note that Q1 is one

to one on span(e1, · · · ,eL) and maps this space onto itself because Q1 maps ek to a nonzeromultiple of ek. Hence its restriction to this subspace has an inverse which does the same. Italso maps all of U1 to span(e1, · · · ,eL). This follows from the definition of Q1 given in theabove sum. For x ∈ Q1/2 (U) , Jx ∈U1 and so

JJ∗Jx = Q1 (Jx) ∈ span(e1, · · · ,eL)

Hence Jx ∈ Q−11 (span(e1, · · · ,eL)) ∈ span(e1, · · · ,eL) . Recall that J is one to one so there

is only one element of J−1x.