65.4. SOME HILBERT SPACE THEORY 2241

H because ∥∥∥∥∥∥ fi

(e j√λ j

,x

)U1

∥∥∥∥∥∥H

≤ ∥ fi∥H1√λ j∥x∥U1

Thus each term in the finite sum of 65.4.6 is in L (U1,H)0 and this proves the lemma.It is interesting to note that Q1/2

1 U1 = J(Q1/2 (U)

).

L

∑j=1

√λ je j

(J∗e j√

λ j,x

)Q1/2(U)

= Jx

and{

J∗e j√λ j

}are an orthonormal set in Q1/2 (U). Therefore, the sum of the squares of(

J∗e j√λ j,x)

Q1/2(U)

is finite. Hence you can define y ∈U1 by

y≡L

∑j=1

(J∗e j√

λ j,x

)Q1/2(U)

e j

AlsoL

∑i=1

√λ iei⊗ ei (y) =

L

∑i=1

√λ iei

(J∗ei√

λ i,x)

Q1/2(U)

=L

∑i=1

ei (ei,Jx)U1= Jx

Now you can show that Q1/21 = ∑

Li=1√

λ iei⊗ ei. You do this by showing that it works andcommutes with every operator which commutes with Q1. Thus Jx = Q1/2

1 y. This showsthat J

(Q1/2 (U)

)⊆ Q1/2

1 (U1). However, you can also turn the inclusion around. Thus ifyou start with y ∈U1 and form

Q1/21 y =

L

∑i=1

√λ iei⊗ ei (y) =

L

∑i=1

√λ iei (y,ei) ,

then the (y,ei)2U1

has a finite sum because the {ei} are orthonormal. Thus you can form

x≡L

∑i=1

(y,ei)U1

J∗ei√λ i∈ Q1/2 (U)

Then since the{

J∗e j√λ j

}are orthonormal,

J (x) =L

∑j=1

√λ je j

(J∗e j√

λ j,x

)Q1/2(U)

=L

∑j=1

√λ je j (y,e j)U1

=L

∑j=1

√λ je j⊗ e j (y) = Q1/2

1 (y)