19.8. STURM LIOUVILLE PROBLEMS 555

where xk ≡ (x,ek). Therefore using Minkowski’s inequality,

∥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

∥T∥L2= ∥x∥∥T∥L2

Therefore, since finite sums of the form ∑nk=1 xkek are dense in H, it follows T ∈L (H,G)

and ∥T∥ ≤ ∥T∥L2Next consider the norm. I need to verify the norm does not depend on the choice of

orthonormal basis. Let { fk} be an orthonormal basis for G. Then for {ek} an orthonormalbasis for H,

∑k∥Tek∥2 = ∑

k∑

j

∣∣(Tek, f j)∣∣2 = ∑

k∑

j

∣∣(ek,T ∗ f j)∣∣2

= ∑j∑k

∣∣(ek,T ∗ f j)∣∣2 = ∑

j

∥∥T ∗ f j∥∥2

.

The above computation makes sense because it was just shown that T is continuous. Thesame result would be obtained for any other orthonormal basis

{e′j}

and this shows thenorm is at least well defined. It is clear that this does indeed satisfy the axioms of anorm.and this proves the above claims.

It only remains to verify L2 (H,G) is a separable Hilbert space. It is clear that it is aninner product space because you only have to pick an orthonormal basis, {ek} and definethe inner product as

(S,T )≡∑k(Sek,Tek) .

This satisfies the axioms of an inner product and delivers the well defined norm so it is awell defined inner product. Indeed, we get it from

(S,T )≡ 14

(∥S+T∥2

L2−∥S−T∥2

L2

)and the norm is well defined giving the same thing for any choice of the orthonormal basisso the same is true of the inner product.

Consider completeness. Suppose then that {Tn} is a Cauchy sequence in L2 (H,G) .Then from 19.8.64 {Tn} is a Cauchy sequence in L (H,G) and so there exists a unique