19.8. STURM LIOUVILLE PROBLEMS 553

Therefore,limn→∞||A−An||= 0

and so A is the limit in operator norm of finite rank bounded linear operators, each of whichis compact. Therefore, A is also compact.

Definition 19.8.8 The trace of a nuclear operator A ∈L (H,H) such that A≥ 0 is definedto equal

∞

∑k=1

(Aek,ek)

where {ek} is an orthonormal basis for the Hilbert space, H.

Theorem 19.8.9 Definition 19.8.8 is well defined and equals ∑∞j=1 λ j where the λ j are the

eigenvalues of A.

Proof: Suppose {uk} is some other orthonormal basis. Then

ek =∞

∑j=1

u j (ek,u j)

By Lemma 19.8.7 A is compact and so

A =∞

∑k=1

λ kuk⊗uk

where the uk are the orthonormal eigenvectors of A which form a complete orthonormalset. Then

∞

∑k=1

(Aek,ek) =∞

∑k=1

(A

(∞

∑j=1

u j (ek,u j)

),

∞

∑j=1

u j (ek,u j)

)

=∞

∑k=1

∑i j(Au j,ui)(ek,u j)(ui,ek)

=∞

∑k=1

∞

∑j=1

(Au j,u j)∣∣(ek,u j)

∣∣2=

∞

∑j=1

(Au j,u j)∞

∑k=1

∣∣(ek,u j)∣∣2 = ∞

∑j=1

(Au j,u j)∣∣u j∣∣2

=∞

∑j=1

(Au j,u j) =∞

∑j=1

λ j

and this proves the theorem.This is just like it is for a matrix. Recall the trace of a matrix is the sum of the eigen-

values.