19.8. STURM LIOUVILLE PROBLEMS 551

19.8.1 Nuclear OperatorsDefinition 19.8.5 A self adjoint operator A ∈ L (H,H) for H a separable Hilbert spaceis called a nuclear operator if for some complete orthonormal set, {ek} ,

∞

∑k=1|(Aek,ek)|< ∞

To begin with here is an interesting lemma.

Lemma 19.8.6 Suppose {An} is a sequence of compact operators in L (X ,Y ) for twoBanach spaces, X and Y and suppose A ∈L (X ,Y ) and

limn→∞||A−An||= 0.

Then A is also compact.

Proof: Let B be a bounded set in X such that ||b|| ≤C for all b ∈ B. I need to verify ABis totally bounded. Suppose then it is not. Then there exists ε > 0 and a sequence, {Abi}where bi ∈ B and ∣∣∣∣Abi−Ab j

∣∣∣∣≥ ε

whenever i ̸= j. Then let n be large enough that

||A−An|| ≤ε

4C.

Then ∣∣∣∣Anbi−Anb j∣∣∣∣ =

∣∣∣∣Abi−Ab j +(An−A)bi− (An−A)b j∣∣∣∣

≥∣∣∣∣Abi−Ab j

∣∣∣∣−||(An−A)bi||−∣∣∣∣(An−A)b j

∣∣∣∣≥

∣∣∣∣Abi−Ab j∣∣∣∣− ε

4CC− ε

4CC ≥ ε

2,

a contradiction to An being compact. This proves the lemma.Then one can prove the following lemma. In this lemma, A≥ 0 will mean (Ax,x)≥ 0.

Lemma 19.8.7 Let A≥ 0 be a nuclear operator defined on a separable Hilbert space, H.Then A is compact and also, whenever {ek} is a complete orthonormal set,

A =∞

∑j=1

∞

∑i=1

(Aei,e j)ei⊗ e j.

Proof: First consider the formula. Since A is given to be continuous,

Ax = A

(∞

∑j=1

(x,e j)e j

)=

∞

∑j=1

(x,e j)Ae j,