590 CHAPTER 22. HILBERT SPACES

22.5.1 Nuclear OperatorsA very useful idea in linear algebra is the trace of a matrix. It is one of the principle invari-ants and is just the sum of the entries along the main diagonal. Thus if A ∈L (Cn,Cn),the trace is ∑ie

Ti Aei where the ei are the standard orthonormal basis vectors, ei having a

1 in the ith position and 0 elsewhere. If you used another orthonormal basis, {vi} the tracewould be the same because the mapping vi → ei will preserve lengths and is therefore aunitary transformation. The two computations would involve a similarity transformation.In infinite dimensions when you have a separable Hilbert space, the notion of trace mightnot make sense. The nuclear operators are those for which it will make sense.

Definition 22.5.9 A self adjoint operator A ∈L (H,H) for H a separable Hilbertspace is called a nuclear operator if for some complete orthonormal set, {ek} ,It followsthat ∑

∞k=1 |(Aek,ek)|< ∞.

We specialize to self adjoint operators because this will ensure that (Ax,x) is real. Tobegin with here is an interesting lemma.

Lemma 22.5.10 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 isalso compact.

Proof: Let D be a bounded set in X such that ∥b∥ ≤C for all b ∈ D. I need to verifyA(B) is totally bounded. Suppose then it is not. Then there exists ε > 0 and an infinitesequence, {Abi} where bi ∈ D 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∥∥−2

ε

4CC ≥ ε

2,

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

Lemma 22.5.11 Let A≥ 0 be a nuclear operator defined on a separable Hilbert spaceH. 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,

the series converging because x = ∑∞j=1 (x,e j)e j. Then also since A is self adjoint,

∑j=1

∑i=1

(Aei,e j)ei⊗ e j (x)≡∞

∑j=1

∑i=1

(Aei,e j)(x,e j)ei =∞

∑j=1

(x,e j)∞

∑i=1

(Aei,e j)ei

590 CHAPTER 22. HILBERT SPACES22.5.1 Nuclear OperatorsA very useful idea in linear algebra is the trace of a matrix. It is one of the principle invari-ants and is just the sum of the entries along the main diagonal. Thus if A € #(C",C”),the trace is Vie; T Ae; where the e; are the standard orthonormal basis vectors, e; having a1 in the i” position and 0 elsewhere. If you used another orthonormal basis, {v;} the tracewould be the same because the mapping v; — e; will preserve lengths and is therefore aunitary transformation. The two computations would involve a similarity transformation.In infinite dimensions when you have a separable Hilbert space, the notion of trace mightnot make sense. The nuclear operators are those for which it will make sense.Definition 22.5.9 4 self adjoint operator A € &(H,H) for H a separable Hilbertspace is called a nuclear operator if for some complete orthonormal set, {e,} ,It followsthat Ye_, |(Aex, ex) | < %.We specialize to self adjoint operators because this will ensure that (Ax,.x) is real. Tobegin with here is an interesting lemma.Lemma 22.5.10 Suppose {Ay} is a sequence of compact operators in & (X,Y) for twoBanach spaces, X and Y and suppose A € £ (X,Y) and limps. ||A — An|| = 0. Then A isalso compact.Proof: Let D be a bounded set in X such that ||b|| < C for all b € D. I need to verifyA(B) is totally bounded. Suppose then it is not. Then there exists € > 0 and an infinitesequence, {Ab;} where bi € D and \|Ab; —Ab All > € whenever i # j. Then let n be largeenough that ||A —A,,|| < 7@.Then|Anbi—Anbj|| = ||Ab; —Abj + (An ADB An — A)b|||2 ||Abi—Abj|| — (II4n—A) Bill + |](An—A) ill)> ||Abi—Adj||-2 gonea contradiction to A, being compact. MiThen one can prove the following lemma. In this lemma, A > 0 will mean (Ax, x) >0.Lemma 22.5.11 Let A > 0 be a nuclear operator defined on a separable Hilbert spaceH. Then A is compact and also, whenever {ex} is a complete orthonormal set,coA=)jHli4:(Ae;,e;) ej @e;.1Proof: First consider the formula. Since A is given to be continuous,ana (x,e;)e a) d| x,e;)Ae;,1 j=lj=the series converging because x = )'7"_; (x,e ;)e;. Then also since A is self adjoint,co»j=li© ©(Ae, e;) ej Bej(x )= yy (Ae;,e;) (x,e;)e =) ( X,e; yy (Ae;,e;) ejj=1li=1 j=lMs1=i"na.