586 CHAPTER 22. HILBERT SPACES

cnk such that limn→∞

∥∥Py−∑nk=1 cn

kuk∥∥ = 0. Then by Corollary 22.4.3 and what was just

observed,

0 = limn→∞

∥∥∥∥∥Py−n

∑k=1

(Py,uk)uk

∥∥∥∥∥= limn→∞

∥∥∥∥∥Py−n

∑k=1

(y,uk)uk

∥∥∥∥∥and so the sum converges to Py as claimed. ■

22.5 Compact Operators in Hilbert SpaceDefinition 22.5.1 Let A ∈L (H,H) where H is a Hilbert space. So the map, x→(Ax,y) is continuous and linear. By the Riesz representation theorem, there exists a uniqueelement of H, denoted by A∗y such that

(Ax,y) = (x,A∗y) .

It is clear y→ A∗y is linear and continuous. It is linear because((x,A∗ (ay+bz)) = ā(Ax,y)+ b̄(Ax,z)

= ā(x,A∗y)+ b̄(x,A∗z) = (x,aA∗y+bA∗z)

)A∗ is called the adjoint of A. A is a self adjoint operator if A = A∗. Thus for a self adjointoperator, (Ax,y) = (x,Ay) for all x,y ∈ H and so (Ax,x) = (x,Ax) = (Ax,x) so (Ax,x)is real. A is a compact operator if whenever {xk} is a bounded sequence, there existsa convergent subsequence of {Axk}. Equivalently, A maps bounded sets to sets whoseclosures are compact.

There is an important observation about the range of a compact operator. It is a generalresult so I will express it in terms of Banach space.

Proposition 22.5.2 Let X be a Banach space and let A ∈L (X ,X) be a compact oper-ator meaning that the image of a bounded set is a pre-compact set. Then A(X) is separable.

Proof: If for every n ∈ N there is a 1/n net for A(B(0,1)) , then A(B(0,1)) would beseparable, and a countable dense subset would be the union of these 1/n nets. It would fol-low then that A(X) is also separable. A countable dense subset would be positive rationalnumbers times these countably many points in A(B(0,1)). Therefore, if A(X) is not sepa-rable, there is some ε > 0 such that there is no ε net, for A(B(0,1)) meaning that there areinfinitely many points which are all ε apart in A(B(0,1)). Let these points be {Auk}∞

k=1 .But now this is a contradiction because there can be no convergent subsequence. ■

The big result is called the Hilbert Schmidt theorem. It is a generalization to arbitraryHilbert spaces of standard finite dimensional results having to do with diagonalizing asymmetric matrix.

Theorem 22.5.3 Let A be a compact self adjoint operator defined on a Hilbertspace H. Then there exists a countable set of eigenvalues, {λ i} and an orthonormal set ofeigenvectors, ui satisfying

λ i is real, |λ n| ≥ |λ n+1| , Aui = λ iui, (22.15)

and either limn→∞ λ n = 0,or for some n, span(u1, · · · ,un) = A(H) . In any case,

span({ui}∞

i=1) is dense in A(H) . (22.16)

and for all x ∈ H, Ax = ∑∞k=1 λ k (x,uk)uk where the sum might be finite.

586 CHAPTER 22. HILBERT SPACESch such that limp ||Py — Dt_ cfux|| = 0. Then by Corollary 22.4.3 and what was justobserved,nPy— YP (Py, ux) uxk=10 = lim= limn—yoo n—yoonPy—Y° (y, ux) uxk=1and so the sum converges to Py as claimed. ll22.5 Compact Operators in Hilbert SpaceDefinition 22.5.1 Let A €¢ Y(H,H) where H is a Hilbert space. So the map, x >(Ax, y) is continuous and linear. By the Riesz representation theorem, there exists a uniqueelement of H, denoted by A*y such that(Ax,y) = (x,A*y).It is clear y — A*y is linear and continuous. It is linear because(x,A* (ay + bz)) = a(Ax,y) + b (Ax, z)= (x,A*y) +b(x,A*z) = (x,aA*y + bA*z)A* is called the adjoint of A. A is a self adjoint operator if A = A*. Thus for a self adjointoperator, (Ax,y) = (x,Ay) for all x,y © H and so (Ax,x) = (x,Ax) = (Ax,x) so (Ax,x)is real. A is a compact operator if whenever {xx} is a bounded sequence, there existsa convergent subsequence of {Ax,}. Equivalently, A maps bounded sets to sets whoseclosures are compact.There is an important observation about the range of a compact operator. It is a generalresult so I will express it in terms of Banach space.Proposition 22.5.2 Let X be a Banach space and let A € & (X,X) be a compact oper-ator meaning that the image of a bounded set is a pre-compact set. Then A (X) is separable.Proof: If for every n € N there is a 1/n net for A(B(0,1)), then A(B(0,1)) would beseparable, and a countable dense subset would be the union of these 1 /n nets. It would fol-low then that A(X) is also separable. A countable dense subset would be positive rationalnumbers times these countably many points in A (B(0,1)). Therefore, if A (X) is not sepa-rable, there is some € > 0 such that there is no € net, for A (B (0, 1)) meaning that there areinfinitely many points which are all € apart in A(B(0,1)). Let these points be {Aug}; -But now this is a contradiction because there can be no convergent subsequence. llThe big result is called the Hilbert Schmidt theorem. It is a generalization to arbitraryHilbert spaces of standard finite dimensional results having to do with diagonalizing asymmetric matrix.Theorem 22.5.3 Let A be a compact self adjoint operator defined on a Hilbertspace H. Then there exists a countable set of eigenvalues, {A} and an orthonormal set ofeigenvectors, uj; SatisfyingA; is real, |An| > |Anszil, Au; = Aim, (22.15)and either limps. An = 0,or for some n, span (uy,-++ ,Un) =A(H). In any case,span ({u;};-,) is dense in A(H). (22.16)and for all x € H, Ax = Vp_) Ax (%, Ug) ux where the sum might be finite.