22.5. COMPACT OPERATORS IN HILBERT SPACE 587

Proof: First note that if you have a self adjoint operator A, then its eigenvalues are real.Here is why:

(Au,u) = λ ∥u∥2 = (u,Au) = λ̄ ∥u∥2 .

If ∥A∥= 0 then pick u ∈ H with ∥u∥= 1 and let λ = 0. Since A(H) = 0 it follows thespan of u is dense in A(H) and the formula for Ax holds.

Assume A ̸= 0. I will show there exists an eigenvector u. From the definition of ∥A∥there exists xn,∥xn∥= 1, and ∥Axn∥→ ∥A∥ ≡ |λ |. Now it is clear that A2 is also a compactself adjoint operator. Consider((

λ2−A2

)xn,xn

)= λ

2 (xn,xn)−(A2xn,xn

)= λ

2−∥Axn∥2→ 0.

Since A is compact, there exists a subsequence of {xn} still denoted by {xn} such that Axn

converges to some element of H. Thus since λ2−A2 satisfies

((λ

2−A2)

y,y)≥ 0 in

addition to being self adjoint, it follows

x,y→((

λ2−A2

)x,y)

satisfies all the axioms for an inner product except for the one which says that (z,z) = 0only if z = 0. Therefore, the Cauchy Schwarz inequality may be used to write∣∣∣((λ

2−A2)

xn,y)∣∣∣≤ ((λ

2−A2)

y,y)1/2((

λ2−A2

)xn,xn

)1/2≤ en ∥y∥ .

where en→ 0 as n→ ∞. Taking the sup over ∥y∥ ≤ 1, limn→∞

∥∥∥(λ2−A2

)xn

∥∥∥= 0. Since

A2xn converges, it follows, since λ ̸= 0 that {xn} is a Cauchy sequence converging to x with∥x∥= 1. Therefore, A2xn→ A2x and so

∥∥∥(λ2−A2

)x∥∥∥= 0. Now this shows that

(λ I +A)(λ I−A)x = 0.

If (λ I−A)x = 0, let u≡ x. If (λ I−A)x = y ̸= 0, let u≡ y∥y∥ . Note that this did not identify

the sign of λ . Also note that since λ ̸= 0,u ∈ A(H).Let A ∈ F mean that A consists of vectors of A(H) ,A is an orthonormal set of

vectors, and for each u ∈ A ,Au = λu for some λ . I claim that A is countable becausefrom the compactness of A,A(H) is separable by Proposition 22.5.2 but these vectors ofA(H) are all at least 1/2 apart. Partially order F by set inclusion. Let C be a maximalchain. Then A∞ ≡ ∪C is a maximal element of F . I need to show its span is dense inA(H) . If spanA∞ fails to contain A(H) , then there is a nonzero vector w≡ Av which is notin spanA∞. Then (w−Pw)/∥w−Pw∥ is a unit vector perpendicular to spanA∞. Is thisvector in A(H)? Is P(Av) = A(Pv)? Using Corollary 22.4.4,

P(w) = P(Av) =N

∑k=1

(Av,uk)uk =N

∑k=1

λ k (v,uk)uk

=N

∑k=1

(v,uk)Auk = A(Pv) ∈ A(H)

so this unit vector w is in(spanA∞

)⊥and A∞∪{w} is larger than the maximal element of

F so it must be the case that spanA∞ ⊇ A(H) after all.

22.5. COMPACT OPERATORS IN HILBERT SPACE 587Proof: First note that if you have a self adjoint operator A, then its eigenvalues are real.Here is why:(Au,u) =A |lul|? = (u,Au) = A jul”.If ||A|| = 0 then pick u € A with ||u|| = 1 and let A = 0. Since A (H) = 0 it follows thespan of u is dense in A (H) and the formula for Ax holds.Assume A # 0. I will show there exists an eigenvector u. From the definition of ||A||there exists x), ||xp|| = 1, and ||Ax,|| > ||A|| = |A|. Now it is clear that A? is also a compactself adjoint operator. Consider(( —A?) nn) =)? (XnsXn) — (A?Xn,xn) =)? \|Axn|| + 0.Since A is compact, there exists a subsequence of {x,} still denoted by {x,} such that Ax,converges to some element of H. Thus since A? — A? satisfies (7 ~ A?) y, y) >0inaddition to being self adjoint, it followssy ((22-42) 59)satisfies all the axioms for an inner product except for the one which says that (z,z) = 0only if z= 0. Therefore, the Cauchy Schwarz inequality may be used to write|((22 a2) sas) < (22a?) v9) (2242) stn) < enwhere e, — 0 as n > ov, Taking the sup over ||y|| < 1, lim). | (2? ~ A) Xn || = 0. SinceA?xn converges, it follows, since A #0 that {x, } is a Cauchy sequence converging to x with||x|| = 1. Therefore, Ax, —> A*x and so | (2? ~ A?) x|| = 0. Now this shows that(AI+A) (AI—A)x=0.If (AI —A)x=0, letu=x. If (AT—A)x=y £40, letu= ine Note that this did not identifythe sign of A. Also note that since A 4 0,u € A(A).Let & € F mean that & consists of vectors of A(H),@ is an orthonormal set ofvectors, and for each u € e&,Au = Au for some A. I claim that ./ is countable becausefrom the compactness of A,A (H) is separable by Proposition 22.5.2 but these vectors ofA(H) are all at least 1/2 apart. Partially order ¥ by set inclusion. Let @ be a maximalchain. Then .% = U@ is a maximal element of .¥. I need to show its span is dense inA(H). If span .&%. fails to contain A (H#), then there is a nonzero vector w = Av which is notin span.%. Then (w— Pw) /||w— Pw] is a unit vector perpendicular to span.%. Is thisvector in A(H)? Is P(Av) =A (Pv)? Using Corollary 22.4.4,P(w) = P(Avy)= y (Av, Ug) Ug = yal V, Ux) U,N= pa VU ) Au, = A(Pv) € A(H)so this unit vector w is in (span Aes) ~ and e. {w} is larger than the maximal element ofF so it must be the case that span .% D A (H) after all.