Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 21I (Linear Analysis)

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

Compactness gives a stronger statement than the one requested. Since $T(B_H)$ has compact closure, finitely many open balls of radius $\delta$ cover it. Choose their centres from $T(B_H)$ and let $E$ be their finite-dimensional linear span. ThenFor the orthogonal projection $P_E$, best approximation in a Hilbert space givesIn particular this holds for every vector $e_n$ in the given orthonormal basis.

The operator $P_{E}T$ has image in $E$ and hence has finite rank, whileThus every compact operator is an operator norm limit of finite-rank operators. This is the finite-rank approximation theorem for compact operators on a Hilbert space.

Put $E=\mathrm{im}S$, a finite-dimensional subspace, and decompose $H=E\oplus E^{\perp }$. The restrictionis injective, because a nonzero vector in its kernel would be a $\lambda$-eigenvector of $S$. By finite-dimensionality it is therefore surjective.

Given $y=y_E+y_{\perp }$, first put $x_{\perp }=-y_{\perp }/\lambda$. Then $S{x}_{\perp }\in E$, so choose $x_E\in E$ such thatFor $x=x_E+x_{\perp }$ we obtain $(S-\lambda I)x=y$. Hence $S-\lambda I$ is surjective. This is the finite-rank Fredholm alternative.

Write $A=T-\lambda I$. We first prove that $\mathrm{im}A$ is dense. Otherwise some nonzero $y$ lies in its orthogonal complement, soBy part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in operator norm. Sincethe operator $S_j^{\ast }-\overline{\lambda }I$ is not injective. Therefore $S_j-\lambda I$ is not surjective; the contrapositive of part (c) makes $\lambda$ an eigenvalue of $S_j$. Choose unit vectors $x_j$ with $S_j{x}_j=\lambda x_j$. ThenCompactness gives a subsequence for which $T{x}_j$ converges. Since $\lambda \ne 0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\mathrm{im}A$ is dense.

Next suppose $A$ were not bounded below, or equivalently that its lower norm were zero. There would be unit vectors $x_j$ with $A{x}_j\to 0$, henceAgain a convergent subsequence of $T{x}_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies

This lower bound makes $\mathrm{im}A$ closed: if $A{x}_j$ converges, then $x_j$ is Cauchy and its limit maps to the same limit. Since the image is both dense and closed, it is all of $H$. We have proved the Fredholm alternative for a compact operator: every nonzero spectral value of a compact operator is an eigenvalue.