The Hilbert–Schmidt Theorem for Compact Self-Adjoint Operators

Derives Lemma A23.2. Suppose \(A\) is compact and let \(\norm{x_{n}}\leq c\) for all \(n\); we may assume \(c>0\). The vectors \(Ax_{n}/c\) lie in the image of the closed unit ball, hence in the set \(K=\overline{A\set{x\mid\norm{x}\leq1}}\), which is compact by hypothesis. A compact subset of the metric space \(\mathcal{H}\) is sequentially compact (Theorem 10.31), so some subsequence \(Ax_{n_{k}}/c\) converges in \(K\), and therefore \(Ax_{n_{k}}\) converges.

Conversely, assume the sequential property and let \((y_{m})\) be a sequence in \(K\). Each \(y_{m}\) is a limit of points \(Ax\) with \(\norm{x}\leq1\), so we may choose \(x_{m}\) with \(\norm{x_{m}}\leq1\) and \(\norm{y_{m}-Ax_{m}}<1/m\). By hypothesis some \(Ax_{m_{k}}\) converges, and then \(y_{m_{k}}\) converges to the same limit, which lies in the closed set \(K\). Hence \(K\) is sequentially compact, hence compact by Theorem 10.31.

Derives Lemma A23.3. A closed subspace of a complete metric space is complete, so \(M\) is a Hilbert space with the inherited inner product. Symmetry of \(A|_{M}\) is inherited, and by Proposition 16.42 a symmetric operator defined on all of the Hilbert space \(M\) is self-adjoint there. Compactness is Lemma A23.2: a bounded sequence in \(M\) is a bounded sequence in \(\mathcal{H}\), so \((Ax_{n})\) has a convergent subsequence, whose limit lies in the closed set \(M\). The norm bound is immediate, the supremum being taken over a smaller set. For the last claim, let \(y\in M^{\perp}\) and \(z\in M\); then \(Az\in M\), so \(\braket{z}{Ay}=\braket{Az}{y}=0\), and \(Ay\in M^{\perp}\).

Derives Lemma A23.4. By Proposition 16.43 there are unit vectors \(x_{n}\) with

Each number \(\braket{x_{n}}{Ax_{n}}\) is real, because \(A\) is self-adjoint (Proposition 16.42). A real sequence whose absolute values converge to \(\norm{A}\) has a subsequence converging to \(+\norm{A}\) or to \(-\norm{A}\): infinitely many of its terms are non-negative or infinitely many are negative, and along the corresponding indices the absolute value carries the sign. Relabel, and let

Expand, using \(\lambda\) real, \(\norm{x_{n}}=1\), and \(\braket{Ax_{n}}{x_{n}}=\braket{x_{n}}{Ax_{n}}\):

Now \(\norm{Ax_{n}}\leq\norm{A}=\abs{\lambda}\), so the right-hand side of Equation (A23.5) is at most \(2\lambda^{2}-2\lambda\braket{x_{n}}{Ax_{n}}\), which tends to \(0\) by Equation (A23.4). Hence

This much uses only self-adjointness; it says that \(\lambda\) is an approximate eigenvalue. Compactness converts the approximation into an eigenvector.

The sequence \((x_{n})\) is bounded, so by Lemma A23.2 there is a subsequence with \(Ax_{n_{k}}\longrightarrow y\) for some \(y\in\mathcal{H}\). Since \(\lambda\neq0\) we may write

by Equation (A23.6). The norm is continuous (Corollary 16.5), so \(\norm{e}=\lim\norm{x_{n_{k}}} =1\); and \(A\) is continuous (Proposition 16.36), so \(Ax_{n_{k}}\longrightarrow Ae\). But \(Ax_{n_{k}}\longrightarrow y=\lambda e\). Limits in a metric space are unique, so \(Ae=\lambda e\).

Derives Lemma A23.5. Set \(\mathcal{H}_{1}=\mathcal{H}\) and \(A_{1}=A\). Suppose \(\mathcal{H}_{N}\) and \(A_{N}\) have been defined, with \(\mathcal{H}_{N}\) a closed \(A\)-invariant subspace and \(A_{N}=A|_{\mathcal{H}_{N}}\) compact and self-adjoint on it (Lemma A23.3). If \(A_{N}=0\) the construction stops, and then every \(x\in\mathcal{H}_{N}\) has \(Ax=0\). If \(A_{N}\neq0\), apply Lemma A23.4 inside the Hilbert space \(\mathcal{H}_{N}\): it produces a unit vector \(e_{N}\in\mathcal{H}_{N}\) and a real \(\lambda_{N}\) with \(\abs{\lambda_{N}}=\norm{A_{N}}\neq0\) and \(Ae_{N}=A_{N}e_{N}=\lambda_{N}e_{N}\).

The vector \(e_{N}\) lies in \(\mathcal{H}_{N}\), hence is orthogonal to \(e_{1},\dots,e_{N-1}\), so the system stays orthonormal. Put \(\mathcal{H}_{N+1}=\mathcal{H}_{N}\cap\set{e_{N}}^{\perp} =\set{e_{1},\dots,e_{N}}^{\perp}\), which is closed (Proposition 16.17). It is \(A\)-invariant: for \(x\in\mathcal{H}_{N+1}\) and \(j\leq N\),

using self-adjointness and \(\lambda_{j}\in\R\); and \(Ax\in\mathcal{H}_{N}\) because \(\mathcal{H}_{N}\) is invariant. Finally \(\norm{A_{N+1}}\leq\norm{A_{N}}\) because \(A_{N+1}\) is a restriction of \(A_{N}\), whence \(\abs{\lambda_{N+1}}\leq\abs{\lambda_{N}}\); the construction therefore delivers a non-increasing sequence of moduli, and it is exactly the statement \(\norm{A_{N}}=\abs{\lambda_{N}}\) that Lemma A23.4 supplies at each stage.

Derives Lemma A23.6. Both statements follow from one computation. Let \((f_{n})\) be an orthonormal sequence with \(Af_{n}=\nu_{n}f_{n}\). For \(n\neq m\), orthogonality gives

Eigenvalues. Apply this to \(f_{n}=e_{n}\), \(\nu_{n}=\lambda_{n}\). If \((\lambda_{n})\) does not tend to \(0\) there are \(\varepsilon>0\) and infinitely many indices with \(\abs{\lambda_{n}}\geq\varepsilon\); along them Equation (A23.9) gives \(\norm{Ae_{n}-Ae_{m}}^{2}\geq2\varepsilon^{2}\), so \((Ae_{n})\) has no Cauchy subsequence, hence no convergent subsequence. Since \((e_{n})\) is bounded, this contradicts Lemma A23.2. As the moduli are non-increasing, \(\lambda_{n}\longrightarrow0\).

Multiplicity. If \(\ker(A-\lambda\identity)\) were infinite dimensional for some \(\lambda\neq0\), Gram–Schmidt (Proposition 16.23) would produce an infinite orthonormal sequence \((f_{n})\) inside it, all with \(\nu_{n}=\lambda\), and Equation (A23.9) would give \(\norm{Af_{n}-Af_{m}}^{2}=2\abs{\lambda}^{2}>0\): the same contradiction.

Proof of Theorem A23.1. Derives Theorem A23.1. Let \(\set{e_{n}}\), \(\set{\lambda_{n}}\) be the system of Lemma A23.5. Fix \(x\in\mathcal{H}\) and put

Then \(\braket{e_{j}}{x_{N}}=\braket{e_{j}}{x}-\braket{e_{j}}{x}=0\) for \(j\leq N\), so \(x_{N}\in\mathcal{H}_{N+1}\), and Bessel's inequality (Proposition 16.27) gives

Applying \(A\) to Equation (A23.10) and using \(Ae_{n}=\lambda_{n}e_{n}\),

the last equality because \(x_{N}\in\mathcal{H}_{N+1}\). Therefore, by Lemma A23.5 and Equation (A23.11),

If the construction stops at stage \(N\) then \(A_{N}=0\) and Equation (A23.12) already gives Equation (A23.1) as a finite sum. If it does not stop, \(\abs{\lambda_{N+1}}\longrightarrow0\) by Lemma A23.6, so the right-hand side of Equation (A23.13) tends to \(0\) and Equation (A23.1) holds with convergence in norm. The eigenvalue relation \(Ae_{n}=\lambda_{n}e_{n}\), the ordering of the moduli and the finiteness of the non-zero eigenspaces are Lemmas A23.5 and A23.6. For \(A\neq0\), \(\norm{A}=\norm{A_{1}}=\abs{\lambda_{1}} =\max_{n}\abs{\lambda_{n}}\), the maximum being attained because the moduli decrease.

It remains to prove Equation (A23.2). Write \(M=\overline{\text{span}\set{e_{n}}}\), a closed subspace, so that \(\mathcal{H}=M\oplus M^{\perp}\) by Theorem 16.18. If \(x\in M^{\perp}\) then \(\braket{e_{n}}{x}=0\) for every \(n\), and Equation (A23.1) gives \(Ax=0\): thus \(M^{\perp}\subseteq\ker A\). Conversely if \(Ax=0\) then for every \(n\),

and \(\lambda_{n}\neq0\) forces \(\braket{e_{n}}{x}=0\), so \(x\in M^{\perp}\); hence \(\ker A=M^{\perp}\), which is Equation (A23.2). Every vector of \(\ker A\) is an eigenvector with eigenvalue \(0\), so an orthonormal basis of \(\ker A\) adjoined to \(\set{e_{n}}\) is an orthonormal system of eigenvectors whose closed span is \(M\oplus\ker A=\mathcal{H}\): an orthonormal basis of \(\mathcal{H}\) in the sense of Definition 16.29.