Compact Operators

Abstract

If $\mathfrak{H}$ is a normed vector space, a linear operator $T: \mathfrak{H} \rightarrow \mathfrak{H}$ is called bounded if there exists a constant C such that $|Tx| \leqslant C|x|$ for all $x \in \mathfrak{H}$. In this case, the smallest such C is called the operator norm of T, and is denoted |T|.

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If $\mathfrak{H}$ is a normed vector space, a linear operator $T: \mathfrak{H} \rightarrow \mathfrak{H}$ is called bounded if there exists a constant C such that |Tx|≤ C|x| for all $x \in \mathfrak{H}$. In this case, the smallest such C is called the operator norm of T, and is denoted |T|. The boundedness of the operator T is equivalent to its continuity. If $\mathfrak{H}$ is a Hilbert space, then a bounded operator T is self-adjoint if

$$\displaystyle{\left \langle Tf,g\right \rangle = \left \langle f,Tg\right \rangle }$$

for all $f,g \in \mathfrak{H}$. As usual, we call f an eigenvector with eigenvalue λ if f≠0 and Tf = λf. Given λ, the set of eigenvectors with eigenvalue λ (together with 0, which is not an eigenvector) is called the λ-eigenspace. It follows from elementary and well-known arguments that if T is a self-adjoint bounded operator, then its eigenvalues are real, and the eigenspaces corresponding to distinct eigenvalues are orthogonal. Moreover, if $V \subset \mathfrak{H}$ is a subspace such that T(V ) ⊂ V, it is easy to see that also $T({V }^{\perp }) \subset {V }^{\perp }$.

A bounded operator $T: \mathfrak{H} \rightarrow \mathfrak{H}$ is compact if whenever $\{x_{1,}x_{2},x_{3},\ldots \}$ is any bounded sequence in $\mathfrak{H}$, the sequence $\{Tx_{1},Tx_{2},\ldots \}$ has a convergent subsequence.

FormalPara Theorem 3.1 (Spectral theorem for compact operators).

Let T be a compact self-adjoint operator on a Hilbert space $\mathfrak{H}$ . Let $\mathfrak{N}$ be the nullspace of $T$ . Then the Hilbert space dimension of ${\mathfrak{N}}^{\perp }$ is at most countable. ${\mathfrak{N}}^{\perp }$ has an orthonormal basis ϕ _i (i = 1,2,3,…) of eigenvectors of T so that $T\phi _{i} = \lambda _{i}\phi _{i}$ . If ${\mathfrak{N}}^{\perp }$ is not finite-dimensional, the eigenvalues λ _i → 0 as i → ∞.

Since the eigenvalues λ _i → 0, if λ is any nonzero eigenvalue, it follows from this statement that the λ-eigenspace is finite-dimensional.

FormalPara Proof.

This depends upon the equality

$$\displaystyle{ \vert T\vert =\sup _{0\neq x\in \mathfrak{H}} \frac{\vert \left \langle Tx,x\right \rangle \vert } {\left \langle x,x\right \rangle }. }$$

(3.1)

To prove this, let B denote the right-hand side. If $0\neq x \in \mathfrak{H}$,

$$\displaystyle{\vert \left \langle Tx,x\right \rangle \vert \leqslant \vert Tx\vert \cdot \vert x\vert \leqslant \vert T\vert \cdot \vert x{\vert }^{2} = \vert T\vert \cdot \left \langle x,x\right \rangle,}$$

so $B \leqslant |T|$. We must prove the converse. Let λ > 0 be a constant, to be determined later. Using $\left \langle {T}^{2}x,x\right \rangle = \left \langle Tx,Tx\right \rangle$, we have

$$\displaystyle\begin{array}{rcl} & & \left \langle Tx,Tx\right \rangle {}\\ & & \quad = \tfrac{1} {4}\left \vert \left \langle T(\lambda x + {\lambda }^{-1}Tx),\lambda x + {\lambda }^{-1}Tx\right \rangle -\left \langle T(\lambda x - {\lambda }^{-1}Tx),\lambda x - {\lambda }^{-1}Tx\right \rangle \right \vert {}\\ & &\quad \leqslant \tfrac{1} {4}\left \vert \left \langle T(\lambda x + {\lambda }^{-1}Tx),\lambda x + {\lambda }^{-1}Tx\right \rangle \right \vert +\left \vert \left \langle T(\lambda x - {\lambda }^{-1}Tx),\lambda x - {\lambda }^{-1}Tx\right \rangle \right \vert {}\\ & &\quad \leqslant \tfrac{1} {4}\left [B\left \langle \lambda x + {\lambda }^{-1}Tx,\lambda x + {\lambda }^{-1}Tx\right \rangle + B\left \langle \lambda x - {\lambda }^{-1}Tx,\lambda x - {\lambda }^{-1}Tx\right \rangle \right ] {}\\ & & \quad = \tfrac{B} {2} \left [{\lambda }^{2}\left \langle x,x\right \rangle + {\lambda }^{-2}\left \langle Tx,Tx\right \rangle \right ]. {}\\ \end{array}$$

Now taking $\lambda = \sqrt{\vert Tx\vert /\vert x\vert }$, we obtain

$$\displaystyle{\vert Tx{\vert }^{2} = \left \langle Tx,Tx\right \rangle \leqslant B\vert x\vert \vert Tx\vert,}$$

so $|Tx| \leqslant B|x|$, which implies that $|T| \leqslant B$, whence (3.1).

We now prove that ${\mathfrak{N}}^{\perp }$ has an orthonormal basis consisting of eigenvectors of T. It is an easy consequence of self-adjointness that ${\mathfrak{N}}^{\perp }$ is T-stable. Let Σ be the set of all orthonormal subsets of ${\mathfrak{N}}^{\perp }$ whose elements are eigenvectors of T. Ordering Σ by inclusion, Zorn’s lemma implies that it has a maximal element S. Let V be the closure of the linear span of S. We must prove that $V = {\mathfrak{N}}^{\perp }$. Let $\mathfrak{H}_{0} = {V }^{\perp }$. We wish to show $\mathfrak{H}_{0} = \mathfrak{N}$. It is obvious that $\mathfrak{N} \subseteq \mathfrak{H}_{0.}$ To prove the opposite inclusion, note that $\mathfrak{H}_{0}$ is stable under T, and T induces a compact self-adjoint operator on $\mathfrak{H}_{0}$. What we must show is that $T\vert \mathfrak{H}_{0} = 0.$ If T has a nonzero eigenvector in $\mathfrak{H}_{0}$, this will contradict the maximality of Σ. It is therefore sufficient to show that a compact self-adjoint operator on a nonzero Hilbert space has an eigenvector.

Replacing $\mathfrak{H}$ by $\mathfrak{H}_{0}$, we are therefore reduced to the easier problem of showing that if T≠0, then T has a nonzero eigenvector. By (3.1), there is a sequence $x_{1},x_{2},x_{3},\ldots$ of unit vectors such that $\vert \left \langle Tx_{i},x_{i}\right \rangle \vert \rightarrow \vert T\vert$. Observe that if $x \in \mathfrak{H}$, we have

$$\displaystyle{\left \langle Tx,x\right \rangle = \left \langle x,Tx\right \rangle = \overline{\left \langle Tx,x\right \rangle }}$$

so the $\left \langle Tx_{i},x_{i}\right \rangle$ are real; we may therefore replace the sequence by a subsequence such that $\left \langle Tx_{i},x_{i}\right \rangle \rightarrow \lambda$, where λ = ±|T|. Since T≠0, λ≠0. Since T is compact, there exists a further subsequence $\{x_{i}\}$ such that Tx _i converges to a vector v. We will show that $x_{i} \rightarrow {\lambda }^{-1}v$.

Observe first that

$$\displaystyle{\vert \left \langle Tx_{i},x_{i}\right \rangle \vert \leqslant \vert Tx_{i}\vert \vert x_{i}\vert = \vert Tx_{i}\vert \leqslant \vert T\vert \vert x_{i}\vert = \vert \lambda \vert,}$$

and since $\left \langle Tx_{i},x_{i}\right \rangle \rightarrow \lambda$, it follows that |Tx _i |→|λ|. Now

$$\displaystyle{\vert \lambda x_{i} - Tx_{i}{\vert }^{2} = \left \langle \lambda x_{ i} - Tx_{i},\lambda x_{i} - Tx_{i}\right \rangle = {\lambda }^{2}\vert x_{ i}{\vert }^{2} + \vert Tx_{ i}{\vert }^{2} - 2\lambda \left \langle Tx_{ i},x_{i}\right \rangle,}$$

and since |x _i | = 1, |Tx _i |→|λ|, and $\left \langle Tx_{i},x_{i}\right \rangle \rightarrow \lambda$, this converges to 0. Since Tx _i → v, the sequence λx _i therefore also converges to v, and $x_{i} \rightarrow {\lambda }^{-1}v$. Now, by continuity, $Tx_{i} \rightarrow {\lambda }^{-1}\,Tv$, so $v = {\lambda }^{-1}\,Tv$. This proves that v is an eigenvector with eigenvalue λ. This completes the proof that ${\mathfrak{N}}^{\perp }$ has an orthonormal basis consisting of eigenvectors.

Now let $\{\phi _{i}\}$ be this orthonormal basis and let λ _i be the corresponding eigenvalues. If ε > 0 is given, only finitely many $\vert \lambda _{i}\vert > \epsilon$ since otherwise we can find an infinite sequence of ϕ _i with |Tϕ _i | > ε. Such a sequence will have no convergent subsequence, contradicting the compactness of T. Thus, ${\mathfrak{N}}^{\perp }$ is countable-dimensional, and we may arrange the $\{\phi _{i}\}$ in a sequence. If it is infinite, we see the $\lambda _{i}\longrightarrow 0$. □

FormalPara Proposition 3.1.

Let X and Y be compact topological spaces with Y a metric space with distance function d. Let U be a set of continuous maps $X\longrightarrow Y$ such that for every x ∈ X and every ε > 0 there exists a neighborhood N of x such that $d\big(f(x),f(x^{\prime})\big) < \epsilon$ for all x′ ∈ N and for all f ∈ U. Then every sequence in U has a uniformly convergent subsequence.

We refer to the hypothesis on U as equicontinuity.

FormalPara Proof.

Let $S_{0} =\{ f_{1},f_{2},f_{3},\ldots \}$ be a sequence in U. We will show that it has a convergent subsequence. We will construct a subsequence that is uniformly Cauchy and hence has a limit. For every n > 1, we will construct a subsequence $S_{n} =\{ f_{n1},f_{n2},f_{n3},\ldots \}$ of S _n−1 such that $\sup _{x\in X}d\big(f_{ni}(x),f_{nj}(x)\big) \leqslant 1/n$.

Assume that S _n−1 is constructed. For each x ∈ X, equicontinuity guarantees the existence of an open neighborhood N _x of x such that $d\big(f(y),f(x)\big) \leqslant \frac{1} {3n}$ for all y ∈ N _x and all f ∈ X. Since X is compact, we can cover X by a finite number of these sets, say $N_{x_{1}},\ldots,N_{x_{m}}$. Since the f _n−1,i take values in the compact space Y, the m-tuples $\big(f_{n-1,i}(x_{1}),\ldots,f_{n-1,i}(x_{m})\big)$ have an accumulation point, and we may therefore select the subsequence $\{f_{ni}\}$ such that $d\big(f_{ni}(x_{k}),f_{nj}(x_{k})\big) \leqslant \frac{1} {3n}$ for all i, j and $1\leqslant K \leqslant m$. Then for any y, there exists x _k such that $y \in N_{x_{k}}$ and

$$\displaystyle\begin{array}{rcl} d\big(f_{ni}(y),f_{nj}(y)\big)& \leqslant & d\big(f_{ni}(y),f_{ni}(x_{k})\big) + d\big(f_{ni}(x_{k}),f_{nj}(x_{k})\big) {}\\ & & +d\big(f_{nj}(y),f_{nj}(x_{k})\big) \leqslant \tfrac{1} {3n} + \tfrac{1} {3n} + \tfrac{1} {3n} = \tfrac{1} {n}. {}\\ \end{array}$$

This completes the construction of the sequences $\{f_{ni}\}$.

The diagonal sequence $\{f_{11},f_{22},f_{33},\ldots \}$ is uniformly Cauchy. Since Y is a compact metric space, it is complete, and so this sequence is uniformly convergent. □

We topologize C(X) by giving it the L ^∞ norm | | _∞ (sup norm).

FormalPara Proposition 3.2 (Ascoli and Arzela).

Suppose that X is a compact space and that U ⊂ C(X) is a bounded subset such that for each x ∈ X and ε > 0 there is a neighborhood N of x such that $\vert f(x) - f(y)\vert \leqslant \epsilon$ for all y ∈ N and all f ∈ U. Then every sequence in U has a uniformly convergent subsequence.

Again, the hypothesis on U is called equicontinuity.

FormalPara Proof.

Since U is bounded, there is a compact interval $Y \subset \mathbb{R}$ such that all functions in U take values in Y. The result follows from Proposition 3.1. □

Exercises

FormalPara Exercise 3.1.

Suppose that T is a bounded operator on the Hilbert space $\mathfrak{H}$, and suppose that for each ε > 0 there exists a compact operator T _ε such that $\vert T - T_{\epsilon }\vert < \epsilon$. Show that T is compact. (Use a diagonal argument like the proof of Proposition 3.1.)

FormalPara Exercise 3.2 (Hilbert–Schmidt operators).

Let X be a locally compact Hausdorff space with a positive Borel measure μ. Assume that L ²(X) has a countable basis. Let $K \in {L}^{2}(X \times X)$. Consider the operator on L ²(X) with kernel K defined by

$$\displaystyle{Tf(x) =\int _{X}K(x,y)\,f(y)\,\mathrm{d}\mu (y).}$$

Let ϕ _i be an orthonormal basis of L ²(X). Expand K in a Fourier expansion:

$$\displaystyle{K(x,y) =\sum _{ i=1}^{\infty }\psi _{ i}(x)\,\overline{\phi _{i}(y)},\psi _{i} = T\phi _{i}.}$$

Show that $\sum \vert \psi _{i}{\vert }^{2} =\int \int \vert K(x,y){\vert }^{2}\mathrm{d}\mu (x)\,\mathrm{d}\mu (y) < \infty$. Consider the operator T _N with kernel

$$\displaystyle{K_{N}(x,y) =\sum _{ i=1}^{N}\psi _{ i}(x)\,\overline{\phi _{i}(y)}.}$$

Show that T _N is compact, and deduce that T is compact.