Compact Operators and Spectra

In finite dimensions, a symmetric matrix has an orthonormal basis of eigenvectors with real eigenvalues (1A.6 Symmetric Matrices and the Spectral Theorem). In infinite dimensions this can fail completely: the multiplication operator $f(x) \mapsto xf(x)$ on $L^2(0, 1)$ is symmetric and has no eigenvectors at all. But for one important class of operators the finite-dimensional picture survives almost unchanged. Compact operators, those that map bounded sets to sets with compact closure, behave like matrices whose entries fade out. For them, the Fredholm alternative holds ("unique solvability follows from uniqueness"), and the self-adjoint ones have an orthonormal basis of eigenvectors with eigenvalues tending to $0$.

The main example is the inverse of the Laplacian on a bounded domain or a closed manifold. Its eigenfunctions are the vibration modes of a drum, the shapes that sand traces on a vibrating plate, and the building blocks of the heat equation on a curved space. And the bottom of the spectrum of an operator of this kind, $-4\Delta + R$, is Perelman's $\lambda$-invariant.

By the end of this chapter you will be able to:

• recognise compact operators (finite rank, Hilbert–Schmidt integral operators, diagonal operators with entries tending to $0$), and explain why they turn weak convergence into strong;
• state the Fredholm alternative and use it;
• prove the spectral theorem for compact self-adjoint operators by maximising a quadratic form;
• describe the Dirichlet eigenvalues and eigenfunctions of the Laplacian, compute them on an interval, a square and a disc, and use the Rayleigh quotient;
• explain what can and cannot be heard about the shape of a drum.

7.1Patterns in sand

7.2Compact operators

Compact operators are automatically bounded. The identity of an infinite-dimensional space is not compact (4A.1 Banach Spaces and Bounded Operators). Examples:

• Finite rank. A bounded operator with finite-dimensional range is compact (bounded sets in $\mathbb{R}^n$ have compact closure).
• Limits of finite rank. If $\|K_n - K\| \to 0$ with each $K_n$ compact, then $K$ is compact (a diagonal argument, Exercise 7.7). Compact operators form a closed subspace of $L(X, Y)$, and a two-sided ideal: $AK$ and $KB$ are compact when $K$ is and $A$, $B$ are bounded.
• Diagonal operators. On $\ell^2$, $D(x_k) = (d_kx_k)$ is compact exactly when $d_k \to 0$ (Exercise 7.6).
• Integral operators. If $k \in L^2([0, 1]^2)$, then $Tf(x) = \int_0^1k(x, y)f(y)\,dy$ is compact on $L^2(0, 1)$ (a Hilbert–Schmidt operator): expanding $k$ in an orthonormal basis of $L^2([0,1]^2)$ and truncating gives finite-rank approximations converging in operator norm (Exercise 7.8). The solution operators of elliptic boundary value problems are of this kind.

The property that makes compact operators useful is the following.

7.3The Fredholm alternative

For a matrix, $Ax = b$ is solvable for every $b$ if and only if $Ax = 0$ only for $x = 0$. In infinite dimensions this fails (the shift on $\ell^2$ is injective and not onto, 4A.1 Banach Spaces and Bounded Operators). It survives for compact perturbations of the identity.

Proof architecture (Brezis, Theorem 6.6). (1) On the kernel, $K$ acts as the identity, so the unit ball of the kernel is compact, and the kernel is finite-dimensional by Riesz's lemma (4A.1 Banach Spaces and Bounded Operators). Closed range: if $(I - K)u_n \to f$, one can arrange $(u_n)$ to be bounded (subtract components in the kernel), and then compactness of $K$ gives a convergent subsequence of $Ku_n$, hence of $u_n$. (2) If $I - K$ is injective but not surjective, the ranges $X_n = (I - K)^n(X)$ form a strictly decreasing sequence of closed subspaces; Riesz's lemma gives unit vectors $x_n \in X_n$ at distance $\geq \tfrac12$ from $X_{n+1}$, and then $\|Kx_n - Kx_m\| \geq \tfrac12$, contradicting compactness. The converse uses duality, and boundedness of the inverse is the bounded inverse theorem (4A.2 Baire Category and Its Consequences).

In words: for an equation "identity minus compact", uniqueness implies existence, with a bounded solution operator. This is how existence is proved for many linear elliptic problems (6A.5 Weak Solutions and Elliptic Regularity): show the problem is of Fredholm type, then show that the homogeneous equation has only the zero solution.

7.4The spectral theorem

An operator $K$ on a Hilbert space is self-adjoint if $\langle Kx, y\rangle = \langle x, Ky\rangle$ for all $x, y$. Its eigenvalues are then real and eigenvectors for different eigenvalues are orthogonal, by the computations of 1A.6 Symmetric Matrices and the Spectral Theorem.

Step 1 is the direct method (4A.6 Weak Convergence and the Direct Method): maximise a quadratic form on the unit sphere, using weak compactness, with the compact operator providing the continuity that weak convergence alone would not. It is also Lagrange multipliers (2B.9 The Inverse and Implicit Function Theorems): the eigenvector equation $Kx = \mu x$ is the condition for a critical point of $\langle Kx, x\rangle$ on the sphere $\|x\| = 1$. In finite dimensions this was 2B.9 The Inverse and Implicit Function Theorems's exercise proving that symmetric matrices have eigenvectors.

7.5The spectrum of the Laplacian

Let $\Omega \subseteq \mathbb{R}^n$ be a bounded open set. For $f \in L^2(\Omega)$, Lax–Milgram (4A.4 Hilbert Spaces and Lax–Milgram) gives a unique weak solution $u = Gf \in H^1_0(\Omega)$ of $-\Delta u = f$, $u = 0$ on $\partial\Omega$. The solution operator $G : L^2 \to L^2$ is bounded, self-adjoint and positive ($\langle Gf, f\rangle = \int|\nabla u|^2 > 0$ for $f \neq 0$). It is also compact, because it maps $L^2$ boundedly into $H^1_0$ and the inclusion $H^1_0(\Omega) \hookrightarrow L^2(\Omega)$ is compact. That last fact is Rellich's theorem, proved in 4A.10 Sobolev Embeddings and Critical Exponents; we use it now and complete the argument there.

Applying the spectral theorem to $G$ and inverting its eigenvalues $\mu_k$ gives:

On simple domains the eigenfunctions can be found by separating variables:

• Interval $(0, \pi)$: $\phi_k = \sqrt{2/\pi}\sin(kx)$, $\lambda_k = k^2$ (Exercise 7.11).
• Square $(0, 1)^2$: $\phi_{m,n} = 2\sin(m\pi x)\sin(n\pi y)$, $\lambda = \pi^2(m^2 + n^2)$, for $m, n \geq 1$.
• Unit disc: $J_m(j_{m,k}r)\cos(m\theta)$ and $J_m(j_{m,k}r)\sin(m\theta)$, $\lambda = j_{m,k}^2$, where $J_m$ is the Bessel function and $j_{m,k}$ its $k$-th positive zero. The lowest is $\lambda_1 = j_{0,1}^2 \approx 2.405^2 \approx 5.78$.

7.5.1The Rayleigh quotient

Step 1 of the spectral theorem, applied to $G$, says the first eigenvalue is a minimum:

the Rayleigh quotient, attained exactly at multiples of $\phi_1$. Higher eigenvalues are minima over orthogonal complements, or, without knowing the earlier eigenfunctions, the min–max values $\lambda_k = \min_{\dim V = k}\max_{u\in V\setminus0}\frac{\int|\nabla u|^2}{\int u^2}$ (the Courant–Fischer principle). Any test function gives an upper bound for $\lambda_1$: a bump on a small ball inside $\Omega$ shows that small domains have large $\lambda_1$, and since $\lambda_1(\Omega)$ decreases when $\Omega$ grows (more test functions), a drum's lowest note goes down as it gets bigger.

7.6History

Ivar Fredholm's 1903 theory of integral equations contained the alternative now named after him. David Hilbert (1904–1910) and Erhard Schmidt (1907) developed the spectral theory of symmetric integral operators, and Frigyes Riesz (1918) gave the abstract theory of compact operators. Hermann Weyl's asymptotic law for eigenvalues appeared in 1911, and Richard Courant's min–max principle around 1920. Chladni's experiments date from 1787, Sophie Germain's prize-winning memoir from 1816, Pearson's principal axes from 1901, Kac's question from 1966 and Gordon, Webb and Wolpert's answer from 1992.

7.7Exercises

Solution

For $\lambda \leq 0$ only $\phi = 0$ satisfies the boundary conditions. For $\lambda = \omega^2 > 0$, $\phi = a\sin\omega x + b\cos\omega x$; $\phi(0) = 0$ forces $b = 0$, and $\phi(\pi) = 0$ forces $\omega \in \mathbb{N}$. So $\lambda_k = k^2$, $\phi_k = \sin kx$. If $\phi = \sum c_k\sin kx$, then $\int\phi'^2 = \frac\pi2\sum k^2c_k^2 \geq \frac\pi2\sum c_k^2 = \int\phi^2$.

Solution

$\int(4|\nabla w|^2 + R_0w^2) \geq R_0\int w^2 = R_0$, with equality iff $\nabla w = 0$, i.e. $w$ constant (on a connected manifold), and constants are allowed.