Consider the Poisson equation on a bounded domain $U \subset \mathbb{R}^n$: \begin{align*} -\Delta u = f \quad \text{in } U, \qquad u = 0 \quad \text{on } \partial U. \end{align*} To analyse this problem rigorously, we must first decide what kind of object $-\Delta$ is. Is it bounded? Invertible? What spaces should it act between? The theory of linear operators on Banach spaces is the language for making these choices precise, and each section of this page is driven by the same question: what do we need to know about $-\Delta$ in order to find and understand its solutions? ## Bounded and Closed Operators The simplest question about $-\Delta$ is whether it is bounded — whether it maps bounded [sets](/page/Set) to bounded sets. [definition:Linear Operator] Let $X$ and $Y$ be real [Banach spaces](/page/Banach%20Space). A **linear operator** from $X$ to $Y$ is a map defined on a linear subspace $D(A) \subseteq X$, called the **domain** of $A$: \begin{align*} A: D(A) &\to Y \\ u &\mapsto Au, \end{align*} satisfying $A(\lambda u + \mu v) = \lambda Au + \mu Av$ for all $u, v \in D(A)$ and $\lambda, \mu \in \mathbb{R}$. [/definition] [definition:Bounded Linear Operator] A linear operator $A: D(A) \subseteq X \to Y$ is **bounded** if there exists $M \ge 0$ such that \begin{align*} \|Au\|_Y \le M\|u\|_X \quad \text{for every } u \in D(A). \end{align*} The space of all bounded linear operators defined on all of $X$ is denoted $\mathcal{L}(X, Y)$, equipped with the **operator norm** \begin{align*} \|A\|_{\mathcal{L}(X,Y)} := \sup\left\{ \|Au\|_Y : u \in X,\; \|u\|_X \le 1 \right\}. \end{align*} With this norm $\mathcal{L}(X,Y)$ is itself a Banach space. [/definition] The operator $-\Delta$ is not bounded on $L^2(U)$: the [functions](/page/Function) $u_k(x) = \sin(k\pi x_1)$ have $\|u_k\|_{L^2} \approx 1$ yet $\|-\Delta u_k\|_{L^2} = k^2\pi^2\|u_k\|_{L^2} \to \infty$. [Differentiation](/page/Derivative) amplifies high-frequency oscillations without bound, and no constant $M$ can control it. However, $-\Delta$ does satisfy a strictly weaker property — closedness — which turns out to be enough for the purposes of PDE theory. [definition:Closed Linear Operator] A linear operator $A: D(A) \subseteq X \to Y$ is **closed** if whenever $\{u_k\}_{k=1}^\infty \subseteq D(A)$ satisfies $u_k \to u$ in $X$ and $Au_k \to v$ in $Y$, it follows that $u \in D(A)$ and $Au = v$. [/definition] The graph of $A$ is the subset of $X \times Y$ defined by \begin{align*} \mathrm{Graph}(A) := \{(u, Au) : u \in D(A)\} \subseteq X \times Y. \end{align*} Equipped with the norm $\|(u, v)\|_{X \times Y} := \|u\|_X + \|v\|_Y$, the product $X \times Y$ is a Banach space. The sequential definition of closedness is equivalent to a [topological](/page/Topology) one: [theorem:Graph Characterisation of Closed Operators] Let $X$ and $Y$ be Banach spaces and let $A: D(A) \subseteq X \to Y$ be a linear operator. Then $A$ is closed if and only if $\mathrm{Graph}(A)$ is a closed subspace of $X \times Y$. [/theorem] This characterisation is often the more convenient one to work with in practice: to verify that $A$ is closed, it suffices to show that $\mathrm{Graph}(A)$ is closed in the product topology, which can be done without reference to specific [sequences](/page/Sequence). Closedness says the operator respects [limits](/page/Limit) when both inputs and outputs converge simultaneously — a much weaker demand than boundedness, which requires outputs to be controlled by inputs alone. Every bounded operator is closed (by [continuity](/page/Continuity)), but the converse requires the domain to be all of $X$: [quotetheorem:217] For $-\Delta$, the domain is $H^2(U) \cap H_0^1(U) \subsetneq L^2(U)$, so the theorem does not force boundedness — and indeed $-\Delta$ is unbounded. The theorem instead becomes a tool in the other direction: once we invert $-\Delta$ to produce $(-\Delta)^{-1}: L^2(U) \to L^2(U)$ defined everywhere, closedness is straightforward to verify and the Closed Graph Theorem gives boundedness of the inverse for free. [example:Differentiation Operator] On $X = C([0,1])$ with uniform norm $\|u\|_\infty = \sup_{x \in [0,1]}|u(x)|$, define \begin{align*} A: C^1([0,1]) &\to C([0,1]) \\ u &\mapsto u'. \end{align*} The sequence $u_k(x) = \sin(k\pi x)$ satisfies $\|u_k\|_\infty = 1$ but $\|Au_k\|_\infty = k\pi \to \infty$, so $A$ is unbounded. However, $A$ is closed: if $u_k \to u$ and $u_k' \to v$ both uniformly, then the classical theorem on differentiation of [uniformly convergent](/page/Uniform%20Convergence) sequences gives $u \in C^1([0,1])$ and $u' = v$. There is no contradiction with the Closed Graph Theorem because $C^1([0,1]) \subsetneq C([0,1])$. [/example] ## Dual Spaces and the Weak Formulation Since $-\Delta$ is unbounded, pointwise solutions to $-\Delta u = f$ are too much to ask for rough data $f$. The standard remedy is to multiply by a [test function](/page/Test%20Function) $\phi \in H_0^1(U)$ and integrate by parts, obtaining the **weak formulation**: \begin{align*} \int_U \nabla u \cdot \nabla \phi \, d\mathcal{L}^n = \int_U f\, \phi \, d\mathcal{L}^n \quad \text{for every } \phi \in H_0^1(U). \end{align*} The right-hand side is a bounded linear functional of $\phi$ — an element of the dual of $H_0^1(U)$. This forces us to develop a theory of dual spaces. [definition:Dual Space] Let $X$ be a Banach space. The **dual space** $X^*$ is the Banach space $\mathcal{L}(X, \mathbb{R})$ of all bounded linear functionals on $X$: \begin{align*} f: X &\to \mathbb{R} \\ u &\mapsto f(u), \end{align*} equipped with the norm $\|f\|_{X^*} = \sup\{|f(u)| : \|u\|_X \le 1\}$. [/definition] The weak formulation of the Poisson problem lives most naturally in a Hilbert space, where every functional has a concrete representation as an inner product. [definition:Hilbert Space] A **Hilbert space** is a real vector space $H$ equipped with a symmetric, bilinear, positive-definite inner product \begin{align*} (\cdot, \cdot)_H: H \times H &\to \mathbb{R} \\ (u, v) &\mapsto (u, v)_H \end{align*} such that $H$ is complete with respect to the induced norm $\|u\|_H := \sqrt{(u,u)_H}$. Every Hilbert space is in particular a Banach space. [/definition] [quotetheorem:221] For the Poisson problem: the right-hand side $\phi \mapsto \int_U f\phi \, d\mathcal{L}^n$ defines a bounded functional on $H_0^1(U)$ when $f \in L^2(U)$, and the Riesz theorem (or Lax-Milgram for asymmetric bilinear forms) produces a unique $u \in H_0^1(U)$ satisfying the weak formulation. This constructs the solution operator $(-\Delta)^{-1}: L^2(U) \to H_0^1(U)$, which is bounded by the [Closed Graph Theorem](/theorems/217). Composing with the inclusion $H_0^1(U) \hookrightarrow L^2(U)$ then gives a bounded operator $K: L^2(U) \to L^2(U)$ — and, crucially, a compact one. ## Weak Convergence Before turning to compact operators we need one more tool: a notion of convergence powerful enough to extract limits from bounded sequences in infinite dimensions. When a Galerkin scheme produces a sequence of approximate solutions $\{u_k\} \subseteq H_0^1(U)$ bounded in energy, norm convergence of a subsequence is not guaranteed — but [weak convergence](/page/Weak%20Convergence) is. A full treatment of weak convergence, weak* convergence, and their roles in PDE theory is given on the [Weak Convergence](/pages/1041) page. We recall the definition and the central compactness result here. [definition:Weak Convergence] Let $X$ be a Banach space. A sequence $\{u_k\}_{k=1}^\infty \subseteq X$ **converges weakly** to $u \in X$, written $u_k \rightharpoonup u$, if $f(u_k) \to f(u)$ for every $f \in X^*$. [/definition] The spaces relevant to PDE theory — $L^p(U)$ and $W^{k,p}(U)$ for $1 < p < \infty$, and all [Hilbert spaces](/page/Hilbert%20Space) — are **reflexive**, meaning the canonical embedding $X \hookrightarrow X^{**}$ is surjective. Reflexivity is precisely the condition that makes the following possible. [quotetheorem:214] Weak convergence is sufficient to pass to the limit in the linear weak formulation: if $u_k \rightharpoonup u$ in $H_0^1(U)$, then $\int_U \nabla u_k \cdot \nabla \phi \, d\mathcal{L}^n \to \int_U \nabla u \cdot \nabla \phi \, d\mathcal{L}^n$ for each fixed $\phi$, since this is exactly the evaluation of a fixed bounded functional. For nonlinear problems, however, weak convergence is not enough — products $u_k^2$ need not converge weakly to $u^2$ — and strong convergence of subsequences is required. This is what compact operators provide. ## Compact Operators [definition:Compact Operator] A bounded linear operator $K \in \mathcal{L}(X, Y)$ is **compact** if every bounded sequence $\{u_k\}_{k=1}^\infty \subseteq X$ has a subsequence $\{u_{k_j}\}_{j=1}^\infty$ such that $\{Ku_{k_j}\}_{j=1}^\infty$ converges strongly in $Y$. The space of compact operators is denoted $\mathcal{K}(X, Y) \subseteq \mathcal{L}(X, Y)$. [/definition] The solution operator $K = (-\Delta)^{-1}: L^2(U) \to L^2(U)$ is compact. The proof is a two-step argument that illustrates the typical structure: $K$ maps $L^2$ boundedly into $H_0^1$ (by the Riesz theorem), and $H_0^1$ embeds compactly into $L^2$ by the following fundamental theorem. [definition:Compact Embedding] Let $X$ and $Y$ be Banach spaces with $X \subseteq Y$ and $\|u\|_Y \le C\|u\|_X$ for all $u \in X$. The embedding is **compact**, written $X \subset\subset Y$, if the inclusion map \begin{align*} i: X &\to Y \\ u &\mapsto u \end{align*} is a compact operator: every sequence bounded in $X$ has a subsequence converging strongly in $Y$. [/definition] [quotetheorem:64] [quotetheorem:213] With $K = (-\Delta)^{-1}$ compact, the eigenvalue problem $-\Delta u = \lambda u$ becomes $u = \lambda Ku$, i.e., $(I - \lambda K)u = 0$. The solvability of the inhomogeneous problem $(I - \lambda K)u = g$ is governed by the following theorem, which is the direct infinite-dimensional analogue of the finite-dimensional result that $Ax = b$ is solvable if and only if $b \perp \ker A^\top$. [quotetheorem:72] ## Spectral Theory [The Fredholm alternative](/page/The%20Fredholm%20Alternative) directly implies the key structural fact about the spectrum of $-\Delta$: its eigenvalues form a discrete sequence tending to infinity. To see why, we first record what the [spectrum of a compact operator](/theorems/220) looks like in general. [definition:Resolvent Set] Let $X$ be a complex Banach space and $T \in \mathcal{L}(X)$. The **resolvent set** of $T$ is \begin{align*} \rho(T) := \{\lambda \in \mathbb{C} : T - \lambda I \text{ is bijective}\}. \end{align*} For $\lambda \in \rho(T)$, the [Closed Graph Theorem](/theorems/217) guarantees $(T - \lambda I)^{-1} \in \mathcal{L}(X)$. [/definition] [definition:Spectrum] The **spectrum** of $T \in \mathcal{L}(X)$ is the complement of the resolvent set: \begin{align*} \sigma(T) := \mathbb{C} \setminus \rho(T). \end{align*} [/definition] [definition:Point Spectrum] The **point spectrum** $\sigma_p(T) \subseteq \sigma(T)$ is the set of eigenvalues: \begin{align*} \sigma_p(T) := \{\lambda \in \mathbb{C} : N(T - \lambda I) \ne \{0\}\}. \end{align*} In infinite dimensions the inclusion $\sigma_p(T) \subseteq \sigma(T)$ can be strict: $T - \lambda I$ may be injective but fail to be surjective. [/definition] [quotetheorem:223] Applying this to $K = (-\Delta)^{-1}$ on $L^2(U)$: the eigenvalues of $K$ are $\{1/\lambda_j\}$ where $\lambda_j$ are the Dirichlet eigenvalues of $-\Delta$. Since $1/\lambda_j \to 0$, we have $\lambda_j \to \infty$. Each eigenspace is finite-dimensional by the theorem, and the [Fredholm Alternative](/theorems/72) confirms that for $\lambda \notin \{\lambda_j\}$ the problem $-\Delta u - \lambda u = f$ has a unique solution for every $f$. The entire spectral structure of $-\Delta$ — discrete eigenvalues of finite multiplicity tending to infinity — is a consequence of the compactness of its inverse. ## Problems [problem] Let $H = \ell^2(\mathbb{N})$ and define \begin{align*} K: H &\to H \\ (x_1, x_2, x_3, \ldots) &\mapsto \left(0,\; x_1,\; \frac{x_2}{2},\; \frac{x_3}{3},\; \ldots\right). \end{align*} 1. Show that $K$ is compact. 2. Show that $N(I - K) = \{0\}$ and deduce that $I - K$ is invertible with bounded inverse. [/problem] [solution] **Part 1: Compactness.** Define the finite-rank truncations \begin{align*} K_n: H &\to H \\ (x_1, x_2, \ldots) &\mapsto \left(0,\; x_1,\; \frac{x_2}{2},\; \ldots,\; \frac{x_{n-1}}{n-1},\; 0,\; 0,\; \ldots\right). \end{align*} Each $K_n$ has finite-dimensional range and is therefore compact. For the tail, \begin{align*} \|(K - K_n)x\|_H^2 = \sum_{j=n}^\infty \frac{|x_j|^2}{j^2} \le \frac{1}{n^2}\sum_{j=n}^\infty |x_j|^2 \le \frac{1}{n^2}\|x\|_H^2, \end{align*} so $\|K - K_n\|_{\mathcal{L}(H)} \le \frac{1}{n} \to 0$. Since $\mathcal{K}(H,H)$ is closed in $\mathcal{L}(H,H)$, $K$ is compact. **Part 2: Trivial kernel.** Suppose $(I - K)u = 0$ for $u = (u_1, u_2, \ldots) \in H$. Reading componentwise: \begin{align*} u_1 = 0, \qquad u_{k+1} = \frac{u_k}{k} \quad \text{for every } k \in \mathbb{N}. \end{align*} Since $u_1 = 0$, induction gives $u_k = 0$ for every $k \in \mathbb{N}$, so $N(I - K) = \{0\}$. Case A of the [Fredholm Alternative](/theorems/72) then guarantees that $I - K$ is bijective with bounded inverse. [/solution] ## References 1. L.C. Evans, *Partial Differential Equations* (2010). 2. H. Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2010). 3. M. Reed and B. Simon, *Methods of Modern Mathematical Physics I: Functional Analysis* (1980).