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**: