Functional analysis Distributions and generalized functions Theorem

Schwartz kernel theorem

Builds on Nuclear space Dirac delta function

Statement

Let $X \subseteq \R^m$ and $Y \subseteq \R^n$ be open sets. Let $\mathcal{D}(X)$ be the space of test functions on $X$ , and let $\mathcal{D}'(X)$ be the space of distributions on $X$ . For $\varphi \in \mathcal{D}(X)$ and $\psi \in \mathcal{D}(Y)$ , let $(\varphi \otimes \psi)(x, y) = \varphi(x) \psi(y)$ .

Each $K \in \mathcal{D}'(X \times Y)$ defines a linear map $A_K \colon \mathcal{D}(Y) \to \mathcal{D}'(X)$ by $\langle A_K \psi, \varphi \rangle = \langle K, \varphi \otimes \psi \rangle$ . It is sequentially continuous: if $\psi_j \to 0$ in $\mathcal{D}(Y)$ , then $\langle A_K \psi_j, \varphi \rangle \to 0$ for every $\varphi \in \mathcal{D}(X)$ .
Conversely, for each linear map $A \colon \mathcal{D}(Y) \to \mathcal{D}'(X)$ that is sequentially continuous in this sense, there is exactly one $K \in \mathcal{D}'(X \times Y)$ with $A = A_K$ .

The distribution $K$ is the Schwartz kernel of $A$ .

Remarks

The hard part is the existence of $K$ in part 2. One proof expands test functions in Fourier series on large cubes and controls the coefficients. Another uses that the spaces of test functions are nuclear spaces, so that continuous bilinear forms on $\mathcal{D}(X) \times \mathcal{D}(Y)$ extend to $\mathcal{D}(X \times Y)$ .

Example: the identity map of $\mathcal{D}(\R)$ , viewed as a map into $\mathcal{D}'(\R)$ , has kernel $\delta(x - y)$ . Indeed $\langle \delta(x - y), \varphi \otimes \psi \rangle = \int \varphi(x) \psi(x)\, dx$ . The operator $\psi \mapsto \psi'$ has kernel $\delta'(x - y)$ , the derivative in $x$ of the Dirac delta on the diagonal.

Going deeper

For a student

The theorem lets one describe operators such as differential operators, pseudodifferential operators and Fourier integral operators through their kernels. Properties of the operator then become properties of $K$ , for example the singular support or the wave front set of $K$ . Function kernels are not enough: the identity operator of $L^2(\R)$ is not an integral operator with a function kernel, so one really needs distributions.

Reading path

To understand this page from the ground up, read these 6 pages in order. The foundations come first.

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Tag	`2ZKWZ0A4MT`
URI	`m2://2ZKWZ0A4MT`
Kind	statement, theorem
Writing	W0 stub
MSC	46F05, 46A11
Status	This copy of the site has no build data. Run `m2 status 2ZKWZ0A4MT`.
Uses	Nuclear space, Dirac delta function
Same as	Wikidata Q1276318 Wikipedia Schwartz kernel theorem
Changed	2 November 2026