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For a compact convex set $K\subset {\mathbb{R}}^n$, letThe Paley–Wiener–Schwartz theorem says that if $u$ is a compactly supported distribution with support in $K$, its Fourier--Laplace transformis entire and, for some $C,N$,Conversely, every entire function satisfying such an estimate is the transform of a distribution supported in $K$.

For the forward direction, compact support lets $u$ act on the exponential after insertion of a cutoff equal to one near $K$. Differentiation in $z$ may be passed under the pairing, proving entire analyticity. The finite-order estimate for $u$ bounds derivatives of the exponential on $K$ by a polynomial in $|z|$ times $e^{H_K(\mathrm{Im}z)}$.

Conversely, restrict the entire function $F$ to ${\mathbb{R}}^n$. Its polynomial growth defines a tempered distribution $u$ by inverse Fourier transform. If a test function is supported outside $K$, separate its compact support from $K$ by a real vector $\eta$. Shifting the Fourier inversion contour from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n+it\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{t{H}_K(\eta )}$ as $t\to \infty$, so the pairing vanishes. Hence $\mathrm{supp}u\subset K$, completing the converse.

If $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ solves $P(D)u=v$, Fourier transformation givesThe Paley–Wiener–Schwartz theorem makes $\hat{u}$ entire, so $\hat{v}/P$ is entire.

Conversely, suppose $F(z)=\hat{v}(z)/P(z)$ is entire. Polynomial division estimates away from the finitely many zeros of $P$, together with the maximum principle on fixed disks around those zeros, show that $F$ retains a Paley--Wiener--Schwartz bound, with only the polynomial exponent changed. The converse theorem therefore gives $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ with $\hat{u}=F$. Then $P(D)u=v$. Thus