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The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

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.

True. The Fourier transform of a compactly supported distribution is a smooth function of at most polynomial growth. A sufficiently negative Sobolev weight makes its square integrable, so every $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$ belongs to $H^t({\mathbb{R}}^n)$ for some $t$.

Let $u$ be a compactly supported distribution. It has some finite order $m$. Choose $k$ so large that the Bessel potential kernel $G_k={\mathcal{F}}^{-1}(1+|\xi {|}^2{)}^{-k}$ has enough continuous derivatives forto be bounded and continuous. Compact support of $u$ makes boundedness uniform under translation. Since $(1-\Delta {)}^k{G}_k={\delta }_0$ distributionally,Expanding each power of $\Delta$ expresses $u$ as a finite sum of derivatives of the bounded continuous function $f$. This proves the structure theorem for compactly supported distributions.