= Solution
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 $N\geq0$,
$$
\boxed{|F(\zeta)|\leq C(1+|\zeta|)^N e^{R|\operatorname{Im}\zeta|}.}
$$
The distribution is unique by <Fourier inversion>. More generally a compact convex support set $K$ replaces $R|\operatorname{Im}\zeta|$ by its <support function> $H_K(\operatorname{Im}\zeta)$.
For the forward implication, a <compactly supported distribution> has finite <order of a distribution>, say $m$, on a fixed neighborhood of its support. Define
$$
F(\zeta)=\langle u,e^{-ix\cdot\zeta}\rangle,
$$
using a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives $\partial_{\zeta_j}F=\langle u,-ix_je^{-ix\cdot\zeta}\rangle$, so $F$ is entire.
To obtain the exact radius $R$, rather than an enlarged radius, choose a <cutoff function> $\chi_\varepsilon$ equal to one near $\overline B_R$ and supported in $B_{R+\varepsilon}$, with derivatives through order $m$ bounded by $C_\gamma\varepsilon^{-|\gamma|}$. Set $\varepsilon=(1+|\zeta|)^{-1}$. The finite-order bound and the <Leibniz rule> give
$$
|F(\zeta)|\leq C(1+|\zeta|)^m e^{(R+\varepsilon)|\operatorname{Im}\zeta|}
\leq Ce(1+|\zeta|)^m e^{R|\operatorname{Im}\zeta|}.
$$
This proves the required growth estimate.
For the converse, $F$ has at most <polynomial growth> on real frequencies, so define its inverse <tempered distribution> by
$$
\langle u,\varphi\rangle=(2\pi)^{-n}\int_{\mathbb R^n}F(\xi)\widehat\varphi(-\xi)\,d\xi.
$$
Let a compactly supported test function lie in a half-space $x\cdot\omega\geq R+\eta$, with $|\omega|=1$ and $\eta>0$. By <contour shifting>,
$$
\langle u,\varphi\rangle=(2\pi)^{-n}\int_{\mathbb R^n}
F(\xi+it\omega)\widehat\varphi(-\xi-it\omega)\,d\xi,\qquad t\geq0.
$$
Here is the decay needed to justify the shift and then let $t$ grow. Repeated <integration by parts>, applying $1-\Delta_x$ to $e^{-t x\cdot\omega}\varphi(x)$, gives for every integer $M$,
$$
|\widehat\varphi(-\xi-it\omega)|
\leq C_M(1+t)^{2M}e^{-(R+\eta)t}(1+|\xi|^2)^{-M}.
$$
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 yields
$$
|\langle u,\varphi\rangle|
\leq C'(1+t)^{N+2M}e^{-\eta t}\longrightarrow0.
$$
Thus $u$ vanishes on such test functions. Every point outside $\overline B_R$ has a neighborhood of this form. A <partition of unity> therefore proves $\operatorname{supp}u\subseteq\overline B_R$, 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> $y=tx$ gives
$$
\int u(tx)\varphi(x)\,dx=t^{-n}\int u(y)\varphi(y/t)\,dy.
$$
The map $\varphi\mapsto\varphi(\mathord\cdot/t)$ is continuous on the <Schwartz space>, so duality extends the <dilation of a distribution> to
$$
\boxed{\langle\delta_tu,\varphi\rangle=t^{-n}\langle u,\delta_{1/t}\varphi\rangle.}
$$
Its Fourier transform obeys $\widehat{\delta_tu}(\zeta)=t^{-n}F(\zeta/t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta|)^Ne^{|\operatorname{Im}\zeta|}$. Hence
$$
|\widehat{\delta_tu}(\zeta)|\leq C_t(1+|\zeta|)^N e^{|\operatorname{Im}\zeta|/t}.
$$
The converse theorem gives the <support law for distributional dilation>
$$
\boxed{\operatorname{supp}(\delta_tu)\subseteq\{x:|x|\leq1/t\}=\{x:|tx|\leq1\}.}
$$
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