= Solution
For a compact convex set $K\subset\mathbb R^n$, let
$$
H_K(\eta)=\sup_{x\in K}x\mathbin\cdot\eta.
$$
The <Paley–Wiener–Schwartz theorem> says that if $u$ is a <compactly supported distribution> with support in $K$, its Fourier--Laplace transform
$$
\widehat u(z)=\langle u(x),e^{-ix\cdot z}\rangle
$$
is entire and, for some $C,N$,
$$
|\widehat u(z)|\leq C(1+|z|)^N
e^{H_K(\operatorname{Im}z)}.
$$
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 $e^{H_K(\operatorname{Im}z)}$.
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 $\mathbb R^n+i t\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{tH_K(\eta)}$ as $t\to\infty$, so the pairing vanishes. Hence $\operatorname{supp}u\subset K$, completing the converse.
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