Paley–Wiener–Schwartz theorem

For a compact convex set $K\subset {\mathbb{R}}^n$, the Fourier transform gives a bijection between distributions supported in $K$ and entire functions satisfyingfor some $C,N$. Here $H_K$ is the support function. In the closed-ball case $K={\overline{B}}_R$, the exponential factor is $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential to infer support in $K$ from this bound: the support function cannot distinguish a nonconvex set from its convex hull.

Nonzero compactly supported distributions with pairwise disjoint supports are linearly independent: a smooth cutoff function isolating one support isolates its coefficient. Injectivity transfers this independence to their Fourier transforms. Exponential-type bounds and the Translation property of the Fourier transform can prove the needed support separation before the distributions themselves are known explicitly.

Let $F$ be an entire function of one variable satisfying $|F(\zeta )|\le C(1+|\zeta |{)}^N{e}^{R|\mathrm{Im}\zeta |}$, and let $P$ be a nonzero polynomial. If $F/P$ is entire, it satisfies a bound of the same exponential type, with a possibly different polynomial power. Indeed, outside a sufficiently large disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^{\mathrm{deg}P}$; inside the disk the extended quotient is bounded. Combining these estimates proves the claim. The same reasoning works with the exact support function of a compact interval. Zeros of $F$ must cancel every root of $P$ with its full multiplicity.

For the forward implication, finite order of a distribution makes $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$ entire. Use a smooth cutoff in an $\epsilon$-neighborhood of the support set, with derivatives of order $j$ bounded by $C_j{\epsilon }^{-j}$. Taking $\epsilon =(1+|\zeta |{)}^{-1}$ in the finite-order estimate gives the polynomial factor and adds at most $e$ to the desired exponential bound. This shrinking cutoff recovers the exact support indicator, instead of an arbitrarily enlarged one.

For the converse, the real restriction of $F$ has polynomial growth and defines an inverse tempered distribution. If a test function $\phi$ is supported in $x\cdot \omega \ge R+\eta$ for a unit vector $\omega$ and $\eta >0$, contour shifting givesRepeated integration by parts gives, for every integer $M$,Choosing $2M>N+n$ justifies the shift by Cauchy integral theorem and bounds the pairing by $C^{\prime }(1+t{)}^{N+2M}{e}^{-\eta t}$, which tends to zero. Half-spaces of this form cover the complement of the ball, and a partition of unity proves the support inclusion. For general compact convex $K$, replace $R$ by $H_K(\omega )$ in each separating direction. Fourier inversion gives uniqueness. A related proof outline appears in Richard Melrose's distribution-theory problem set.

For an entire function of polynomial growth and exponential type $H_K$, multiply it by $\hat{\rho }(\epsilon z)$ with a unit-mass mollifier. The product becomes rapidly decreasing in real frequency, so its inverse Fourier transform has an absolutely convergent contour integral. Shifting in a separating direction forces that smooth inverse to vanish outside $K+\epsilon {\overline{B}}_1$. Letting $\epsilon$ tend to zero recovers a distribution supported in $K$.