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Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, with inverse factor $(2\pi {)}^{-n}$. The Paley–Wiener–Schwartz theorem can be stated with sharp convex support: for a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let its support function be $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$. Then $F$ is the Fourier transform of a unique distribution supported in $K$ if and only if it is entire andfor some $C$ and nonnegative integer $M$. For $K={\overline{B}}_R$, this is the familiar bound $C(1+|z|{)}^M{e}^{R|\mathrm{Im}z|}$. Allowing some $R$ characterizes all compactly supported distributions. Here the extension of the Fourier transform is $F(z)=⟨u,e^{-ix\cdot z}⟩$, interpreted with a cutoff equal to one near the support of a distribution.

First suppose $\mathrm{supp}u\subset K$. Fix one smooth cutoff function equal to one near $K$. Pairing the resulting compactly supported exponential with $u$ shows that $F$ is an entire function: differentiation with respect to $z_j$ inserts $-i{x}_j$, and the power series converges in the test-function seminorms uniformly on compact subsets of ${\mathbb{C}}^n$.

To keep the exponential type exactly $H_K$, rather than that of a fixed larger neighborhood, use a shrinking-cutoff exponential-type estimate. There are cutoffs ${\chi }_{\epsilon }$ equal to one on $K+\epsilon B_1$, supported in $K+3\epsilon {\overline{B}}_1$, and satisfying $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. One construction convolves the indicator of $K+2\epsilon B_1$ with a unit-mass mollifier supported in $\epsilon B_1$. Continuity of $u$ on a fixed compact neighborhood gives a finite order of a distribution $q$ there. By the product rule,Taking $\epsilon =(1+|z|{)}^{-1}$ proves the required bound with $M=2q$, since the extra exponential factor is at most $e^3$. The value of the pairing is independent of the chosen cutoff because all cutoffs agree near $\mathrm{supp}u$. Thus the forward direction has the exact asserted support function, without an unproved estimate on derivatives restricted only to $K$.

Conversely, suppose the entire $F$ has the stated bound. Its restriction to real frequency has polynomial growth, so define a tempered distribution $u$ byThe Schwartz space decay makes this integral absolutely convergent and continuous; by Fourier inversion, $\hat{u}=F$ on real frequency. It remains to prove the support assertion by mollifier regularization for contour recovery of support.

Choose a nonnegative unit-mass mollifier $\rho$ supported in $B_1$, and setOn real frequency, $F_{\epsilon }$ decays faster than any polynomial after enough applications of integration by parts to $\hat{\rho }$; hence $h_{\epsilon }$ is smooth, by differentiation under the integral sign. More generally, for every integer $L$ there is $C_{\epsilon ,L}$ such thatTo obtain this estimate, write the transform of ${\rho }_{\epsilon }$ as the real-frequency transform of $e^{x\cdot \eta }{\rho }_{\epsilon }(x)$ and integrate by parts; each derivative introduces at most one factor of $|\eta |$.

Fix a unit vector $v$ and take $L>M+n+1$. A contour-shift proof of the Paley–Wiener–Schwartz theorem moves the inverse-transform contour to ${\mathbb{R}}^n+itv$:Here is a justification of the shift. Rotate coordinates so $v$ is the first coordinate direction, apply the Cauchy integral theorem on a rectangle in the first complex variable, and integrate over the other real variables. For fixed $t$, the vertical sides at real part $\pm A$ have an integrated bound proportional to $A^{M-L+n-1}$, and hence vanish as $A\to \infty$. The same decay bounds give absolute convergence on the horizontal sides. No contour shift of an unregularized polynomially growing integral is needed.

Positive homogeneity of the support function now givesIf $x\notin K+\epsilon {\overline{B}}_1$, the Hahn-Banach separation theorem supplies a unit vector $v$ with $x\cdot v>H_K(v)+\epsilon$. Letting $t\to \infty$ proves $h_{\epsilon }(x)=0$. Therefore $h_{\epsilon }\in C_c^{\infty }$ and $\mathrm{supp}{h}_{\epsilon }\subset K+\epsilon {\overline{B}}_1$.

Since $\hat{\rho }(\epsilon \xi )\to 1$ and its modulus is at most one on real frequency, the dominated convergence theorem in the formula for $⟨u,\phi ⟩$ gives $h_{\epsilon }\to u$ as tempered distributions, and thus as distributions. A test function supported outside $K$ has positive distance from $K$, so its pairing with $h_{\epsilon }$ is zero for all sufficiently small $\epsilon$. This proves $\mathrm{supp}u\subset K$. The Fourier transform of a compactly supported distribution constructed in the forward direction equals $F$ on ${\mathbb{R}}^n$; applying the one-variable identity theorem successively in each coordinate extends the equality to ${\mathbb{C}}^n$. Injectivity of the Fourier transform of a tempered distribution proves uniqueness and completes both directions.

For the independence application, put $g_m(z)=e^{iz\cdot y_m}{f}_m(z)$ and $r_m=1/(m+1)$. The ball version of the Paley–Wiener–Schwartz theorem supplies a nonzero distribution $v_m$ supported in ${\overline{B}}_{r_m}(0)$ with ${\hat{v}}_m=g_m$. By the Translation property of the Fourier transform,where $⟨{\tau }_{y}v,\phi ⟩=⟨v,\phi (\cdot +y)⟩$. Distinct integer points are at least distance one apart. For distinct positive indices $j,k$, the largest possible radius sum is $1/2+1/3=5/6<1$. Thus these closed balls, and hence the distribution supports, are pairwise disjoint.

If ${\sum }_{m=1}^N{c}_m{f}_m=0$, injectivity of the Fourier transform gives ${\sum }_m{c}_m{w}_m=0$. For each $j$, choose a smooth cutoff function equal to one near its ball and zero near all other balls. Multiplying the distributional identity by this cutoff isolates $c_j{w}_j=0$. Since $f_j$ is not identically zero, $w_j\ne 0$, so $c_j=0$. Consequently $f_1,\dots ,f_N$ are linearly independent over $\mathbb{C}$. This Fourier independence from disjoint distribution supports uses the quantitative exponential types to establish support separation.