= Solution
Use the <Fourier transform> convention $\widehat\varphi(\xi)=\int e^{-ix\cdot\xi}\varphi(x)\,dx$, 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 \b[$F$ is the <Fourier transform> of a unique <distribution> supported in $K$ if and only if it is entire and]
$$
\boxed{|F(z)|\leq C(1+|z|)^M e^{H_K(\operatorname{Im}z)}\quad(z\in\mathbb C^n)}
$$
for some $C$ and nonnegative integer $M$. For $K=\overline B_R$, this is the familiar bound $C(1+|z|)^M e^{R|\operatorname{Im}z|}$. Allowing some $R$ characterizes all <compactly supported distributions>. Here the extension of the <Fourier transform> is $F(z)=\langle u,e^{-ix\cdot z}\rangle$, interpreted with a cutoff equal to one near the <support of a distribution>.
First suppose $\operatorname{supp}u\subset K$. 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 $-ix_j$, 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 $\chi_\varepsilon$ equal to one on $K+\varepsilon B_1$, supported in $K+3\varepsilon\overline B_1$, and satisfying $|\partial^\alpha\chi_\varepsilon|\leq C_\alpha\varepsilon^{-|\alpha|}$ for $0<\varepsilon\leq1$. One construction convolves the indicator of $K+2\varepsilon B_1$ with a unit-mass <mollifier> supported in $\varepsilon B_1$. Continuity of $u$ on a fixed compact neighborhood gives a finite <order of a distribution> $q$ there. By the product rule,
$$
|F(z)|=|\langle u,\chi_\varepsilon e^{-ix\cdot z}\rangle|
\leq C\varepsilon^{-q}(1+|z|)^q
\exp\big(H_K(\operatorname{Im}z)+3\varepsilon|\operatorname{Im}z|\big).
$$
Taking $\varepsilon=(1+|z|)^{-1}$ 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 $\operatorname{supp}u$. 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> $u$ by
$$
\langle u,\varphi\rangle=(2\pi)^{-n}\int_{\mathbb R^n}F(\xi)\widehat\varphi(-\xi)\,d\xi,
\qquad\varphi\in\mathcal S(\mathbb R^n).
$$
The <Schwartz space> decay makes this integral absolutely convergent and continuous; by <Fourier inversion>, $\widehat u=F$ 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 set
$$
F_\varepsilon(z)=F(z)\widehat\rho(\varepsilon z),\qquad
h_\varepsilon(x)=(2\pi)^{-n}\int_{\mathbb R^n}e^{ix\cdot\xi}F_\varepsilon(\xi)\,d\xi.
$$
On real frequency, $F_\varepsilon$ decays faster than any polynomial after enough applications of <integration by parts> to $\widehat\rho$; hence $h_\varepsilon$ is smooth, by <differentiation under the integral sign>. More generally, for every integer $L$ there is $C_{\varepsilon,L}$ such that
$$
|\widehat\rho(\varepsilon(\xi+i\eta))|
\leq C_{\varepsilon,L}(1+|\xi|)^{-L}(1+|\eta|)^L e^{\varepsilon|\eta|}.
$$
To obtain this estimate, write the transform of $\rho_\varepsilon$ as the real-frequency transform of $e^{x\cdot\eta}\rho_\varepsilon(x)$ 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 $\mathbb R^n+itv$:
$$
h_\varepsilon(x)=(2\pi)^{-n}\int_{\mathbb R^n}e^{ix\cdot(\xi+itv)}F_\varepsilon(\xi+itv)\,d\xi.
$$
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 gives
$$
|h_\varepsilon(x)|\leq C_{\varepsilon,L}(1+t)^{M+L}
\exp\big(-t[x\cdot v-H_K(v)-\varepsilon]\big).
$$
If $x\notin K+\varepsilon\overline B_1$, the <Hahn-Banach separation theorem> supplies a unit vector $v$ with $x\cdot v>H_K(v)+\varepsilon$. Letting $t\to\infty$ proves $h_\varepsilon(x)=0$. Therefore $h_\varepsilon\in C_c^\infty$ and $\operatorname{supp}h_\varepsilon\subset K+\varepsilon\overline B_1$.
Since $\widehat\rho(\varepsilon\xi)\to1$ and its modulus is at most one on real frequency, the <dominated convergence theorem> in the formula for $\langle u,\varphi\rangle$ gives $h_\varepsilon\to u$ as <tempered distributions>, and thus as <distributions>. A <test function> supported outside $K$ has positive distance from $K$, so its pairing with $h_\varepsilon$ is zero for all sufficiently small $\varepsilon$. This proves $\operatorname{supp}u\subset K$. 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 $\widehat v_m=g_m$. By the <Translation property of the Fourier transform>,
$$
f_m(z)=e^{-iz\cdot y_m}\widehat v_m(z)=\widehat w_m(z),\qquad
w_m=\tau_{y_m}v_m,\qquad
\operatorname{supp}w_m\subset\overline B_{r_m}(y_m),
$$
where $\langle\tau_y v,\varphi\rangle=\langle v,\varphi(\,\cdot+y)\rangle$. 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 $\sum_{m=1}^N c_mf_m=0$, injectivity of the <Fourier transform> gives $\sum_m c_mw_m=0$. 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 $c_jw_j=0$. Since $f_j$ is not identically zero, $w_j\ne0$, so $c_j=0$. Consequently \b[$f_1,\ldots,f_N$ are linearly independent over $\mathbb C$]. This <Fourier independence from disjoint distribution supports> uses the quantitative exponential types to establish support separation.
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