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Use $D=-i{\partial }_x$. For a nonzero one-variable polynomial $P$ and $v\in {\mathcal{E}}^{\prime }(\mathbb{R})$, a compactly supported solution to $P(D)u=v$ exists exactly when $v$ annihilates every smooth solution of the transposed equation $P(-D)\phi =0$. For simple roots ${\alpha }_j$, these test solutions span the exponentials $e^{-i{\alpha }_{j}x}$, so the condition is $\hat{v}({\alpha }_j)=0$. Dividing the entire transform by $P$ and applying polynomial division preservation of exponential type and the Paley–Wiener–Schwartz theorem constructs the compactly supported solution. It is unique because an entire function killed by a nonzero polynomial must vanish identically. For a root of multiplicity $r$, the conditions become ${\hat{v}}^{(k)}(\alpha )=0$ for $0\le k<r$, corresponding to $x^k{e}^{-i\alpha x}$.

For real wave speed $c$ and $t\ge 0$, define the complex-frequency cosine through the entire series $C_t(z)={\sum }_{j\ge 0}(-1{)}^j(ct{)}^{2j}(z\cdot z{)}^j/(2j)!$. The expression is independent of the square-root branch even though the square root itself need not be entire. Writing $z=a+ib$ gives $|\mathrm{Im}\sqrt{z\cdot z}|\le |b|$, and hence $|C_t(z)|\le e^{|c|t|\mathrm{Im}z|}$. The Paley–Wiener–Schwartz theorem therefore makes its inverse Fourier transform a distribution supported in the radius-$|c|t$ ball. Multiplication by $C_t$ evolves initial displacement with zero initial velocity in the wave equation, giving finite propagation speed for compactly supported initial distributions.

For $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$, its Fourier transform is the smooth functionIt has at most polynomial growth, and its extension to complex frequency is controlled more precisely by the Paley–Wiener–Schwartz theorem.

Paley–Wiener theorems relate spatial support or decay to analytic continuation and growth of the Fourier transform. Different versions apply to square-integrable functions or to distributions. The Paley–Wiener–Schwartz theorem is the distributional version.

For $u\in {\mathcal{E}}^{\prime }$, choose a cutoff function $\chi =1$ on a neighborhood of its support. Its extension to smooth functions makesindependent of the choice of cutoff. For real frequency, the definition of the distributional Fourier transform gives $F(\lambda )=\hat{u}(\lambda )$: interchange the pairing with the integral of a Schwartz function, using the finite-order estimate on the cutoff's compact support. On each compact set of complex frequencies the exponential and all its $x$ derivatives have uniformly convergent power series. Continuity of the distribution allows termwise differentiation, withConsequently $F$ is entire on ${\mathbb{C}}^n$.

For the precise exponential type, a fixed enlarged support would give an unnecessarily enlarged radius. Instead use the shrinking-cutoff exponential-type estimate. On one fixed compact neighborhood of the closed radius-$\delta$ ball, $u$ has finite order of a distribution $m$. Choose ${\chi }_{\epsilon }=1$ near that ball, supported in the radius-$(\delta +\epsilon )$ ball, with $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1/2$. The finite-order estimate givesTake $\epsilon =[2(1+|z|){]}^{-1}$. Its extra exponential is bounded by $e^{1/2}$, while the derivative cost is polynomial. ThusThe entire function was defined with a fixed cutoff; only its bound uses a frequency-dependent cutoff, so no holomorphic dependence is lost. This proves the required estimate with some $N\ge 0$ without enlarging $\delta$.

For the converse, set $V(z)=e^{iz\cdot y}U(z)$. Its restriction to real frequencies has polynomial growth. Define the inverse tempered distribution $v$ byRapid decay of the Schwartz function transform proves convergence and continuity, and Fourier inversion gives $\hat{v}=V$ on real frequencies.

We prove its support directly by contour shifting, rather than assuming the support conclusion of the Paley–Wiener–Schwartz theorem. Let $\omega$ be a real unit vector and take a test function $\varphi$ supported in $x\cdot \omega \ge \delta +ϵ$ for some $ϵ>0$. For every integer $M\ge 0$, integration by parts in the real-frequency Fourier integral givesIndeed, move $(1-{\Delta }_x{)}^M$ from the oscillatory exponential to $e^{-t\omega \cdot x}\varphi$; its derivatives supply at most $2M$ powers of $t$.

The product $V(z)\hat{\varphi }(-z)$ is entire. Rotate coordinates so that $\omega$ is the first coordinate vector and apply Cauchy integral theorem to a rectangle in that one complex coordinate, integrating the remaining real coordinates afterwards. For fixed $t$, choose $2M>N+n+2$; the displayed decay estimate, uniformly on the intervening imaginary segment, makes the vertical faces vanish as the real rectangle width tends to infinity. The horizontal integrals are absolutely convergent. HenceUsing $(1+|\xi +it\omega |{)}^N\le C(1+t{)}^N(1+|\xi |{)}^N$ and the growth hypothesis bounds this pairing byIt tends to zero as $t\to \infty$, so the pairing vanishes. Every point outside the closed radius-$\delta$ ball lies in such a separating half-space; a finite partition of unity for the support of a test function outside the ball proves $\mathrm{supp}v\subset \overline{B}(0,\delta )$.

Translate this compactly supported distribution by $y$: define $⟨u,\varphi ⟩=⟨v,\varphi (\cdot +y)⟩$. The Translation property of the Fourier transform givesThe compact-support entire extension agrees with $U$ everywhere by the identity theorem, applied successively in the complex coordinates. Fourier inversion also gives uniqueness. This completes both directions of the ball version of the Paley–Wiener–Schwartz theorem.

For the wave equation, take a real constant $c\ne 0$. The Fourier transform method for the wave equation givesThis inverse tempered distribution is twice differentiable in $t$: time derivatives introduce polynomial frequency factors, still integrable against every Schwartz function. It has the specified initial displacement and zero initial velocity and satisfies the equation distributionally.

To apply the support theorem, replace the real norm by the entire wave cosine multiplierThis series is entire and equals $\cos (ct\sqrt{z\cdot z})$ regardless of the square-root choice. Write $z=a+ib$ and $q=\sqrt{z\cdot z}$. Since $|z\cdot z|\le |a{|}^2+|b{|}^2$,Together with $|\cos w|\le e^{|\mathrm{Im}w|}$, this yields $|C_t(z)|\le e^{|c|t|\mathrm{Im}z|}$. The forward estimate for the translated initial support now givesThe converse therefore proves finite propagation with the stated speed:In particular, $\sqrt{z\cdot z}$ itself need not be entire; it is the even cosine series that provides the required entire extension.

For completeness this construction identifies the distributional Cauchy solution even without initially assuming spatial temperedness. The zero-displacement wave multiplier is $S_{\tau }(z)={\int }_0^{\tau }{C}_s(z)ds$, entire with bound $|S_{\tau }(z)|\le \tau e^{|c|\tau |\mathrm{Im}z|}$. Given a compactly supported smooth $\varphi$ and final time $T$, the backward solution $w(s)={\mathcal{F}}^{-1}[S_{T-s}\hat{\varphi }]$ is smooth, has support in one compact ball for $0\le s\le T$ by the same support theorem, and satisfies $w(T)=0$, $w_s(T)=-\varphi$. For the difference $h$ of two distributional solutions with zero initial data,All pairings use compactly supported test functions. The expression vanishes initially and equals $⟨h(T),\varphi ⟩$ finally. Hence $h(T)=0$, establishing uniqueness in the usual time-differentiable distributional solution class.

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.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

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 $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,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 yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, 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 $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear partial differential operator $P(D)$ on ${\mathbb{R}}^n$ has a fundamental solution of a linear differential operator: there is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ such that $P(D)E={\delta }_0$.

Write $D=-i\partial$. After an orthogonal change of coordinates and multiplication by a nonzero constant, its polynomial symbol may be written as a monic polynomial in the last frequency,For each real ${\mu }^{\prime }$, this polynomial has $M$ complex roots counted with multiplicity. Among a fixed finite collection of horizontal lines at bounded heights, one can choose a line that stays a positive distance from all those roots. Continuity of the roots preserves the choice on a neighborhood $N({\mu }^{\prime })$. Take a countable locally finite cover by such neighborhoods, refine it to a measurable disjoint partition ${\mathbb{R}}^{n-1}={⨆}_j{\Delta }_j$, and let $c_j$ be the chosen height on ${\Delta }_j$. The resulting Hörmander staircasehas bounded heights and may be chosen so that $|P({\xi }^{\prime },s+i{c}_j)|\ge 1$ on each step.

For a test function $\phi$, defineThe Paley–Wiener–Schwartz theorem gives rapid decay in the real frequency directions and at most a fixed exponential factor in the bounded imaginary direction. Together with $|P|\ge 1$, this proves that the integral defines a continuous distribution. Applying $P(D)$ cancels the denominator. The remaining integrand is entire in $z$, so the Cauchy integral theorem shifts every horizontal contour to the real axis; the partition then recombines into ${\mathbb{R}}^{n-1}$. The Fourier inversion theorem giveswhich proves the theorem.

For a compact convex set $K\subset {\mathbb{R}}^n$, letThe Paley–Wiener–Schwartz theorem says that if $u$ is a compactly supported distribution with support in $K$, its Fourier--Laplace transformis entire and, for some $C,N$,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 $e^{H_K(\mathrm{Im}z)}$.

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 ${\mathbb{R}}^n+it\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{t{H}_K(\eta )}$ as $t\to \infty$, so the pairing vanishes. Hence $\mathrm{supp}u\subset K$, completing the converse.

If $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ solves $P(D)u=v$, Fourier transformation givesThe Paley–Wiener–Schwartz theorem makes $\hat{u}$ entire, so $\hat{v}/P$ is entire.

Conversely, suppose $F(z)=\hat{v}(z)/P(z)$ is entire. Polynomial division estimates away from the finitely many zeros of $P$, together with the maximum principle on fixed disks around those zeros, show that $F$ retains a Paley--Wiener--Schwartz bound, with only the polynomial exponent changed. The converse theorem therefore gives $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ with $\hat{u}=F$. Then $P(D)u=v$. Thus

For $t>0$, the support of a distribution satisfiesThe duality formula $⟨{\delta }_{t}u,\phi ⟩=t^{-n}⟨u,\phi (\cdot /t)⟩$ proves one inclusion by transporting test functions; applying the inverse dilation proves the other. For a compact support, the Paley–Wiener–Schwartz theorem gives another proof: $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}\hat{u}(\zeta /t)$ rescales the exponential growth indicator by $1/t$.