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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 $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$, 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.

Suppose $\phi$ is continuous at $x$ and, for some $T>0$,For $0<t<T$, its convolution with the heat kernel $K_t(s)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$ is smooth and solves the heat equation. On every compact set in $\mathbb{R}\times (0,T)$, spatial or time derivatives of the heat kernel are bounded by the displayed integrable weight times a constant: their polynomial factors are absorbed by the strictly narrower Gaussian. This justifies differentiation under the integral sign.

Moreover, $\int K_t=1$, soInside $|\xi -x|<\delta$, continuity makes the difference uniformly small. Outside that interval, for $t\le T/4$, the inequality ${\xi }^2\le 2(\xi -x{)}^2+2x^2$ givesThe right side tends to zero, proving that the weighted tail contribution vanishes; the constant term $\phi (x)$ is controlled by the vanishing Gaussian tail mass. Thus $(K_t\ast \phi )(x)\to \phi (x)$ as $t\downarrow 0$. In particular, continuous data of polynomial growth satisfy this hypothesis, even when they are unbounded. This extends the Gaussian approximate identity argument beyond bounded data.

Use the primitive real-analytic Eisenstein series with exactly the normalization in the PDF. Its coset expression uses ${\Gamma }_{\infty }=\{\pm \left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right):n\in \mathbb{Z}\}$, so primitive bottom rows differing by sign represent the same coset. The factor $1/2$ in the primitive-pair sum accounts for this sign identification. Unlike the sum over all nonzero integer pairs, this Eisenstein series has no additional factor $2\zeta (2w)$. Write $w$ for the Rankin–Selberg integral parameter and $s$ for the Dirichlet series parameter to keep their shift visible.

First establish an initial region where all rearrangements are absolute. The invariant norm of a modular form $y^{k/2}|f(\tau )|$ is bounded: it is $\Gamma$-invariant, continuous on a truncated fundamental domain, and tends to zero at the cusp. Integrating one period of the Fourier series givesWith $y=1/n$, this proves the Fourier coefficient bound for a cusp form $a_n=O(n^{k/2})$, and similarly $b_n=O(n^{k/2})$. Hence $D(f,g,s)$ is initially absolutely convergent for $\mathrm{Re}s>k+1$.

For $\mathrm{Re}w>1$, the primitive real-analytic Eisenstein series is locally uniformly convergent and holomorphic in $w$. On the standard fundamental domain it has at most polynomial growth in $y$, locally uniformly in this half-plane. These are the analytic properties of the Eisenstein series needed here; they can also be checked directly from its defining sum. For example, for real $\sigma >1$ and $y\ge 1$, majorize the primitive sum by the sum over all nonzero pairs and useThis bound follows by comparing a shifted one-dimensional sum with its integral and a largest summand. Summing over $m\ne 0$ gives $E(\tau ,\sigma )=O(y^{\sigma }+y^{1-\sigma })$. For complex $w$, $|E(\tau ,w)|\le E(\tau ,\mathrm{Re}w)$, and the estimates are uniform on compact parameter sets. Since $f$ and $g$ decay exponentially on the cusp end, the integral $I(f,g,w)$ is therefore holomorphic throughout $\mathrm{Re}w>1$ by dominated integration.

The product $f(\tau )\overline{g(\tau )}{y}^k$ is $\Gamma$-invariant: the two modular factors cancel the transformation $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$. With the hyperbolic area $d\mu =dxdy/y^2$, Rankin–Selberg unfolding gives, initially for $\mathrm{Re}w>2$,Absolute unfolding is justified by replacing $f\overline{g}$ by $|fg|$ and using the same finite fundamental-domain integral with $E(\tau ,\mathrm{Re}w)$. The subsequent termwise integrations can be justified particularly economically by Cauchy-Schwarz inequality: on each horizontal line,The coefficient bounds make the right side $O(y^{-k-1})$ as $y\downarrow 0$, and it decays exponentially as $y\to \infty$. Multiplication by $y^{\mathrm{Re}w+k-2}$ is integrable at zero when $\mathrm{Re}w>2$. Thus the integral of the inner Fourier series may be evaluated first. Orthogonality of the exponentials givesThe Mellin transform of each exponential isIn the same initial region the sum of the absolute values of these integrals is finite by $a_n\overline{b_n}=O(n^k)$. Consequently the exact Rankin–Selberg integral identity isIn particular the two parameters are not equal: the Dirichlet series argument is shifted by $k-1$.

The holomorphy of $I$ on $\mathrm{Re}w>1$ now gives the required continuation:The right side agrees with the original Dirichlet series on $\mathrm{Re}s>k+1$ and is holomorphic on the larger half-plane. No continuation of the Eisenstein series beyond $\mathrm{Re}w>1$ is needed.

It is important to supply a further argument for absolute convergence: analytic continuation alone does not imply it. For real $w>1$, both $E(\tau ,w)$ and $|f(\tau ){|}^2$ are nonnegative, and their fundamental-domain integral is finite by the cusp estimates above. Tonelli theorem therefore unfolds $I(f,f,w)$ even before knowing convergence of the Dirichlet series. On each fixed horizontal line, the Fourier series converges absolutely and uniformly, so integration in $x$ givesA second use of Tonelli theorem, now in $y$, proves the positive unfolding of a cusp-form square:Since every real $s>k$ has $s=w+k-1$ with $w>1$, we have absolute convergence of $D(f,f,s)$ for $\mathrm{Re}s>k$. Absolute convergence for complex $s$ follows by taking its real part. In fact Cauchy-Schwarz inequality also gives absolute convergence of $D(f,g,s)$ there from the two corresponding square sums.

Finally let $\sigma >(k+1)/2$ and choose $k<\beta <2\sigma -1$. The preceding square-coefficient bound and Cauchy-Schwarz inequality giveThis proves absolute convergence of cusp-form L-series from square coefficients, and henceEquivalently, the suggested elementary inequality works by choosing $k-\sigma <\alpha <\sigma -1$: after multiplication by $n^{-\sigma }$ its two bounds sum to a convergent $\sum n^{-(\sigma -\alpha )}$ and a convergent $\sum |a_n{|}^2{n}^{-(\sigma +\alpha )}$. The existence of this interval is precisely $2\sigma >k+1$.

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.

We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The Schwartz space consists of smooth functions with finite seminormsfor every pair of multi-indices. Convergence means convergence in each seminorm. The tempered distribution space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual space, with weak convergence tested against every Schwartz function. A continuous functional satisfies a bound by finitely many of these seminorms, equivalently by one sufficiently large weighted derivative seminorm.

Use the Fourier transform conventionDifferentiation under the integral and integration by parts express ${\xi }^{\alpha }{\partial }_{\xi }^{\beta }\hat{f}$ as a constant of modulus one times the Fourier transform of ${\partial }_x^{\alpha }(x^{\beta }f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz seminorms, using the integrable weight $(1+|x|{)}^{-s}$. Thus $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

For completeness, Fourier inversion follows by inserting $e^{-\epsilon |\xi {|}^2}$ in the inverse integral and using Fubini's theorem. The result is convolution with the Gaussian approximate identityIt tends to $f$, while integrability of $\hat{f}$ allows the damping factor to be removed by dominated convergence theorem. Consequently ${\mathcal{F}}^{2}f=(2\pi {)}^{n}f(-\cdot )$. Reflection preserves every Schwartz seminorm, so the inverse transform is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the Fourier transform of a tempered distribution by transposition,The Schwartz-space continuity just proved makes this a tempered distribution. Its inverse is $(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$, where $⟨\mathcal{R}u,\psi ⟩=⟨u,\psi (-\cdot )⟩$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the strong dual topology, because the Schwartz-space maps take bounded sets to bounded sets.

The convolution of a tempered distribution with a Schwartz function isSmooth dependence of translated Schwartz functions gives ${\partial }^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$. The finite-seminorm estimate and $1+|y|\le (1+|x|)(1+|x-y|)$ show that each derivative has at most polynomial growth. In particular, $u\ast \phi$ is a smooth function defining a tempered distribution. It need not itself be a Schwartz function; for example $1\ast \phi =\int \phi$.

Writing $\check{\phi }(y)=\phi (-y)$, its distributional pairing is $⟨u\ast \phi ,\psi ⟩=⟨u,\check{\phi }\ast \psi ⟩$. The inner convolution is a Schwartz function, and this identity follows by integration in the Schwartz topology, justified by the weighted seminorm estimates. A direct Fubini's theorem calculation gives $\check{\phi }\ast \hat{\psi }=\mathcal{F}(\hat{\phi }\psi )$. HenceMultiplication is well defined because multiplication by $\hat{\phi }$ acts continuously on $\mathcal{S}$.

Now write the Hilbert transform as convolution with $K={\pi }^{-1}\mathrm{pv}(1/x)$, the principal-value reciprocal distribution. The given Heaviside function transform, together with ${\mathcal{F}}^{2}H=2\pi H(-\cdot )$, yieldsThe value of the sign function at zero is irrelevant to its regular distribution. Thus the Hilbert-transform Fourier multiplier isApplying Plancherel theorem, whose normalization here is $\Vert f{\Vert }_2^2=(2\pi {)}^{-1}\Vert \hat{f}{\Vert }_2^2$, proves the isometry:The principal-value integral agrees with this convolution: near its singular point subtract $\phi (x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The Hilbert transform has domain $\mathcal{S}$, but generally does not take values in $\mathcal{S}$, as the tail in part (c) demonstrates.

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

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.