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For $D_j=-i{\partial }_j$, the Fourier transform of a tempered distribution satisfiesThe base cases follow from $⟨D_{j}u,\phi ⟩=i⟨u,{\partial }_j\phi ⟩$ and integration by parts in the Fourier integral. Iterating gives the multi-index formulas. The $D$ convention matters: for the unscaled derivative, $\widehat{{\partial }_{j}u}=i{\xi }_j\hat{u}$.

A tempered distribution is a continuous linear functional $u:\mathcal{S}(\mathbb{R})\to \mathbb{C}$. Equivalently, for some integer $N$ and constant $C$,The Fourier transform of a tempered distribution is defined byThe Fourier bounds for Schwartz functions make the right side a continuous linear functional on the Schwartz space, so $\hat{u}$ is again a tempered distribution.

For an integrable function $f$, the regular distribution $u_f$ acts by $⟨u_f,\varphi ⟩=\int f\varphi$. The estimate $|⟨u_f,\varphi ⟩|\le \Vert f{\Vert }_1{p}_{0,0}(\varphi )$ proves it is a tempered distribution. Since $\iint |f(x)\varphi (\xi )|dxd\xi <\infty$, Fubini's theorem givesThus the distributional Fourier transform agrees with the ordinary Fourier transform.

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 $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, 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.

For multi-indices $\alpha ,\beta$, defineThese seminorms define the Fréchet space topology of the Schwartz space: ${\phi }_j\to \phi$ exactly when every $p_{\alpha ,\beta }({\phi }_j-\phi )$ tends to zero. An equivalent increasing family isA tempered distribution is a continuous linear functional on this space. Equivalently, $|⟨T,\phi ⟩|\le C{q}_m(\phi )$ for some $C,m$. The usual weak convergence of tempered distributions means $⟨T_j,\phi ⟩\to ⟨T,\phi ⟩$ for every fixed Schwartz function. The strong dual topology instead requires uniform convergence on every subset of the Schwartz space that is a bounded set in a topological vector space; the Fourier maps below are continuous in both topologies.

Using $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, differentiation under the integral and integration by parts giveThe omitted coefficient has modulus one. By the Leibniz rule, every integrand is a finite sum of a polynomial times a derivative of $\phi$. Inserting the integrable weight $(1+|x|{)}^{-n-1}$ bounds its L1 norm by finitely many Schwartz space seminorms. In particular $q_m(\hat{\phi })\le C_m{q}_{m+n+1}(\phi )$. Thus the Fourier transform maps $\mathcal{S}$ continuously into itself.

For completeness, the Fourier inversion theorem follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi {|}^2}$ is $g_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. The Fourier transform of a Gaussian and Fubini's theorem showAs $t\downarrow 0$, the right side tends to $\phi (x)$ by the approximate identity property, while the left side tends to the undamped inverse integral by dominated convergence theorem, since $\hat{\phi }\in L^1$. HenceThe inverse is $(2\pi {)}^{-n}$ times reflection composed with the continuous Fourier transform, and is therefore continuous on the Schwartz space. This proves a continuous linear isomorphism with continuous inverse.

Define the Fourier transform of a tempered distribution byThe continuous map on Schwartz space makes this a tempered distribution; the transpose of ${\mathcal{F}}^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the strong dual topology, the transform of a bounded set of Schwartz functions is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.

Now let $R\in SO(n)$ be a rotation matrix in the special orthogonal group and write ${\rho }_{R}T=T\circ R$. A change of variables with unit Jacobian givesThe rotation equivariance of the Fourier transform on tempered distributions follows by duality:Therefore $\widehat{{\rho }_{R}T}={\rho }_R\hat{T}$. Applying this identity and the inverse Fourier transform gives the two directions:This is precisely preservation of radial tempered distributions. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary radial function.

For an orthogonal matrix $R$, a unit-Jacobian change of variables gives $\widehat{\phi \circ R^t}=\hat{\phi }\circ R^t$. The dual definition of the Fourier transform of a tempered distribution consequently givesSince the Fourier transform is invertible on the Schwartz space and its dual, a tempered distribution is invariant under a rotation group exactly when its transform is invariant. This includes radial tempered distributions.