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Use complex-linear distributions, so the pairing contains no complex conjugation. The space of smooth functions is $\mathcal{E}(X)=C^{\infty }(X)$, with seminormsThus $f_j\to f$ means uniform convergence of every derivative on every compact subset of $X$. A cofinal compact exhaustion with $K_j\subset \mathrm{int}{K}_{j+1}$ makes $p_j=p_{K_j,j}$ an increasing defining family. Such an exhaustion exists for every open set; for example, use bounded sets staying a positive distance from its complement and enlarge them slightly. This is a Fréchet space.

The compactly supported distribution space is the continuous dual space ${\mathcal{E}}^{\prime }(X)$. We use its weak-star topology: $u_j\to u$ precisely when $⟨u_j,f⟩\to ⟨u,f⟩$ for every $f\in \mathcal{E}(X)$. A stronger standard choice is the strong dual topology, which requires uniform convergence on bounded subsets of $\mathcal{E}(X)$; specifying the weak convention avoids conflating these definitions.

A continuous linear functional necessarily takes null sequences to zero. Conversely, suppose a linear functional $u$ fails to be continuous. For each $j$, it is unbounded on $\{f:p_j(f)\le 1\}$; otherwise scaling would give a continuity estimate. We can therefore choose $f_j$ withFor each fixed $l$, $p_l(f_j)\le p_j(f_j)\to 0$ once $j\ge l$. This contradicts the assumed null-sequence property. Hence sequential continuity characterizes the continuous dual:This is the sequential continuity criterion in a metrizable vector space applied to a linear functional.

Continuity also yields one compact $K⋐X$, an integer $m$ and a constant $C$ such that $|u(f)|\le C{p}_{K,m}(f)$. Thus $u$ has compact support in $K$ and finite order of a distribution. Conversely, a compactly supported distribution extends to $\mathcal{E}(X)$ by $u(f)=u(\eta f)$, where a cutoff function $\eta$ equals one near its support of a distribution. The local finite-order estimate makes this extension continuous and independent of $\eta$. This identifies the dual definition with compactly supported distributions.

Here is an explicit compact continuous-derivative representation of a distribution. Extend $u$ by zero to ${\mathbb{R}}^n$ using a cutoff function inside $X$, retaining finite order $m$. Put $r=m+2$ andThis locally $C^m$ function satisfies ${\partial }^{\gamma }E={\delta }_0$. Indeed, each one-dimensional factor has $r$th distributional derivative equal to the Dirac delta distribution. The convolution $F=u\ast E$ is continuous: the finite-order estimate extends $u$ to $C^m$ functions near its compact support, and translated $E$ varies continuously in their $C^m$ norms. Equivalently, mollify $E$ and use the estimate to obtain local uniform convergence of the convolved functions. Distributional differentiation gives ${\partial }^{\gamma }F=u$.

Choose $\chi \in C_c^{\infty }(X)$ equal to one near $\mathrm{supp}u$. Since $\chi u=u$, repeated Leibniz rule givesEvery coefficient function on the right is continuous and compactly supported in $X$. Thus the required representation is finite:The equality first holds on test functions and then on all smooth functions after inserting a common cutoff, so it holds in ${\mathcal{E}}^{\prime }(X)$.

A general distribution need not admit one finite such sum, even if compact support is not required of the continuous functions. On $X=\mathbb{R}$, considerThe sum is locally finite, so it defines a distribution. If it were a finite sum ${\sum }_{l=0}^M{\partial }^l{g}_l$ with all $g_l$ continuous, then on every fixed compact set its action would be bounded by test derivatives through order $M$. Near an integer $j>M$, however, it is exactly ${\delta }_j^{(j)}$, whose order is $j$. To see the contradiction directly, choose a test function $\psi$ with ${\psi }^{(j)}(0)\ne 0$ and put ${\psi }_{\epsilon }(x)={\epsilon }^j\psi ((x-j)/\epsilon )$. All derivative norms through order $M$ tend to zero, while $⟨{\delta }_j^{(j)},{\psi }_{\epsilon }⟩=(-1{)}^j{\psi }^{(j)}(0)$. This is a distribution of unbounded order. In dimensions $n>1$, the same example uses point masses at $(j,0,\dots ,0)$ and derivatives in the first coordinate.

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.

The multiplication of a distribution by a smooth function is $⟨\psi T,\phi ⟩=⟨T,\psi \phi ⟩$. If $\psi$ is a Schwartz function, the Leibniz rule shows that $\phi \mapsto \psi \phi$ is a continuous map $\mathcal{S}\to \mathcal{S}$, so $\psi T$ is a tempered distribution. When both $\psi$ and $T$ are radial, $\psi \circ R^t=\psi$, andThus Multiplication by a radial Schwartz function preserves radial tempered distributions.

For the convolution of a tempered distribution with a Schwartz function, setTranslations of $\psi$ depend smoothly on $x$ in the Schwartz space, so this is a smooth function with ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T_y,{\partial }^{\alpha }\psi (x-y)⟩$. The bound for $T$ by finitely many seminorms, together with $1+|y|\le (1+|x|)(1+|x-y|)$, provesThus the function also defines a tempered distribution. For a radial function $\psi$, put $g_x(y)=\psi (x-y)$. Then $\psi (Rx-y)=\psi (x-R^{t}y)=g_x(R^{t}y)$, soHence Convolution with a radial Schwartz function preserves radial tempered distributions.

Smoothing alone need not give a Schwartz function: for the constant tempered distribution $T=1$ and a Schwartz function with integral one, $T\ast \psi =1$. The radial Schwartz approximation of tempered distributions therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial mollifier $\rho \in C_c^{\infty }$, supported in the unit ball with integral one, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one on the unit ball. PutEach $f_j$ is a smooth function of compact support, hence a Schwartz function, and the two invariance calculations above make it radial.

It remains to prove convergence, including the simultaneous changes of both scales. With $\check{\rho }(x)=\rho (-x)$, the distributional convolution pairing isFor every fixed $m$, the Leibniz rule, rapid decay outside the radius-$j$ ball, and the chain rule for $\chi (x/j)$ giveConvolution by ${\check{\rho }}_{\epsilon }$ is uniformly bounded in $q_m$ for $0<\epsilon \le 1$, because its shifts have size at most one. The mean value theorem applied to $\phi (x-y)-\phi (x)$ similarly givesSplitting the error into the convolved cutoff error and the approximate identity error provesIf $|⟨T,\theta ⟩|\le C{q}_m(\theta )$, this yieldsConsequentlyweakly, and even in the strong dual topology, since $q_{m+1}$ is uniformly bounded on every subset of the Schwartz space that is a bounded set in a topological vector space. No assertion that $T\ast {\rho }_{1/j}$ itself is rapidly decreasing is needed.

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.

Choose a radial mollifier $\rho \in C_c^{\infty }$ with integral one and support in the unit ball, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one there. For a radial tempered distribution $T$, the functionsare radial Schwartz functions. Smoothness comes from convolution of a tempered distribution with a Schwartz function, and compact support comes from the cutoff. Pairing with $\phi$ gives $⟨f_j,\phi ⟩=⟨T,{\check{\rho }}_{1/j}\ast (\chi (\cdot /j)\phi )⟩$. With $q_m(\phi )={\max }_{|\beta |\le m}{\sup }_x(1+|x|{)}^m|{\partial }^{\beta }\phi (x)|$, the cutoff tail and the mean value theorem giveThe continuity estimate for $T$ therefore proves $f_j\to T$ even in the strong dual topology. Inserting a cutoff is essential because convolution alone need not give rapid decay.

For a locally convex space $E$, the strong dual topology on its continuous dual space $E^{\prime }$ is generated by seminorms $p_B(T)={\sup }_{\phi \in B}|⟨T,\phi ⟩|$, where $B$ runs through the bounded sets in a topological vector space $E$. Convergence therefore means uniform convergence of pairings on every such set. For the Schwartz space, the defining boundedness is boundedness of every Schwartz seminorm, rather than one norm or one common compact support.