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Use complex-linear distribution pairings, without conjugation. Write $\mathcal{D}=C_c^{\infty }({\mathbb{R}}^n)$ for the space of test functions. The smoothing convolution with a test function isOn a compact set of $x$ values, all the translated test functions have support in one compact set. Continuity of the distribution therefore permits differentiation in $x$, giving ${\partial }_x^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$ for every multi-index. In particular this convolution is a smooth function, even if $u$ is not tempered.

For the first associativity identity, integration against $\psi$ and the distribution pairing can be interchanged: the integrand has a common compact support in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. ConsequentlyThe common-support argument matters: a general distribution cannot be paired with arbitrary noncompact functions.

For convolution of distributions with a compactly supported factor, first take $v$ with compact support $K$ and defineThe inner pairing is interpreted using a cutoff function equal to one near $K$. It is smooth in $x$ and has support in $\mathrm{supp}\chi -K$. To see continuity, restrict $\chi$ to a fixed compact support $A$. The order of a distribution estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ givesThus $w$ is a distribution, not merely a formal iterated pairing.

Choose an additional cutoff function in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the tensor product of distributions permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives $u\ast v=v\ast u$. If $u$ rather than $v$ has compact support, use the same construction with the roles reversed; pairing $u$ against a smooth function is then legitimate.

Evaluating the resulting smoothing convolution with a test function givesIf $v$ has compact support, $v\ast \phi$ is itself a test function; if $u$ has compact support, its action on the smooth inner convolution uses a cutoff. This explains the meaning of the formula in either case. It also proves uniqueness: $(w\ast \phi )(0)=⟨w,\phi (-\cdot )⟩$, and reflection runs through all test functions. When both factors have compact support, the same definition gives $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$, their Minkowski sum.

The Schwartz space consists of smooth functions for which every seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }\varphi (x)|$ is finite. A tempered distribution is a continuous linear functional on this space. Fix the angular-frequency Fourier transform conventionThe Fourier transform isomorphism of the Schwartz space makes the dual definition continuous. For a compactly supported distribution, a fixed cutoff function $\chi =1$ near its support extends the action to smooth functions by $⟨u,g⟩=⟨u,\chi g⟩$. A finite-order estimate controls this by finitely many Schwartz space seminorms, so both compactly supported factors are tempered.

Their Fourier transform of a compactly supported distribution is the smooth function $\hat{u}(\lambda )=⟨u,e^{-i\lambda \cdot x}⟩$. Applying the compact-support convolution definition to the exponential givesThe convolution theorem has no extra factor with this normalization.

For the spherical surface measure convolution, put $k=|\lambda |$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area givesAt $k=0$ the removable value is $4\pi a^2$, the total sphere area. Hence $\widehat{u_a\ast u_b}(\lambda )=16{\pi }^{2}ab\sin (ak)\sin (bk)/k^2$.

For $r=|x|>0$, angular integration in Fourier inversion now givesThis conditional integral can be made rigorous by first inserting $e^{-\epsilon k}$ and then taking $\epsilon \downarrow 0$ in tempered distributions. The supplied sine identity gives an integral of $\pi /4$ when $|a-b|<r<a+b$, and zero off that interval. Thusas a regular distribution. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use ${\int }_0^{\infty }{e}^{-\epsilon k}\sin (sk)dk/k=\arctan (s/\epsilon )$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a locally integrable function in three dimensions. Dominated convergence theorem therefore identifies the distributional limit with the displayed density.

Changing the two endpoint sphere values does not change the regular distribution; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,exactly the product of the original sphere areas.

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.