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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.

For the geometric surface delta distribution on the radius-$a$ sphere in ${\mathbb{R}}^3$, $⟨u_a,\varphi ⟩=a^2{\int }_{S^2}\varphi (a\omega )d\sigma (\omega )$. Angular integration gives the Fourier transform ${\hat{u}}_a(\xi )=4\pi a\sin (a|\xi |)/|\xi |$, with removable value $4\pi a^2$ at zero. The convolution of distributions with a compactly supported factor gives the regular distributionThis density has total mass $16{\pi }^2{a}^2{b}^2$. Values on endpoint spheres do not affect the distribution. When $a=b$, the inverse-distance singularity is locally integrable in three dimensions and is not an additional point mass. The annulus expresses the triangle inequality for the sum of two vectors of fixed lengths.