Correct definition of convolution of distributions?

Question

Wikipedia states, that the definition of convolution of function $f$ with a distribution $T$ is $$\langle T\ast f,\varphi\rangle=\langle T,\tilde{f}\ast\varphi\rangle$$ where $\langle T,f\rangle=T(f(x))$ and $\tilde{f}=\mathrm{d}_{-1}f(x)=f(-x)$ dilation of $f$ and this should hold $\forall\varphi\in\mathscr{S}$.

Then, convolution of distributions is defined by $$(T\ast S)\ast \varphi=T\ast(S\ast\varphi)$$ and $T\ast \varphi$ is supposed to be a function.

My question is, how is this possible if $T\ast\varphi=T(\tilde{\varphi}\ast\phi)$ and $\tilde{\varphi}\ast\phi$ is a function and a distribution acting on a function is a number: $$T:\mathscr{S}(\mathbb{R}^n)\to\mathbb{R}$$ then a convolution defined like this should be a number and convolution defined as $$T\ast(S\ast\varphi)=T(S(\tilde{\varphi}\ast\phi))$$ should be a distribution acting on a number, which is nonsense?

It is stated, that an equivalent definition is $$T\ast \varphi=\langle T,\tau_{-x}\varphi\rangle$$ how can that be an equivalent definition?

Answer 1

This is rather fishy. Convolution corresponds via Fourier transform to pointwise multiplication. You can multiply a tempered distribution by a test function and get a tempered distribution, but in general you can't multiply two tempered distributions and get a tempered distribution. See e.g. the discussion in Reed and Simon, Methods of Modern Mathematical Physics II: Fourier Analysis and Self-Adjointness, sec. IX.10.

For example, with $n=1$ try $f = 1$. $$\widetilde{f} \star \phi(x) = \int_{\mathbb R} \phi(x-t)\; dt = \int_{\mathbb R} \phi(t)\; dt$$ is a constant function, not a member of $\mathscr S$ unless it happens to be $0$. So in general you can't define $T \star f$ for this $f$ and a tempered distribution $T$. What you can define is $T \star f$ for $f \in \mathscr S$. Then it does turn out that the tempered distribution $T \star f$ corresponds to a polynomially bounded $C^\infty$ function (Reed and Simon, Theorem IX.4). But, again, in general you can't make sense of the convolution of this with a tempered distribution.

EDIT: When I say that a tempered distribution $T$ "corresponds to a function" $g$, I mean $T(\phi) = \int g(x)\; \phi(x)\; dx$.

Answer 2

Disclaimer: these are my musings about what's going on, without actually having seen anything that properly explains things.

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First the stuff I do know. Let $V^*$ denote the space of all linear functionals on a vector space $V$.

An important part of multilinear algebra is the tensor product. You can look this up, but the key idea is that $V \otimes W$ is the target space for the most general way for multiplying vectors from $V$ with vectors from $W$ to get a result that is still a vector space, and such that the corresponding tensor product of vectors $\otimes : V \times W \to V \otimes W$ is a bilinear function.

If $V$ and $W$ are finite dimensional, and $v_i$ and $w_j$ are bases, then a basis for $V \otimes W$ would be given by the set $v_i \otimes w_j$.

The odd thing about multilinear algebra is that things can be combined in a lot of ways. For example, a linear functional $T : V \to \mathbf{R}$ can be used to construct a map $V \otimes W \to W$, defined on a generating set by the formula

$$ T(v \otimes w) = T(v) w $$

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Now, the stuff I don't know.

I assume $\mathscr{S}(\mathbf{R}^n)$ denotes the space of test functions. Since the ordinary product of a test function in $x$ and a test function in $y$ is a bilinear map, there is a corresponding linear transformation

$$ \mathscr{S}(\mathbf{R}) \otimes \mathscr{S}(\mathbf{R}) \to \mathscr{S}(\mathbf{R}^2) $$

which replaces the tensor product with the ordinary product. I believe this map is continuous, injective, and has dense image.

For two linear functionals $S$ and $T$ on $\mathscr{S}(\mathbf{R})$, their tensor product acts on the space of tensor products of test functions, given by the formula on a generating set:

$$ (S \otimes T)(f \otimes g) = S(f) T(g) $$

We can thus extend $S \otimes T$ by continuity to be a partial linear functional on $\mathscr{S}(\mathbf{R}^2)$.

And this is about where my musings peter out. Maybe $S \otimes T$ is always a totally defined functional? In any case, a key point is that I'm not trying to convolve two arbitrary distributions on $\mathbf{R}^2$: instead, I'm trying to find a decomposition where I can split the problem into separate variables so that the two distributions are univariate.

This would all be nicer with Hilbert spaces; above when I say "tensor product", I mean the tensor product of the vector space structure. I think the tensor product of the Hilbert space structure works out to be nicer, so that we actually have an isomorphism $\mathscr{L}(\mathbf{R}) \otimes \mathscr{L}(\mathbf{R}) \cong \mathscr{L}(\mathbf{R}^2)$ as well as an isomorphism $H^* \cong H$, and all the facts I know about multilinear algebra still apply too.