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?
Best Answer
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 $T(\phi) = \int g(x)\; \phi(x)\; dx$.