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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?

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    $\begingroup$ Why the thumbs down? If you know the answer, then please answer, don't just thumb down! $\endgroup$ – user74200 Dec 26 '14 at 19:29
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    $\begingroup$ In my answer [math.stackexchange.com/questions/781419/convolution-questions/… I gave a motivation for the standard convolution of distributions. If you accept this commonly used definition, it is not difficult to see how it correspondends with the tensor product mentioned by hurkyl below. $\endgroup$ – Vobo Dec 28 '14 at 11:27
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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 $T(\phi) = \int g(x)\; \phi(x)\; dx$.

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  • $\begingroup$ OK, let me clear my question a little: In my lectures of mathematics, we had a convolution of a (tempered) distribution and a function defined as wikipedia states. My problem is, that if $\langle T\ast f,\varphi\rangle=\langle T,\tilde{f}\ast\varphi\rangle=T(\tilde{f}\ast\varphi)$, then $(\tilde{f}(x)\ast\varphi(x))(y):\mathbb{R}^n\to\mathbb{R}^n$ by definition. Now $T$ by definition acts $T:\mathscr{S}\to\mathbb{R}$, so $T(\tilde{f}\ast\varphi)$ should be a number, if we consider switching integral and distribution, then $T(\tilde{f}\ast\varphi)$ is a distribution. But then we can not define $\endgroup$ – user74200 Dec 26 '14 at 20:49
  • $\begingroup$ we can not define $T\ast S$ by the sense of acting on some function $T\ast S\ast\varphi$, because $S\ast\varphi$ ought to be at most a distribution if not a number(a constant/number does not make a tempered distribution, as you pointed out), then we define something by the problem we want to define, which is a circular problem. However, you and wiki state $T\ast f$ corresponds to a function. My question is - how? What is the correct definition of $T\ast f$ for $f\in\mathscr{S}$? And what is a correct definition of $T\ast S$, $T,S\in\mathscr{S}'$? $\endgroup$ – user74200 Dec 26 '14 at 20:53
  • $\begingroup$ In my book of mathematics $T\ast S$ is defined as $T\ast S\ast\varphi=T(S(\varphi(x+y))$ and $T$ is acting on $y$ and $S$ is acting on $x$. But I want to make a sense of what was said on lectures(i cannot ask my lecturer). $\endgroup$ – user74200 Dec 26 '14 at 20:59
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    $\begingroup$ @user74200: That sounds more like a tensor product of two linear functionals $\mathscr{S}(\mathbb{R})\to\mathbb{R}$ $\endgroup$ – Hurkyl Dec 27 '14 at 0:31
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Disclaimer: these are my musings about what's going on, without actually having seen anything that properly explains things.


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

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


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.

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