Convolution of Distribution with smooth function

Question

I meet an exercise in my homework: Let $f\in \mathcal{D}'(\mathbb{R}^n)$ and $g\in \mathcal{D}(\mathbb{R}^n)$, show that the biliner map $(f,g)\mapsto f*g $ is continuous in $f$ and $g$, respectively.

I have proved that for any fixed $f\in \mathcal{D}'(\mathbb{R}^n)$, if $g_{i}\in \mathcal{D}(\mathbb{R}^n)$ such that $g_{i}\to 0$ in $\mathcal{D}(\mathbb{R}^n)$, then $f*g_{i}\to 0$ in $\mathcal{E}(\mathbb{R}^n)$, but I don't know how to prove that for any fixd $g\in \mathcal{D}(\mathbb{R}^n)$ and $f_{i}\to 0$ in $\mathcal{D'}(\mathbb{R}^n)$ then $f_{i}*g\to 0$ in $\mathcal{E}(\mathbb{R}^n)$. Is this conclusion correct?

Here is my approach: To prove $f_{i}*g\to 0$ in $\mathcal{E}(\mathbb{R}^n)$, we need to show that, for any compact set $K\subset \mathbb{R}^n$, and any multi-index $\alpha$, $$\sup_{x\in K}|\partial^{\alpha}(f_{i}\ast g)|=\sup_{x\in K}|f_{i}\ast \partial^{\alpha}g| =\sup_{x\in K}|\langle f_{i}(y),\partial_{x}^{\alpha}g(x-y)\rangle|\to 0$$ holds for fixed $g\in \mathcal{D}(\mathbb{R}^n)$ as $i\to +\infty$. But we only have $f_{i}\to 0$ in $\mathcal{D}'(\mathbb{R}^n)$, it seems we can only show that, for any fixed $x\in K$, $$ |\langle f_{i}(y),\partial_{x}^{\alpha}g(x-y)\rangle|\to 0$$ for $i\to +\infty$.

Can someone help me with the question above? Thank you very much!

Answer 1

Too long for a comment so posted as answer.

Although artificial, this is an excellent question.

The convolution map $\mathcal{D}'\times\mathcal{D}\rightarrow \mathcal{E}$ is of course bilinear but is not continuous. It is only hypocontinuous. Here $\mathcal{D}$ has its standard topology whose explicit definition textbooks avoid. $\mathcal{E}$ is given the Frechet topology as in the OP. Next, $\mathcal{D}'$ must be given the strong topology, not the weak-$\ast$. Finally, $\mathcal{D}'\times\mathcal{D}$ is given the product topology. Then for fixed $g$ and $\alpha$, the test functions $y\mapsto \partial^{\alpha}g(x-y)$, where $x$ ranges over the compact $K$, form a bounded set in $\mathcal{D}$. Therefore $$ \rho(f)=\sup_{x\in K} |\langle f(y), \partial^{\alpha}g(x-y)\rangle_y| $$ is a continuous seminorm on $\mathcal{D}'$ by the very definition of the correct (strong) topology. So $\rho(f_i)\rightarrow 0$ is included in the package of saying that $f_i$ converges to zero.

Now if one follows textbooks, and puts the weak-$\ast$ topology on $\mathcal{D}'$, then one indeed gets into the kind of trouble pinpointed by the OP.

For the topology of $\mathcal{D}$ see: Doubt in understanding Space $D(\Omega)$

Answer 2

The conclusion you are trying to draw will not hold in the weak-* topology on the space of distributions and Abdelmalek Abdesselam has correctly pointed out the issue: the convolution mapping is discontinuous. While their post gives the reason for continuity in a stronger topology, a counterexample for discontinuity in the setting of the weaker topology is not given.

This topic has been the subject of study in

J. Larcher, Multiplications and Convolutions in L. Schwartz' Spaces of Test Functions and Distributions and their Continuity, arXiv:1209.4174.

A counterexample is given in Proposition 2, namely to the test space $\mathcal{D} := C^\infty_c$ of smooth functions with compact support, paired with the space $\mathcal{E}'$ of distributions with compact support, with convolution mapping $*_{\mathcal{D}}:C^\infty_c \times \mathcal{E}' \to C^\infty_c$. This is stronger than the case of $*_{\mathcal{E}} : (\mathcal{E} = C^\infty(\mathbb{R}^n), \mathcal{E}') \to \mathcal{E}$, wherein $\mathcal{E}'$ carries the topology of seminorms $p_{m,K}(f) := \sup_{x \in K, |\alpha| \leq m} |\partial^\alpha f(x)|$, for $K \subset \mathbb{R}^n$ compact, since we have continuous embeddings $\mathcal{D} \hookrightarrow \mathcal{E}$ and $\mathcal{D} \times \mathcal{E}' \hookrightarrow \mathcal{E} \times \mathcal{E}'$. It's also stronger to consider $*_{\mathcal{D}}$ because the seminorms on $\mathcal{D}$ are continuous norms, while $\mathcal{E}$ does not carry a continuous norm (see the discussion at the beginning of section 3.1).

An alternate argument to deal with $*_{\mathcal{E}}$ can therefore be given more quickly and is stated after Proposition 2: if $T := \delta_0 \in \mathcal{E}'$, then $f *_{\mathcal{E}} \delta_0$ is the evaluation map, but this is only continuous when $\mathcal{E}$ is normed, which we know is not the case for $\mathcal{E} = C^\infty$ with the above seminorms.

The manuscript tabulates the various combinations of spaces and multiplication and convolution operations among them and either proves continuity or discontinuity, along with providing many references.