# Continuity of convolution on $\mathcal{D}'_+$

Let $$\mathcal{D}'_+:=\{T\in \mathcal{D}'(\mathbb{R}): \textrm{supp}(T)\subset [0,\infty)\}$$. Here $$\mathcal{D}'(\mathbb{R})$$ is the usual space of distributions on $$\mathbb{R}$$, equipped with the weak$$\ast$$-topology induced by $$\mathcal{D}(\mathbb{R})$$, and $$\mathcal{D}_+'$$ is given the subspace topology induced from $$\mathcal{D}'(\mathbb{R})$$.

Question: Is convolution $$\ast:\mathcal{D}'_+ \times \mathcal{D}'_+\rightarrow \mathcal{D}'_+$$ separately continuous?

• @Lucia: It's better to use the strong topology instead of the weak-$\ast$ especially if you want nice continuity properties for this kind of multilinear constructions. – Abdelmalek Abdesselam Jun 8 at 20:08

The convolution on $$\mathcal{D}'_+$$ can be defined as follows. Fix a $$C^\infty$$ function $$\psi$$ such that $$\psi(x) = 1$$ for $$x \geqslant 0$$ and $$\psi(x) = 0$$ for, say, $$x \leqslant -1$$. If $$\phi \in \mathcal{D}$$, define $$\tilde\phi(x, y) = \phi(x + y) \psi(x) \psi(y) .$$ Note that $$\tilde{\phi} \in \mathcal{D}(\mathbb{R}^2)$$. The convolution of $$f, g \in \mathcal{D}'_+$$ is given by $$\langle f * g, \phi \rangle = \langle f \otimes g , \tilde\phi\rangle ,$$ where $$f \otimes g$$ is the tensor product of $$f$$ and $$g$$.
(Edited after Abdelmalek's comment; thanks!) The tensor product is a separately continuous function from $$\mathcal{D}' \times \mathcal{D}'$$ into $$\mathcal{D}'(\mathbb{R}^2)$$ (with weak-* topologies), and hence the convolution is separately continuous. (In fact, the tensor product is jointly sequentially continuous, and hence the same is true for the convolution in $$\mathcal{D}'_+$$.)
• If I remember correctly the tensor product is continuous for the strong topology but not for the weak-$\ast$ mentioned by the OP. But other than that +1 – Abdelmalek Abdesselam Jun 8 at 20:07
• I think you mean to say that $\tilde{\phi} \in \mathcal{D}(\mathbb{R}^2)$, not $\mathcal{D}'$? – Nate Eldredge Jun 8 at 20:52
• @AbdelmalekAbdesselam: Can you please see if it is OK now? I am not used to working with the weak-* topology on $\mathcal{D}'$. – Mateusz Kwaśnicki Jun 8 at 21:53