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?
Yes, it is.
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 $\mathcal{D}'_+$.)
