Does $A \colon H^m \to C^\infty$ imply smoothness of the Schwartz kernel?

Question

The Schwartz kernel theorem states, in particular, that for any continuous linear map $A \colon \mathcal D(\mathbb R^d) \to \mathcal D'(\mathbb R^d)$ there exists the unique kernel $K \in \mathcal D'(\mathbb R^d \times \mathbb R^d)$ such that for any $u$, $v \in \mathcal D(\mathbb R^d)$ $$ \langle A u,v\rangle = \langle K,u\otimes v\rangle. $$ It is known that if $A(\mathcal D(\mathbb R^d)) \subset\mathcal D(\mathbb R^d)$ and $A$ extends to a continous linear map $A \colon \mathcal E'(\mathbb R^d) \to C^\infty(\mathbb R^d)$, then its Schwartz kernel $K \in C^\infty(\mathbb R^d \times \mathbb R^d)$. But what if $A$ extends to a continous linear map $A \colon H^m(\mathbb R^d) \to C^\infty(\mathbb R^d)$? Does this weaker assumption imply that $K \in C^\infty(\mathbb R^d \times \mathbb R^d)$?