Suppose $H_1$ and $H_2$ are Hilbert spaces which lie between test-functions and distributions, i.e. $$ \mathscr{D}(\mathbb{R}) \subset H_i \subset\mathscr{D}'(\mathbb{R}), $$ such that the inclusions are continuous. The derivative $d/dx$ can be viewed as continuous linear maps $D\colon\mathscr{D}(\mathbb{R}) \rightarrow \mathscr{D}(\mathbb{R})$ and $\tilde D\colon\mathscr{D}'(\mathbb{R}) \rightarrow \mathscr{D}'(\mathbb{R})$ respectively and with this notation $\tilde D\restriction_{\mathscr{D}(\mathbb{R})}=D$.
Question: If $T\colon H_1\rightarrow H_2$ is bounded and $T\restriction_{\mathscr{D}(\mathbb{R})} = D$, does it follow that $T=\tilde D\restriction_{H_1}$?
An example for this situation would be the Sobolev spaces $H_1=W^{2,k}(\mathbb{R})$ and $H_2=W^{2,k-1}(\mathbb{R})$ and there the assertion is true.
I've tried to look at it abstracty, so to forget about the derivative and ask $D$, $\tilde D$ to be arbitrary continuous linear maps with $\tilde D\restriction_{\mathscr{D}(\mathbb{R})}=D$. Then the only (?) thing we can use is the density of $\mathscr{D}(\mathbb{R}) \subset \mathscr{D}'(\mathbb{R})$. But this does not seem to be enough.
Surely if also $\mathscr{D}(\mathbb{R}) \subset H_i $ was dense the situation would be clear. While this is the case for the Sobolev spaces above, I don't want to assume that because it fails if $\mathbb{R}$ is replaced by $(0,1)$. Another sufficient condition would be to ask $C^\infty(\mathbb{R})\cap H_i \subset H_i$ to be dense. This is true for Sobolev spaces on $(0,1)$ but it is not true for Sobolev spaces on badly behaved domains in $\mathbb{R}^n$.