Denote by $\mathscr{D}^\prime$ and $\mathscr{D}^\prime\_b$ the space of distributions on $\mathbb{R}^n$ equipped with the weak and the strong topology, respectively. Because the topology of $\mathscr{D}^\prime\_b$ has more open sets than the topology of $\mathscr{D}^\prime$, we have
$$L(\mathscr{D}^\prime, F) \subseteq L(\mathscr{D}^\prime\_b, F)$$
for any topological vector space $F$. If $F = \mathbb{C}$, then we have equality (this holds for every locally convex space instead of $\mathscr{D}$). If $F = \mathscr{D}^\prime\_b$, the left-hand-side is a proper subset because the identity is not continuous from $\mathscr{D}^\prime$ to $\mathscr{D}^\prime\_b$ as the latter topology contains strictly more open sets (but the identity is continuous from each topological space to itself).


**Question:** What if $F = \mathscr{D}$, the space of test functions or $F = \mathscr{D}^\prime$, the space of distributions with the weak topology? Is this a strict inclusion or do we happen to have inequality? 


If we don't have equality, what would be an example of a linear map that is contained in the latter space, but not in the first?


I didn't specify the domain yet. Is the answer different when we take the domain of functions in $\mathscr{D}$ to be a compact subset of $\mathbb{R}^n$ or a compact manifold instead of $\mathbb{R}^n$? What if we change the space $\mathscr{D}$ to the spaces $\mathscr{S}$ of Schwartz functions or $\mathscr{E}$ of all smooth functions, with there usual (Fréchet) topologies?



Linear operators on distributions with different topologies
The answer to all of your questions is no. This follows from the simple fact that if a linear mapping from a locally convex space $E$ into a normed space $F$ is continuous for the weak topology on $E$ and the norm on $F$, then its range is finite dimensional.


