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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?

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up vote 4 down vote accepted

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.

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Thank you for this comment! Indeed, take any smooth integral kernel $p$ such that the induced operator $P: \mathscr{D}^\prime \longrightarrow \mathscr{D}$ has infinite rank, then $P$ is only continuous in the strong topology and not in the weak topology. So almost every operator is a counterexample^^. – Matthias Ludewig May 1 '14 at 17:15
The answer is yes for $F=\mathscr D'$ with the weak* topology. – Jochen Wengenroth May 2 '14 at 13:55

If $F$ is any space endowed with its weak topology (or any other inital topology with respect to a family of linear functionals) then every strongly continuous linear operator with values if $F$ is weakly continuous (transitivity of initial topologies). If $F=\mathscr D$ is endowed with its usual LF-space topology there are many strongly continuous operators wich fail to be weakly continuous. I would try somthing like $\mathscr D' \to \mathscr D$, $u\mapsto \phi * (\phi u)$ with a non-zero test function $\phi\in\mathscr D$.

EDIT. The correct argument that every continuous linear operator from $\mathscr D'_b$ a space $F$ carrying the initial topology with respect to a family of linear functionals $\phi_\alpha:F\to \mathbb C$ is weakly continuous uses the reflexivity of $\mathscr D$: If $T:\mathscr D'_b \to F$ is continuous you have to show the coninuity of all $\phi_\alpha \circ T: \mathscr D'_\sigma \to \mathbb C$. But this is implied by $\phi_\alpha \circ T \in (\mathscr D'_b)' = \mathscr D$.

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You forgot the words " continuous,..." before the phrase "...uses the reflexivity...", didn't you? – Matthias Ludewig May 5 '14 at 6:53
@Kofi Thanks. I fixed this. – Jochen Wengenroth May 5 '14 at 7:00

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