$r:\mathscr{S'}\rightarrow\mathscr{D'}$, $u\mapsto u|_{\mathscr{D}}$ is not a topological embedding

Question

Show that $r:\mathscr{S'}\rightarrow\mathscr{D'}$, $u\mapsto u|_{\mathscr{D}}$ is not a topological embedding. For this problem, would it suffice to construct a sequence $\{u_n\}$ in $\mathscr{D'}$ such that $u_n\rightarrow 0$ in $\mathscr{D'}$, but $r(u_n)\nrightarrow 0$ in $\mathscr{S'}$. In that case, what would be an example of such a sequence?

Answer 1

First of all, this depends on the topologies, I assume you mean the weak$^*$ topologies $\sigma(\mathscr S',\mathscr S)$ and $\sigma(\mathscr D',\mathscr D)$. Note that $r$ has dense range (it is the transposed of the inclusion $\mathscr D \hookrightarrow \mathscr S$) and $r$ is injective (because the inclusion $\mathscr D \hookrightarrow \mathscr S$ has dense range). If $r$ were a topological embedding this would imply that $\mathscr S'$ and $\mathscr D'$ have the same dual. But, for any locally convex space $X$, we have $(X',\sigma(X',X))'=X$. Hence $\mathscr D \neq \mathscr S$ yields a contradiction.

EDIT. Here is a "concrete" argument as you wanted to have: Define $u_n\in \mathscr S'$ by $u_n(f)=e^n f^{(n)}(n)$. These are distributions with compact support hence they are tempered. For all $\varphi \in\mathscr D$ you have $u_n(\varphi)=0$ for all $n$ large enough (larger than an upper bound of the support of $\varphi$), hence $u_n\to 0 $ in $\mathscr D'$.

Using a smooth "cut-off" function you easily construct $f\in \mathscr S$ with $f(x)=e^{-x}$ for all $x\ge 0$. But then $u_n(f)=(-1)^n \not\to 0$, hence $u_n$ does not converge to $0$ in $\mathscr S'$.

Of course, the existence of such a sequence implies that $r$ is not a topological embedding. But one has to be careful with the converse implication: Two locally convex topologies on a vector space may have the same convergent sequences although they are different. A classical example is $\ell^1$ endowed with the norm and the weak topology ("Schur's lemma").