# Is the Schwartz topologically emebedded in space of tempered distributions?

Let $$g\mapsto ( \cdot, g)_2$$ denote the map from the Schwartz space $$S$$ into its dual space $$S'$$ where $$(f,g)_2$$ is the inner product in $$L^2$$. Then is this a linear topological embedding ($$S'$$ is endowed with the weak* topology)? Can anyone provide a reference or a simple proof?

• This is not a topological embedding, because seeing $S$ as a subset of $S'$, the natural topology of $S$ is much stronger than the one induced from $S'$ with the weak-$\ast$ or even the more appropriate strong topology. This is why $S$ can afford the luxury of being dense in $S'$ while being complete for its own metrizable topology. – Abdelmalek Abdesselam Feb 8 at 22:59

$$\left|\int_{\mathbb{R}} f(x)g(x)dx \right| \le \left\|\frac{f}{1+x^2}\right\|_{L^1} \|(1+x^2)g\|_{L^\infty}\le \pi \|f\|_{L^\infty}(\|g\|_{L^\infty}+\|x^2g\|_{L^\infty})$$
$$\|.\|_{L^\infty}$$,$$\|x^2.\|_{L^\infty}$$ are in the semi-norms used to construct the Schwartz space.
• @AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $\|.\|_{L^\infty}+\|x^2.\|_{L^\infty}$ and as I shown that one embeds in its strong dual. – reuns Dec 3 '18 at 4:09
• I mean the inverse of $g\mapsto (\cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable. – Andrew Yuan Dec 3 '18 at 6:09
• @AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $\sum_{|\alpha| \le K} \| (1+|x|^M) \partial_\alpha f\|_{L^\infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces. – reuns Dec 3 '18 at 6:32
• Define precisely the topologies in $S,S'$ then $(g,g) = \|g\|_{L^2}^2 > ...$ will make it obvious that the map $g \to \iota(g) \in S'$ is bi-continuous. – reuns Dec 3 '18 at 6:42