Is the Schwartz topologically emebedded in space of tempered distributions?

Question

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?

Answer 1

$$\left|\int_{\mathbb{R}} f(x)g(x)dx \right| \le \left\|\frac{f}{1+x^2}\right\|_{L^1} \|(1+x^2)g\|_{L^\infty}\le \pi \|f\|_{L^\infty}(\|g\|_{L^\infty}+\|x^2g\|_{L^\infty}) $$

$\|.\|_{L^\infty}$,$\|x^2.\|_{L^\infty}$ are in the semi-norms used to construct the Schwartz space.

Answer 2

To make the comment raised by Abdelmalek Abdesselam precise: $S$ is a complete topological vector space and the inclusion map $\iota : S \to S'$ where $S'$ is equipped with either the strong or weak dual topology is linear and continuous, hence also uniformly continuous. Thus, if $\iota$ was a homeomorphism onto its image $\iota(S)$ would be complete in its subspace topology. However, as is well-known, $\iota(S)$ is actually dense in $S'$ (with either topology) which is a contradiction.