# Is a weak* limit of a sequence of tempered distributions indeed a tempered distribution?

The question is as in the title.

Let $$\{ T_n \}$$ be a sequence of tempered distributions such that $$\{ T_n(f) \}$$ converges for every Schwartz function $$f$$. Let us denote the "pointwise" or weak* limit as $$T(f)$$.

Then, is $$T$$ a tempered distribution? Of course it is a linear functional on the Schwartz space, but I cannot see how to show temperedness. It seems quite confusing and nontrivial. Could anyone please clarify?

• With the Hermite functions you can replace the question about $S(\Bbb{R})$ and its dual by the similar question about the rapidly decreasing sequences and its dual (the polynomially bounded sequences). Jan 9, 2023 at 21:45

If $$E$$ is a barrelled lcs and $$f_n \in E'$$ is a pointwise convergent sequence of continuous linear functionals on $$E$$, then $$f(x) = \lim f_n(x)$$ is continuous. Indeed, by Banach-Steinhaus theorem the sequence $$f_n$$ is equicontinuous, i.e. there is a neighborhood of zero $$U \subset E$$, such that $$|f_n(x)| \le 1$$ for all $$n \in \mathbb N$$ and $$x \in U$$. Thus, $$|f(x)| \le 1$$ for all $$x \in U$$, which ensures the continuity of $$f$$.
In order to apply this to your particular case, observe that Schwartz space $$\mathscr S(\mathbb R^d)$$ is a Frechet space and, therefore, it is barrelled.
The same argument can be applied to all types of distributions (as far as I remember), since spaces $$\mathscr D$$, $$C^\infty$$, Gelfand-Shilov spaces, etc., are all barrelled.