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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?

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  • $\begingroup$ 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). $\endgroup$
    – reuns
    Jan 9, 2023 at 21:45

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The answer is yes. I suppose that you can prove this fact by some concrete argument, but in the theory of locally convex spaces there are general methods to prove such things. Here we need the following observation:

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.

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First, emphatically, this is a reasonable question to ask, and is often glossed over in introductory discussions of distributions and/or tempered distributions.

In addition to @Matsmir's good answer, we can see the question as asking whether the weak-star dual of a Frechet space is sequentially complete. In fact, the weak-star dual is "quasi-complete", meaning that every (TVS-sense) bounded Cauchy net converges. The general proof does use Banach-Steinhaus and so on, as @Matsmir says.

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