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?
Best Answer
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.