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I have a series of the form $$ F=\sum_{n=1}^\infty f_n, $$ where each $f_n$ is a tempered distribution, and I know that for each $f_n$ there exists a tempered distribution $g_n$ such that $|f_n|\leq |g_n|$ (in the sense that $|⟨f_n,\varphi⟩|\leq |⟨g_n,\varphi⟩|$ for all $\varphi\in \mathcal S(\mathbb R^k)$) and which sum to a tempered distribution $G=\sum_{n=1}^\infty g_n$ (in the sense that for all $\varphi\in \mathcal S(\mathbb R^k)$ you have $$ \sum_{n=1}^N g_n(\varphi) \stackrel{N\to\infty}{\longrightarrow} G(\varphi) $$ for $G$ a tempered distribution.

I suspect that this is plenty to show that the series sums to (define) a tempered distribution $F$, in a sort of dominated convergence theorem for tempered distributions, but I've been unable to locate such a proof. Is it easy to show directly? If not, where can I find it?

My specific context is showing that $F(x)=\sum_{n=-\infty}^\infty e^{in^2 a}e^{inx}$ is a tempered distribution for $a\in\mathbb R$, where the series is dominated by the Dirac comb, but I think this thread will be more useful for a broader public if it is kept general. (That said, if the result requires stronger hypotheses that still apply to this example, then that's quite OK.)

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    In your particular case you could use the Fourier transform: the series $\sum_{|k|\le n} a_k \delta_k$ converges in $S'$ as long as the sequence $a_k$ is bounded; then apply the inverse FT. – 2017-01-18
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    If $T$ is a tempered distribution then $T \ast e^{-a x^2}$ is a Schwartz function. – 2017-01-19
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    I think the general statement could be a bit fishy. I tried, but I think you need better convergence for $\sum g_n$, pointwise seems a bit too weak. Maybe one could use Banach-Steinhaus at this point. Nuclearity of $\mathcal{S}$ could also be helpful. – 2017-01-19
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    @Daniel I'm happy to consider stronger hypotheses (so long as they don't completely break the example). – 2017-01-19
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    I also think that your general statement is incorrect. Take for example the case when $g_n =(-1)^n/n$ and $f_n = 1/n$. Then clearly $f_n$ and $g_n$ are tempered distributions, and $\sum_{n=1}^N g_n$ converges to one as well when $N \to \infty$, but $\sum_{n=1}^N f_n$ does not. – 2017-01-25
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    +1 to @Rastapopoulos. If you want some "dominated" convergence, then you should assume positivity of $g_n$ in the sense that $\langle g_n,\varphi\rangle \ge 0$ for any non-negative $\varphi\in \mathcal{S}(\mathbb{R}^k)$. – 2017-01-26
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    @Rastapopoulos yes, I see that. The question then becomes whether there is a related result, with stronger hypotheses, that does hold. However, asking for positivity is too strong (as I do want it to hold for the specific example). – 2017-01-26
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    For positive distributions, this should be a consequence of the correspondence between positive distributions and measures. – 2017-01-28

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