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Suppose $h\in S(\mathbb{R}^d)$ (Schwartz space) and a family $\mathcal{F}=\{f(\,\cdot\,;s)\}_{s\in\mathbb{R}^d}\subseteq S^\prime(\mathbb{R}^d)$ of tempered distributions. Then, for each fixed $s$ we have $f(h;s)\in\mathbb{R}$.

Now, assume that the map $s\mapsto f(h;s)$ is again in $S(\mathbb{R}^d)$ and take $g\in S^\prime(\mathbb{R}^d)$. Denote the action of $g$ over $s\mapsto f(h;s)$ by $\langle g_\cdot,f(h;\,\cdot\,)\rangle$.

Is there some theorem/propertie that guarantees that there is a continuous function $M\{g,\mathcal{F}\}:\mathbb{R}^d\to\mathbb{R}$ such that for all $h\in S(\mathbb{R}^d)$ we have $$\langle g_\cdot,f(h;\,\cdot\,)\rangle=\int_{\mathbb{R}^d}h(s)M\{g,\mathcal{F}\}(s)\,\mathrm{d}s\,?$$

EDIT Boris'answer is correct. I'll change the question:

Is there some tempered distribution $M\{g,\mathcal{F}\}$ such that $\langle g_\cdot,f(h;\,\cdot\,)\rangle=M\{g,\mathbb{F}\}(h)$?

In Boris answer, the tempered distribution $M\{g,\mathbb{F}\}$ is $g$.

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  • $\begingroup$ Can't give a detailed answer now but essentially the answer is yes modulo suitable continuity hypotheses on the family, e.g., the map $h\mapsto (s\mapsto f(h;s))$ being continuous from Schwartz space to itself. The relevant theory has to do with the nuclear theorem and the sequels to Schwartz books: numdam.org/item/AIF_1957__7__1_0 and numdam.org/item/AIF_1958__8__1_0 $\endgroup$ – Abdelmalek Abdesselam May 14 at 23:43
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Let $f(\cdot;s)=\delta(x-s)$ and $g=\delta(x)$, then $f(h;\cdot)=h(\cdot)$ and $\langle g_\cdot,f(h;\cdot)\rangle=h(0)$, so there is no such function, that $$ h(0)=\int_{\mathbb{R}^d}h(s)M\{g,\mathcal{F}\}(s)\mathrm{d}s $$

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  • $\begingroup$ Thanks. I changed the question for a new one. Indeed your counterexample is the basis of my original doubt, but I did not ask in the correct sense. $\endgroup$ – sinbadh May 13 at 8:02

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