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Let $T:S(\mathbb{R}^d)\to S(\mathbb{R}^d)$, a continuous linear operator, where $S(\mathbb{R}^d)$ is the Schwartz space. There is a result that guarantees that the family $F=\{\delta_s\circ T\}_{s\in\mathbb{R}^d}$ of functionals is a family of distribution and an $^S$family. That is, for each $u\in S(\mathbb{R}^d)$, the map $s\mapsto (\delta_s\circ T)(u)$ is in $S(\mathbb{R}^d)$.

Now, let $\varphi\in S(\mathbb{R}^{d(n+1)})$. For each $(s_1,...,s_n)\in\mathbb{R}^d$ we have $\varphi(\,\,,s_1,...,s_n)\in S(\mathbb{R}^d)$. Thus, let us take $h:\mathbb{R}^{d(n+1)}\to\mathbb{C}$ given by $$h(s,s_1,...,s_n)=(\delta_s\circ T)(\varphi(\,\,,s_1,...,s_n))=(T\varphi(\,\,,s_1,...,s_n))(s).$$

It is true that $h\in S(\mathbb{R}^d)$?

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  • $\begingroup$ Possible typo: it looks to me like your "$(s_1,\dots,s_n)\in \mathbb{R}^d$" should have been either "$s_1,\dots,s_n\in \mathbb{R}^d$" or "$(s_1,\dots,s_n)\in (\mathbb{R}^d)^n$" given what happens next. $\endgroup$
    – DCM
    Jun 9, 2019 at 9:56
  • $\begingroup$ Isn't this question much the same as asking whether any $\varphi:\mathbb{R}^k\times \mathbb{R}^l\to \mathbb{C}$ which is `separately Schwartz' (in the sense that $\varphi(\bullet,t)\in S(\mathbb{R}^k)$ and $\varphi(a,\bullet)\in S(\mathbb{R}^l)$ for any $a$ and $b$) belongs to $S(\mathbb{R}^k\times \mathbb{R}^l)$? I also don't see how $T$ plays any essential role here. Sorry if I'm missing something! $\endgroup$
    – DCM
    Jun 9, 2019 at 10:27
  • $\begingroup$ @DCM Yes. I was thinking and basically that is the question. $\endgroup$
    – ksoriano
    Jun 10, 2019 at 23:28

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