Does anyone knows if there is a relationship between $\mathcal{S}(\mathbb{R^n})$ and $\mathcal{S}(\mathbb{R})$? That is, if a function is in Schwartz space for each variable, can i prove that it is on Schwartz space of the cartesian?
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I'm not quite sure what you mean by 'each variable' here. If you mean that fixing all variables but $1$, the resulting function is Schwarz on $\mathbb{R}$, I would guess that this is true, and maybe isn't so hard to prove with the mean value theorem. – A. Thomas Yerger Jun 28 '23 at 19:15
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It is the opposite: if $f(x_1,\cdot, ..., \cdot)$ is in Schwarz on $R$, $f(\cdot,x_2, ..., \cdot)$ is in Schwarz on $R$, ...., can i say that $f$ is in Schwarz on $R^n$? – TDg1 Jun 28 '23 at 19:25