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May 6, 2019 at 13:05 comment added paul garrett Well, I was specifically commenting about having such an action on tempered distributions, but, yes, the same idea would apply to Schwartz functions, sure: $u=\widehat{\varphi}*\widehat{v}$, etc.
May 5, 2019 at 23:38 comment added Goulifet Agree when the support of $\widehat{u}$ does not contain 0. For your second example, to define $\mathrm{D}^\gamma u$ in this case, you use that $(\mathrm{i} \omega)^{\gamma} \varphi(\omega)$ is in $\mathcal{S}$, the smoothness coming from the conditions on $\varphi$, am I right?
May 5, 2019 at 21:12 history answered paul garrett CC BY-SA 4.0