An operateor $T$ is globally hypoelliptic if : $u\in S'(\Bbb R^n),Tu\in S(\Bbb R^n)$ imply $u\in S(\Bbb R^n)$. My question why if $u\in L^2(\Bbb R^n): Tu =\lambda u$. Then $u\in S(\Bbb R^n)$. where $\lambda\in\Bbb R^n$, $S(\Bbb R^n)$ is the Schwartz space and $S'(\Bbb R^n)$ is the space of all tempered distributions
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1$\begingroup$ Umm, for $T = Id$, that is $Tu \equiv u$ it is clearly globally hypoelliptic, but your condition is not satisfied... $\endgroup$– Aleksei KulikovCommented Jun 8 at 14:57
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$\begingroup$ when T=Id, If $u\in S'(R^n)$ and $Tu=u\in S(R^n)$. Then $u\in S(R^n)$ $\endgroup$– zoran VicovicCommented Jun 8 at 15:24
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$\begingroup$ Yes, and yet there exists $u\in L^2(\mathbb{R}^n)$ for which $Tu = u$ but which is not Schwartz, e.g. literally any $L^2$-function. $\endgroup$– Aleksei KulikovCommented Jun 8 at 16:33
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