The Schwartz space of rapidly decreasing function (as well as their derivatives) on $\mathbb R^n$ is a Fréchet space, whose (metric complete) topology is given by the usual countable family of semi-norms $(p_k)_{k\in \mathbb N}$ $$ p_k(\phi)=\max_{\vert \alpha\vert, \vert \beta\vert\le k}\Vert x^\alpha\partial_x^\beta \phi\Vert_{L^\infty(\mathbb R^n)}. $$ Is there a simple proof of the fact that this topological space is not normable?
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6$\begingroup$ Well, bounded sets are relatively compact (follows from Ascoli), something which cannot happen in infinite dimensional normed spaces. Not sure if you would call that simple. $\endgroup$– prielCommented Sep 11, 2015 at 9:39
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$\begingroup$ Complete overkill: $S(\mathrm{R}^n)$ is a nuclear space (in the sense of Grothedieck) and therefore it can not be infinite dimensional Banach space. $\endgroup$– Adrián González PérezCommented Aug 8, 2019 at 17:23
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$\begingroup$ @priel could you please elaborate a little more? I mean each of the functions are bounded, but why is the whole family bounded? And to conclude rel. comapctness don't we need equicontinuity as well? $\endgroup$– user1110Commented Apr 15, 2022 at 18:57
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More elementary than Ascoli:
If it was normable, it would mean that there exists a norm $n$ which is continuous, hence for some $k$, $n(\phi) \leqslant C\ p_k(\phi)$ and which defines the topology, i.e. such that for all $k$, $p_k(\phi) \leqslant C_k\ n(\phi)$.
This would imply that all the norms $p_k$ are equivalent for $k$ large enough, which is not the case: for example $\exp(-|x|^2) \sin(N x_1) /N^k$ tends to $0$ for the first $k$ norms but not for the following.
answered Sep 11, 2015 at 11:24
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1$\begingroup$ A continuous map need not be controlled by a single semi-norm. The usual metric on $\mathcal S$ is a counterexample. $\endgroup$ Commented Sep 12, 2015 at 0:13
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1$\begingroup$ But a continuous Norm, in fact a continuous at 0 homogenous map has to be. well.. controlled by a finite familly of semi-norm, but in this case the family of semi-norm is directed so a single one is enough. $\endgroup$ Commented Sep 12, 2015 at 9:37