Timeline for Finite dimension implies regularity

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Dec 6, 2013 at 7:53	comment	added	Jochen Wengenroth		How do you prove that $X \subseteq C^\infty$? The question only gives the definition of $\tau_h$ for functions but it can of course be extended to distributions (and then, e.g. $\mathscr D'$ itself is invariant under all $\tau_h$). However, I think this part of the argument is not necessary: The closure of $X$ is invariant under differentiation (since the distributional derivative is the limit in $\mathscr D'$ of $(\tau_h(u)-u)/h$) and since $X$ is finite dimensional it is closed. As you said, all elements of $X$ satisfy a linear ODE and this implies $X\subseteq C^\infty$.
Dec 6, 2013 at 0:04	comment	added	Pietro Majer		Not only: having shown (by a convolution regularization argument) that $X\subset C^\infty$, it also follows that $X$ is invariant by derivation (for $\phi\in X$, $(\tau_h\phi -\phi)/h\in X$, and converges to $\phi$ uniformly on bounded sets, but all TVS topologies on $X$ coincide, as it is finite-dimensional, so $\phi' \in X$). Since $X$ has finite dimension $m$, any $\phi\in X$ satisfies a linear ODE with constant coefficients, actually the same ODE.
Dec 5, 2013 at 23:13	answer	added	paul garrett		timeline score: 2
Dec 5, 2013 at 22:47	history	asked	smyrlis	CC BY-SA 3.0