Let $\mathscr{D}'(\mathbb R)$ be the set of distributions on $\mathbb R$ and $X$ be a linear subspace of $\mathscr{D}'(\mathbb R)$, which is closed under translations, i.e., if $\varphi\in X$ and $h\in\mathbb R$, then $\tau_h\varphi\in X$, where $$ (\tau_h\varphi)(x) = \varphi(x+h) \quad \text{in the sense of distributions}. $$ If $X$ is finite dimensional, then show that $X\subset\mathrm{C}^\infty(\mathbb R)$.
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ 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. $\endgroup$– Pietro MajerCommented Dec 6, 2013 at 0:04
-
$\begingroup$ 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 $X\subseteq C^\infty$. $\endgroup$– Jochen WengenrothCommented Dec 6, 2013 at 7:53
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
The smooth vectors in a repn of a real Lie group on a quasi-complete locally convex topological vector space are dense, by averaging against compactly-supported smooth functions on the group. Thus, all vectors in a finite-dimensional repn space are smooth. In this example, the Lie group is $\mathbb R$, ...
answered Dec 5, 2013 at 23:13
-
$\begingroup$ One possibly must puzzle out the potential seeming-paradox of the fact that distributions are (distributionally) infinitely-differentiable, so in certain circumstances can be "smooth vectors". But this (correct) argument, when examined in detail, does not subvert the application to the question. $\endgroup$ Commented Dec 5, 2013 at 23:31
-
-
$\begingroup$ Ah, here it's just that the map $\mathbb R \times X \to X$ by $r\cdot f\to f$-translated-by-$r$ is (jointly) continuous, where the "Lie group" is $\mathbb R$ and $X$ is the TVS. $\endgroup$ Commented Dec 6, 2013 at 23:12