3
$\begingroup$

Let $F(=(C_c^\infty)^\ast,S^\ast)$ be the space of (tempered) distributions. Let $B\hookrightarrow F$ be a compactly embedded Banach subspace. Is it true that $B\subset C^{-\alpha}$ for some $\alpha>0$?

Improve this question
$\endgroup$
7
  • $\begingroup$ I'm a little confused: surely that $(C^\infty_c)^*$ does not truly belong in your description of $F$? I'd understand it more easily if were just that $S$ were Schwartz functions and $S^*$ its dual, etc. Or am I misunderstanding what you're asking? $\endgroup$
    – paul garrett
    Commented Jun 11, 2021 at 20:28
  • 1
    $\begingroup$ Also, what Frechet space structure do you put on tempered distributions? The Frechet space structure on Schwartz functions is standard, but/and what on its dual? $\endgroup$
    – paul garrett
    Commented Jun 11, 2021 at 20:36
  • $\begingroup$ @paulgarrett I'm sorry, I will drop the Frechet condition. You are right it is not Frechet under strong topology. $\endgroup$
    – user282470
    Commented Jun 11, 2021 at 20:37
  • $\begingroup$ Ok, and/but now you need to specify the topology... since, at least a-priori, the answer could depend on the choice of topology on the dual. After all, there is a range of reasonable (!) topologies on a dual... $\endgroup$
    – paul garrett
    Commented Jun 11, 2021 at 20:40
  • 2
    $\begingroup$ No, because finite dimensional subspaces are compactly embedded. $\endgroup$
    – Christian Remling
    Commented Jun 11, 2021 at 20:41

1 Answer 1

Reset to default
1
$\begingroup$

The space $\mathscr S'$ of tempered distributions is the dual of the Fréchet spaces $\mathscr S$ and the usual strong topology on $\mathscr S'$ of uniform convergence on bounded (= relatively compact) subsets of $\mathscr S$ is the inductive (or colimit) of Banach spaces $X_n=\{u\in\mathscr S': |\langle u,f\rangle|\le c p_n(f) \text{ for some } c>0\}$ where $$p_n(f)=\sup\{(1+|x|)^n|\partial^\alpha f(x)|: |\alpha|\le n, x\in \mathbb R^d\}.$$

Grothendieck's factorization theorem implies that every continuous linear operator from a Banach space into $\mathscr S'$ factorizes through some $X_n$. This theorem is, e.g., in the book Introduction to Functional Analysis of Meise and Vogt. Since the inclusions $X_n\hookrightarrow X_{n+1}$ are compact, compactness of the operator into $\mathscr S'$ is then a consequence.

Improve this answer
$\endgroup$

You must log in to answer this question.