0
$\begingroup$

I asked this question in the MathStackExchange, but I think I'm not get any answer.

I'm trying to find a reference for the following result:

Theorem: Let $\Omega$ be a open subset of $\mathbb{R}^{d}$ and $k\geq 0$. Then $C_0^{k}(\Omega)$ it is (not) a (semi-)Montel space.

I tried to find in the classical references, Trèves, Horváth, but without success. The best result I found, it is the the Exercise 34.4 in Trèves book, which ensures that $C^k(\Omega)$ it is not a Montel space. Can we conclude from this exercise the Theorem above?

I'm trying to prove the following result, but I don't know if this result is valid.

A proof of this result would also be welcome.

Improve this question
$\endgroup$
3
  • $\begingroup$ At least for $k = \infty$ $C_0^\infty(\Omega)$ is a Montel space (Bourbaki, Topological Vector Spaces IV, Example 4 on p. IV.18). $\endgroup$
    – Dieter Kadelka
    Commented Mar 27 at 14:13
  • $\begingroup$ @DieterKadelka Yes, this result is also found in Trèves and Horváth's books. My question is when $0\leq k <\infty$. $\endgroup$
    – Math
    Commented Mar 27 at 15:02
  • $\begingroup$ Just to be sure : $C^k_0(\Omega)$ is the set of compactly (in $\Omega$) $C^k(\Omega)$ functions that you equip with the inductive limit topology for which $(\varphi_n)_n\rightarrow \varphi$ means existence of $K$ compact subset of $\Omega$ containing supports of all $\varphi_n$ and $\varphi$ function, with uniform convergence on $K$ up to the $k$-th derivative ? $\endgroup$
    – Ayman Moussa
    Commented Mar 27 at 16:05

1 Answer 1

Reset to default
1
$\begingroup$

In the following I write $C^k_K(\Omega)$ for the space of all $C^k(\Omega)$ functions having support included in $K$, equipped with its "natural" norm $\|\cdot\|_{K,k}$ (uniform norm on $K$ of all derivatives up to the order $k$). I also fix a countable exhaustion $(K_n)_n$ of $\Omega$. I assume that $C^k_0(\Omega)$ is equipped with the inductive limit topology $\tau$ which satisfies in particular :

$(i)$ For all $n$, the embedding $C^k_{K_n}(\Omega)\hookrightarrow C^k_0(\Omega)$ is continuous

$(ii)$ The trace topology of $\tau$ on $C^k_{K_n}(\Omega)$ is precisely the one of $\|\cdot\|_{K_n,k}$.

In this setting, I think that $C^k_0(\Omega)$ is not Montel, for finite $k$.

Indeed : $C^k_{K_n}(\Omega)$ is a normed space of infinite dimension : it cannot be Montel (this argument fails when $k=\infty$ obviously), for instance the unit ball of $C^k_{K_n}(\Omega)$ is bounded and not compact. Because of $(i)$, this unit ball is also bounded in $C^k_0(\Omega)$. But because of $(ii)$, it cannot be compact for the $\tau$ toplogy, otherwise this would imply compactness in $C^k_{K_n}(\Omega)$.

Improve this answer
$\endgroup$
11
  • 2
    $\begingroup$ A thing missing here is the fact that $E=C^k_0(\Omega)$ is a strict (lc) inductive limit and therefore it induces on $E_n=C^k_{K_n}(\Omega)$ its original topology for all $n$. This is only true for strict (lc) inductive limits. Moreover, since in this case (also not generally true for all (lc) inductive limits) $E_n$ is closed in $E$ for all $n$, it follows that $E$ is a regular inductive limit, that is, any bounded subset of $E$ is contained and bounded in $E_n$ for some $n$. All that together implies that a compact subset of $E$ must be contained and compact in $E_n$ for some $n$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Mar 27 at 17:12
  • $\begingroup$ Otherwise you cannot argue that if the unit ball of $E_n$ is compact in $E$, then it's compact in $E_n$ as well. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Mar 27 at 17:13
  • $\begingroup$ @PedroLauridsenRibeiro So the conclusion is that there is problem in the above proof? $\endgroup$
    – Math
    Commented Mar 27 at 17:17
  • 1
    $\begingroup$ I rephrased the answer, I hope it's clearer now ! $\endgroup$
    – Ayman Moussa
    Commented Mar 27 at 21:16
  • 1
    $\begingroup$ @AymanMoussa your definition of (lc) inductive limit topology is correct, but that by itself doesn't guarantee that $E$ induces the topology of $E_n$ for all $n$. There are plenty of examples of (lc) inductive limits which are not strict. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Mar 27 at 23:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .