0
$\begingroup$

Let $\Omega\subset \mathbb R^N$ be open. Given a weight function $v\geq 1$ such that $v\in L^1_{\text{loc}}(\Omega)$ and $l.s.c$. Also supposethere exists a Lipschitz continuous sequence $v_n$ such that $1\leq v_n\leq v$ and $v_n\nearrow v$ for all $x\in\Omega$. Note that $v$ may not bounded above.

Define $C_c(\Omega,v):=\{u\in C(\Omega), u/v\in C_c(\Omega)\}$.

My question: do we have $$ C_c(\Omega,v)\subset\bigcup_{n=1}^\infty C_c(\Omega,v_n), $$ or, in another words, for any $f\in C_c(\Omega,v)$ there exists a $n_0$ such that $f\in C_c(\Omega,v_{n_0})$?

I realize that the above argument is trivial as I set $v\in L^1_{\text{loc}}$. To make question more interesting, I delete the assumption that $v\in L^1_{\text{loc}}$ and hence $v$ could be $+\infty$ over a positive measure set. Now $u/v\in C_c(\Omega)$ no longer makes $u$ is compact supported. Now, is my claim still hold?

Improve this question
$\endgroup$
9
  • $\begingroup$ Your last sentence asks about the $\subset$ direction of the equality, but the $\supset$ direction need not hold. Consider something like $\Omega = \mathbb{R}$, $v = 1+1_{(0, +\infty)}$. You can take $v_n$ to be 1 on $(-\infty, 0]$, 2 on $[1/n, +\infty)$, and linear in between. If you take $f \in C_c(\mathbb{R})$ which is 1 in a neighborhood of 0, then $f \in C_c(\Omega, v_n)$ for every $n$, but $f \notin C_c(\Omega, v)$. $\endgroup$
    – Nate Eldredge
    Commented Aug 18, 2015 at 2:23
  • $\begingroup$ I do not get the question: Apparently $C_c(\Omega)$ is the space of continuous functions with compact support. But $u$ and $u/v$ have the same support so that $C_c(\Omega,v)=C_c(\Omega)$??? $\endgroup$
    – Jochen Wengenroth
    Commented Aug 18, 2015 at 9:01
  • $\begingroup$ @JochenWengenroth: you are right. Actually we we have $C_c(\Omega,\nu)\subset C_c(\Omega)$. Think, as Nate point out above, the case that $\nu$ is not continuous. Then $u/v$ maybe compact supported but not continuous. I modified my post a bit, although it looks trivial now... $\endgroup$
    – JumpJump
    Commented Aug 18, 2015 at 12:30
  • $\begingroup$ @NateEldredge I modified my post a bot. Well, as $v$ has to be a.e. finite since $v\in L^1_{\text{loc}}$, it makes $u$ has to be compactly supported if $u\in C_c(\Omega,\nu)$ and hence $u\in C_c(\Omega)$ as Jochen point out. Hence, my claim holds trivially. Am I right? $\endgroup$
    – JumpJump
    Commented Aug 18, 2015 at 12:33
  • $\begingroup$ @NateEldredge I also update my post a bit. Welcome to have a look! $\endgroup$
    – JumpJump
    Commented Aug 18, 2015 at 12:39

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .