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There isn't much context to give but I've tried searching for this a couldn't find any result. In what circumstances is $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$ dense in $l^p(\mathbb{Z},\omega) \quad 1 < p < \infty \quad$, where $\omega$ is a weigth. Is there such a codition?

If you know a "simple" proof or a book where I can find one that would be great.

Thanks in advance.

EDIT (15/04): Sorry I just realized I made a mistake in the question. I wrote $l^2(\mathbb{Z},\omega) \cap l^p(\mathbb{Z},\omega)$ instead of $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$. If I'm not mistaken this changes the answer since one can create a sequence $\varphi$ with an infinite term coinciding with a term where the weight is zero. So at a certain $n \in \mathbb{Z}$ we would have $\varphi_n = \infty$ and $\omega = 0$ which with convention $0*\infty =0$ would not imply that $\varphi \notin l^p(\mathbb{Z},\omega)$. However, $\varphi \notin l^2(\mathbb{Z})$ and is not the limit of any sequence of elements of $l^2(\mathbb{Z})$, so that would implie that $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$ is not dense in $l^p(\mathbb{Z},\omega)$. Is this arguement correct? And is it enough to force the weight to be allways different than zero to have the density?

Thanks

user6767509
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  • If the weight is allowed to take the value $0$, then you either have to have to quotient out the subspace of sequences supported on $\omega^{-1}(0)$ or $\ell^p(\mathbb{Z},\omega)$ is only a seminormed space. In any case, this does not change the answer. – MaoWao Apr 15 '20 at 12:04
  • OK, I get It now, thanks. – user6767509 Apr 15 '20 at 14:47

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The space of finitely supported sequences $c_c(\mathbb{Z})$ is contained in $\ell^2(\mathbb{Z},\omega)\cap \ell^p(\mathbb{Z},\omega)$ and dense in $\ell^p(\mathbb{Z},\omega)$ for all $p\in [1,\infty)$.

MaoWao
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