# When is the weighted space $\ell^p(\mathbb{Z},\omega)$ a Banach algebra ($p>1$)?

Let $\omega:\mathbb{Z}\to (0,\infty)$ and let $1\leq p<\infty$. Consider the space $\ell^p(\mathbb{Z},\omega)$ of complex valued sequences $f=(a_n)_{n \in \mathbb{Z}}$ such that $$\|f\|=\|f\|_{\ell^p(\mathbb{Z},\omega)}:=\left(\sum_{n\in\mathbb{Z}}|a_n|^p\omega(n)^p\right)^{1/p}<\infty.$$

Next, given two complex sequences $f=(a_n)_{n \in \mathbb{Z}}$ and $g=(b_n)_{n \in \mathbb{Z}}$ their formal convolution is defined by $f*g=(c_n)_{n\in\mathbb{Z}}$ where $c_n=\sum_{k\in\mathbb{Z}}a_kb_{n-k}$.

The problem is to find necessary and sufficient conditions on $\omega$ such that $\ell^p(\mathbb{Z},\omega)$ is a Banach algebra. In other words, if $f,g\in\ell^p(\mathbb{Z},\omega)$ then $f*g\in \ell^p(\mathbb{Z},\omega)$ and $\|f*g\|\leq\|f\|\cdot\|g\|$.

For $p=1$ the condition is $\omega(n+k)\leq\omega(n)\omega(k)$.

--- Reformulated the problem so that $f\in\ell^p(\omega)$ is the same as $f\omega\in\ell^p$.

I believe this is an open problem, there are however sufficient conditions: $\omega^{-p'}*\omega^{-p'}\leq \omega^{-p'}$ where $1/p̈́'+1/p=1$ (the history of the condition is hard to tell but it is given as Lemma 8.11 in Acta Mathematica Volume 174, Number 1, 1-84, "Completeness of translates in weighted spaces on the half-line" by Alexander Borichev and Håkan Hedenmalm). The proof is based on Hölder's inequality: $$\|f*g\|^p =\sum_n \left|\sum_k a_kb_{n-k}\right|^p\omega(n)^p\leq$$ $$\leq\sum_n \left(\sum_k |a_k|^p|b_{n-k}|^p\omega(n-k)^p\omega(k)^p\right)\left(\sum_k\frac{1}{\omega(n-k)^{p'}\omega(k)^{p'}}\right)^{\frac{p}{p'}}\omega(n)^p\leq$$ $$\qquad\leq\|f\|^p\|g\|^p$$

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@Kaestur Hakarl Thanks. – AD. Aug 24 '10 at 10:15
@YemonChoi What was your edit? Did I miss something? – AD. Jan 8 '12 at 10:55
Yemon only added the (banach-algebras) tag. In general, you can see what was done in an edit by clicking edited xx time ago above an editor's name. – t.b. Jan 8 '12 at 13:33
@t.b. Great thanks! – AD. Jan 8 '12 at 13:38

Thanks, yes I think so too. There is however an easy necessary condition, let $e_k=(\delta\_n(k))\_n$ (Kronecker delta $\delta\_k(k)=1$ and $\delta\_n(k)=0$ for other $n\not=k$), then (an easy calculation shows) $e\_n*e\_k=e\_{n+k}$ which leads to $\omega(n+k)\leq\omega(n)\omega(k)$. That is any such $\omega$ must be sub-multiplicative. – AD. Aug 27 '10 at 5:12