$f,g \in l^2(\mathbb{Z})$ such that $\sum_{n \in \mathbb{Z}} |f(n)|^2 < \infty, \sum_{n \in \mathbb{Z}} |g(n)|^2 < \infty$.
Show that $(f \star g)(n) = \sum_{m \in \mathbb{Z}} f(m)g(-m+n)$ is convergent and a bounded function of n.
I am not quite sure how to show this but think the first step would be to write $(f \star g)(n)$ as an inner product. Can anyone help with this and also is there an example of $f \star g$ that isn't in $l^2(\mathbb{Z})$?
We will answer the question at the very end of OP. The problem becomes simpler if we may use the Fourier series. For $F\in L^1(-\pi,\pi)$ let $$\widehat{F}(n)={1\over 2\pi}\int\limits_{-\pi}^\pi F(x)e^{-inx}\,dx$$ The space $L^2(-\pi,\pi)$ is unitarily isomorphic to $\ell^2(\mathbb{Z})$ by the mapping $U(F)= \{\widehat{F}(n)\}_{n=-\infty}^\infty.$ If $F,G\in L^2(-\pi,\pi)$ then $FG\in L^1(-\pi,\pi).$ It can be shown that $\widehat{F}*\widehat{G}=\widehat{FG}.$ Hence the question is equivalent to the following: find $F,G\in L^2$ such that $FG\notin L^2.$ Let $F(x)=G(x)=|x|^{-1/4}.$ Then $F\in L^2,$ but $F^2\notin L^2.$ The corresponding sequences are given by $$f(n)=g(n)={1\over 2\pi}\int\limits_{-\pi}^\pi |x|^{-1/4}e^{-inx}\,dx$$