Csdrhrt

Given $f, g in ell_p(mathbb{Z})$, we define the convolution of $f$ and $g$ as follows :
$$(fast g)(x) :=~ sumlimits_{y = -infty}^infty f(y)g(x-y).$$

It is easy to prove that :

• If $p = 1$ then $(fast g)(x)$ converges absolutely for all $x in mathbb{Z}$ and that $|f ast g|_1 leq |f|_1 |g|_1$.


• If $p = 2$ then, by Cauchy-Schwarz inequality, $(fast g)(x)$ converges absolutely for all $x in mathbb{Z}$.


• If $1 < p < 2$ then, by the Hölder's inequality, $(fast g)(x)$ converges absolutely for all $x in mathbb{Z}$ since
  $$ begin{equation}begin{aligned} Big| (f ast g)(x)Big| ~&=~ left|sumlimits_{y in G} f(y)g(x-y)right| \&leq~ sumlimits_{y in G}big| f(y)g(x-y)big| \&overset{text{H}}{leq}~ left( sumlimits_{y in G} big| f(y)big|^pright)^{1/p}left( sumlimits_{y in G}big|g(x-y)big|^{p/(p-1)} right)^{(p-1)/p} \&overset{dagger}{leq}~ |f|_p |g|_{p/(p-1)} \&leq~ | f|_p |g|_p.end{aligned}end{equation},$$
  where $(dagger)$ is a consequence of the following implication :
  $$1 leq r < s leq infty ~~~~~Rightarrow~~~~~ |h|_s leq |h|_r.$$
  

I can't think of an easy way to prove that $(fast g)(x)$ converges uniformly for all $x in mathbb{Z}$ if $f, g in ell_p(mathbb{z})$ and if $p > 2$. Is it even true ?

The classical result is that if $p,q,r ge 1$ and $dfrac 1p + dfrac 1q = 1 + dfrac 1r$ then $|f ast g|_r le |f|_p |g|_q$.

If $p,q > 2$ then there is no such $r ge 1 $ satisfying the above relation, so the result you want will almost certainly fail.

For an example, a simple sequence that belongs to $ell_p(mathbb Z)$ for $p > 2$ but not to $ell_2(mathbb Z)$ is $f(k) = (1 + |k|)^{-1/2}$. Then
$$f ast f(k) = sum_{ell} f(k-ell)f(ell) = sum_{ell} (1 + |k-ell|)^{-1/2} (1 + |ell|)^{-1/2}$$
but this sum is easily seen to diverge for every $k$ - a quick estimate shows
$$f ast f(k) ge frac 13 sum_{|ell| > |k|} |ell|^{-1} = infty.$$