Given $f, g \in \ell_p(\mathbb{Z})$, we define the convolution of $f$ and $g$ as follows :
$$(f\ast g)(x) :=~ \sum\limits_{y = -\infty}^\infty f(y)g(x-y).$$
It is easy to prove that :
- If $p = 1$ then $(f\ast g)(x)$ 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, $(f\ast g)(x)$ converges absolutely for all $x \in \mathbb{Z}$.
- If $1 < p < 2$ then, by the Hölder's inequality, $(f\ast g)(x)$ converges absolutely for all $x \in \mathbb{Z}$ since
$$ \begin{equation}\begin{aligned} \Big| (f \ast g)(x)\Big| ~&=~ \left|\sum\limits_{y \in G} f(y)g(x-y)\right| \\&\leq~ \sum\limits_{y \in G}\big| f(y)g(x-y)\big| \\&\overset{\text{H}}{\leq}~ \left( \sum\limits_{y \in G} \big| f(y)\big|^p\right)^{1/p}\left( \sum\limits_{y \in G}\big|g(x-y)\big|^{p/(p-1)} \right)^{(p-1)/p} \\&\overset{\dagger}{\leq}~ \|f\|_p \|g\|_{p/(p-1)} \\&\leq~ \| f\|_p \|g\|_p.\end{aligned}\end{equation},$$
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 $(f\ast g)(x)$ converges uniformly for all $x \in \mathbb{Z}$ if $f, g \in \ell_p(\mathbb{z})$ and if $p > 2$. Is it even true ?
Best Answer
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 $$f \ast f(k) \ge \frac 13 \sum_{|\ell| > |k|} |\ell|^{-1} = \infty.$$