# discrete convolution $f*g$ belongs to $\ell_\infty$, i.e. the sup norm is finite

Definitions. Fix any $\phi\in(0,1)$ and $\theta\in(0,1)$, and let us define functions \begin{equation}f(n)=\left\{\begin{array}{ll}n^{-\phi},&\text{ if }n\geq 1\\0,&\text{ otherwise}\end{array}\right.\;\;\;\text{ and }\;\;\;g(n)=\left\{\begin{array}{ll}n^{-\theta},&\text{ if }n\geq 1\\0,&\text{ otherwise.}\end{array}\right.\end{equation} We define the convolution $f*g$ by the rule \begin{equation}(f*g)(n)=\sum_{m=-\infty}^\infty f(m)g(n-m)\end{equation} and its sup norm \begin{equation}\|f*g\|_\infty=\sup_{n\in\mathbb{Z}}|(f*g)(n)|.\end{equation} Note that $f$, $g$, and $f*g$ can all be viewed as positive sequences in $\mathbb{N}$, and that \begin{equation}\|f*g\|_\infty=\sup_{n\in\mathbb{N}}\sum_{m=-\infty}^\infty f(m)g(n-m)=\sup_{n\geq 2}\sum_{m=1}^{n-1}m^{-\phi}(m-n)^{-\theta}.\end{equation} Note also that $f$ and $g$ are bounded, and hence belong to the Banach space $\ell_\infty=\ell_\infty(\mathbb{N})$. (In fact, $f\in\ell_p$ ang $g\in\ell_q$ for any $p>1/\phi$ and $q>1/\theta$.)

Conjecture. $\|f*g\|_\infty<\infty$. In other words, $f*g\in\ell_\infty$ (the space of bounded sequences).

Discussion. Young's inequality only seems to work for certain choices of $\phi,\theta$, but I need the sup norm to be finite for arbitrary $\phi,\theta$. By breaking up the sum into two separate sums (with some "wisely-chosen" breaking point $j$), the above conjecture follows from this other conjecture, which I asked about yesterday. But I thought I might be more likely to get a response if I phrased it in the language of convolutions. As explained in the link, my motivation for this is to prove that a certain Banach space I constructed fails to be superreflexive.

Thanks guys!

• for $f \ast g$ to be bounded, you need for example $f \in l_1$ and $g \in l_\infty$. you know the implication $l_p$ implies $l_{p+a}$ for all $a \ge 0$ (this is true only for sequences, not for functions). – reuns Jan 11 '16 at 18:47

You can bound $(f\ast g)_n$ below by

$$\sum_{m = 1}^{n-1} m^{-\phi} (n-m)^{-\psi} \geqslant \sum_{m = 1}^{n-1} (n-1)^{-\phi}(n-1)^{-\psi} = (n-1)^{1 - \phi - \psi}.$$

So a necessary condition for $f\ast g \in \ell_{\infty}$ is $\phi + \psi \geqslant 1$. That is also sufficient, as can be seen by splitting the sum at $n/2$:

\begin{align} \sum_{m = 1}^{n-1} m^{-\phi}(n-m)^{-\psi} &\leqslant (n/2)^{-\psi}\sum_{m = 1}^{\lfloor n/2\rfloor} m^{-\phi} + (n/2)^{-\phi} \sum_{m = \lfloor n/2\rfloor+1}^{n-1} (n-m)^{-\psi}\\ &\leqslant C(n/2)^{-\psi}\cdot (n/2)^{1-\phi} + C (n/2)^{-\phi}\cdot (n/2)^{1-\psi}\\ &= K\cdot n^{1-\phi - \psi}, \end{align}

using the asymptotic behaviour

$$\sum_{m = 1}^k m^{-\alpha} \in \Theta(k^{1-\alpha})$$

for $\alpha < 1$.

• Ah, okay cool. Thank you! – Ben W Jan 11 '16 at 18:57