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!

  • $\begingroup$ 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). $\endgroup$ – 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 $\alpha < 1$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.