Sequence of smooth functions whose image under a maximal operator diverges in $L_p$ norm

Question

For functions $f: \mathbb{R} \rightarrow \mathbb{C}$, define $$ M f(x) = \sup_{t >0} \frac{1}{2}| f(x+t) + f(x-t) |. $$

Given $p \geq 1$, I want to construct a sequence of smooth functions $f_n$ such that $||f_n||_p \leq 1$, $|| M f_n ||_p < \infty$, but $|| M f_n ||_p \rightarrow \infty$.

Is it even possible? What if the smoothness condition is relaxed?

In an earlier version of the question, I forgot the $|| M f_n ||_p < \infty$ condition.

Answer 1

This is not possible (updated question). If $0\lt p\lt \infty$, $f$ is in $L^p$, and $Mf$ is in $L^p$, then $f$ is zero a.e.. In other words, if $0\lt\int_\mathbb{R}|f|^p\lt\infty$, then $\int_\mathbb{R}|Mf|^p=\infty$ (and in fact, |Mf| has a positive lower bound outside of some bounded interval).

Suppose that $0\lt\|f\|_p\lt\infty$. There is some bounded interval $[a,b]$ with $\int_a^b|f|^p\gt0$. WLOG (rescale and shift) assume $[a,b]=[-1,0]$. Thus the essential supremum of $|f|$ on $[-1,0]$ is positive, so there is a positive number $c$ and a set $E\subseteq[-1,0]$ of positive measure $m$ such that $|f(x)|\gt c$ for all $x\in E$. There is a $K\gt 0$ such that the measure of the set $\{x>K:|f(x)|\geq \frac{c}{2}\}$ is less than $\frac{m}{2}$. For $x\gt K/2$, as $t$ ranges over the interval $[x,x+1]$, $x-t$ ranges over $[-1,0]$ while $x+t$ is contained in $(K,\infty)$. Thus the set of such $t$ with $|f(x-t)|>c$ and $|f(x+t)|<\frac{c}{2}$ has measure greater than $\frac{m}{2}\gt 0$ (and in particular is nonempty, but this would also work if you changed the sup to an essential sup). Therefore $|Mf(x)|\gt \frac{c}{4}$ for all $x\gt K/2$.

Answer 2

hmm... $M$ doesn't even map $L^p$ to $L^p$, never mind continuously. Take $f$ to be the characteristic function of $[-1/2,1/2]$. For any $p$, its $L^p$ norm is 1. $Mf(x) = 1$ if $x\in (-1/2,1/2)$ and equals $1/2$ other wise. It it not in $L^p$ for any $p$ other than infinity. By definition $\|Mf(x)\|_\infty < \|f\|_\infty$.

In fact, for any function of compact support, $Mf$ is not in any $L^p$ except possibly $L^\infty$. Are you sure you have the definition correct?