Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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 $x\gt K/2$.

share|cite|improve this answer
That's a really nice argument! I tried a little in this direction but I couldn't get it to work. – t.b. Feb 11 '11 at 22:32

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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