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

Find the Hardy-Littlewood maximal function $Mf$ of the $\Bbb R^\Bbb R$ function $f=\chi_{[-1,1]}$.

How do we find $Mf(x)$ for $|x| > 1$?
I see that it should decrease like $1/x$, but I can't find the formula.

share|cite|improve this question
For future reference, it might be valuable to specify that this is the Hardy-Littlewood maximal function. There are actually many maximal functions that are studied in harmonic analysis. – Christopher A. Wong Nov 13 '12 at 3:41
up vote 4 down vote accepted

Since you're on $\mathbb{R}$, the maximal function looks like:

$$ Mf(x)=\sup_{B(x,\epsilon)}\frac{1}{2\epsilon}\int_{B(x,\epsilon)}\vert f\vert dx\\ =\sup_{\epsilon}\frac{1}{2\epsilon}\lambda(B(x,\epsilon)\cap [-1,1]) $$ where $\lambda$ is the Lebesgue measure.

If $\vert x\vert>1$, the sup should (though this would require an argument) occur when you take your ball to include all of $[-1,1]$ and nothing more - i.e. take $\epsilon=\vert x\vert+1$. Your maximal function will then be $Mf(x)=\frac{1}{\vert x\vert+1}$.

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.