$f$ is increasing implies $Mf$ is increasing ???

Question

For a given function $f:\mathbb{R} \rightarrow \mathbb{R} $, Hardy–Littlewood maximal function of $f$ defined by $$Mf(x)=\sup_{r>0}\frac{1}{2r}\int\chi_{[x-r,x+r]}|f|~d\lambda=\sup_{r>0}\frac{1}{2r}\int^{x+r}_{x-r}|f|~d\lambda$$ where $x\in \mathbb{R}$ and $\lambda$ is the Lebesgue.

Is it ture that if $f$ is increasing then $Mf$ is also increasing?

Answer 1

No, because of the absolute values. Say $f = -2$ on $(-\infty,0]$ and $f = 0$ on $(0,\infty)$. Then $Mf(-1) = 2$ while $Mf(1) = 0$.

However, if $f$ is non-negative, then the result is true: If $y < x$, then for each $r > 0$, $\frac{1}{2r}\int_{y-r}^{y+r} f(t)dt \le \frac{1}{2r}\int_{x-r}^{x+r} f(t)dt$. Now $\sup_{r > 0}$ both sides.