control of an integral using maximal function Let $I$ be a compact interval with center $c(I)$ and N be a large positive integer. It seems to me that there exists a constant $C$ such that for any good function $f$ (e.g. Schwartz function) we have
$$
\frac{1}{|I|}\int\frac{|f(x)|}{(1+\frac{|x-c(I)|}{|I|})^N} dx\le C\inf_{x\in I}Mf(x)
,$$ where $Mf$ is the Hardy-Littlewood maximal function of $f$.
I was unable to prove it. I know that $\int\frac{1}{(1+\frac{|x-c(I)|}{|I|})^N} dx$ is almost the same as $\int_I\frac{1}{(1+\frac{|x-c(I)|}{|I|})^N} dx$ because of the fast decay outside $I$. But when there is an $f(x)$ on the top, I'm worried about the situation $f$ is supported far away from $I$. Any help is appreciated. 
 A: Let us first establish the case $I=\left[-1,1\right]$. This yields
$\lambda\left(I\right)=2$ and $c\left(I\right)=0$. Now, we use a
dyadic partitition of $\mathbb{R}$, i.e. we write
\begin{eqnarray*}
\int\frac{\left|f\left(x\right)\right|}{\left(1+\frac{\left|x-c\left(I\right)\right|}{\lambda\left(I\right)}\right)^{N}}\,{\rm d}x & \lesssim & \int_{\left[-1,1\right]}\frac{\left|f\left(x\right)\right|}{\left(1+\left|x\right|\right)^{N}}\,{\rm d}x+\sum_{n=1}^{\infty}\int_{2^{n-1}\leq\left|x\right|\leq2^{n}}\frac{\left|f\left(x\right)\right|}{\left(1+\left|x\right|\right)^{N}}\,{\rm d}x\\
 & \lesssim & \int_{\left[-1,1\right]}\left|f\left(x\right)\right|\,{\rm d}x+\sum_{n=1}^{\infty}\frac{1}{2^{\left(n-1\right)N}}\int_{2^{n-1}\leq\left|x\right|\leq2^{n}}\left|f\left(x\right)\right|\,{\rm d}x,
\end{eqnarray*}
where all implied constants only depend on $N\geq0$.
Now, let $y\in I=\left[-1,1\right]$ be arbitrary. For $2^{n-1}\leq\left|x\right|\leq2^{n}$,
this yields
$$
\left|x-y\right|\leq\left|x\right|+\left|y\right|\leq2^{n}+1\leq2^{n+1}
$$
and hence $\left\{ x\in\mathbb{R}\,\mid\,2^{n-1}\leq\left|x\right|\leq2^{n}\right\} \subset B_{2^{n+1}}\left(y\right)$
(up to sets of measure zero) and thus
\begin{eqnarray*}
\int_{2^{n-1}\leq\left|x\right|\leq2^{n}}\left|f\left(x\right)\right|\,{\rm d}x & \leq & \int_{B_{2^{n+1}}\left(y\right)}\left|f\left(x\right)\right|\,{\rm d}x\\
 & \leq & \lambda\left(B_{2^{n+1}}\left(y\right)\right)\cdot\left(Mf\right)\left(y\right)\\
 & = & 2^{n+2}\cdot\left(Mf\right)\left(y\right).
\end{eqnarray*}
For the first term, we get $\left[-1,1\right]\subset B_{2}\left(y\right)$
(up to sets of measure zero) and hence similarly
$$
\int_{\left[-1,1\right]}\left|f\left(x\right)\right|\,{\rm d}x\leq4\cdot\left(Mf\right)\left(y\right).
$$
Putting everything together, we arrive at
\begin{eqnarray*}
\int\frac{\left|f\left(x\right)\right|}{\left(1+\frac{\left|x-c\left(I\right)\right|}{\lambda\left(I\right)}\right)^{N}}\,{\rm d}x & \lesssim & 4\cdot\left(Mf\right)\left(y\right)+\sum_{n=1}^{\infty}\frac{2^{n+2}}{2^{\left(n-1\right)N}}\cdot\left(Mf\right)\left(y\right)\\
 & = & \left(Mf\right)\left(y\right)\cdot\left[4+2^{2+N}\sum_{n=1}^{\infty}\left(2^{\left(1-N\right)}\right)^{n}\right]\\
 & \lesssim & \left(Mf\right)\left(y\right)
\end{eqnarray*}
for $N>1$. Again, all implied constant only depend on $N>1$. As
$y\in I=\left[-1,1\right]$ was chosen arbitrarily, this establishes
the claim for $I=\left[-1,1\right]$ and $N>1$.

The problem is now in generalizing this result (or the proof) to the
case of arbitrary intervals $I$. My approach here would be to define
$$
f_{a,b}\left(x\right):=f\left(a\left(x-b\right)\right)
$$
for $a>0$ and $b,x\in\mathbb{R}$. A direct calculation using the
change of variables formula shows that
$$
\left(Mf_{a,b}\right)\left(x\right)=\left(Mf\right)\left(a(x-b)\right).
$$
Furthermore, we have
$$
\max\left\{ 1,\left(\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)^{N}\right\} \leq\left(1+\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)^{N}\leq\left[2\cdot\max\left\{ 1,\frac{\left|x-c\right|}{\lambda\left(I\right)}\right\} \right]^{N}=2^{N}\cdot\max\left\{ 1,\left(\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)^{N}\right\} 
$$
and hence
\begin{eqnarray*}
\frac{1}{\lambda\left(I\right)}\cdot\int\frac{\left|f\left(x\right)\right|}{\left(1+\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)}\,{\rm d}x & \asymp & \frac{1}{\lambda\left(I\right)}\cdot\int\frac{\left|f\left(x\right)\right|}{\max\left\{ 1,\left(\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)^{N}\right\} }\,{\rm d}x\\
 & = & \frac{1}{\lambda\left(I\right)}\cdot\int\frac{\left|f\left(\frac{x-c}{\lambda\left(I\right)}\cdot\lambda\left(I\right)+c\right)\right|}{\max\left\{ 1,\left(\frac{\left|x-c\right|}{\lambda\left(I\right)}\right)^{N}\right\} }\,{\rm d}x\\
 & \overset{\omega=\frac{x-c}{\lambda\left(I\right)}}{=} & \int\frac{\left|f\left(\lambda\left(I\right)\omega+c\right)\right|}{\max\left\{ 1,\omega^{N}\right\} }\,{\rm d}\omega\\
 & \asymp & \int\frac{\left|f\left(\lambda\left(I\right)\left(\omega-\frac{-c}{\lambda\left(I\right)}\right)\right)\right|}{\max\left\{ 1,\left(\frac{\left|\omega-0\right|}{\lambda\left(\left[-1,1\right]\right)}\right)^{N}\right\} }\,{\rm d}\omega\\
 & = & \int\frac{\left|f_{\lambda\left(I\right),-c/\lambda\left(I\right)}\left(\omega\right)\right|}{\max\left\{ 1,\left(\frac{\left|\omega-0\right|}{\lambda\left(\left[-1,1\right]\right)}\right)^{N}\right\} }\,{\rm d}\omega\\
 & \overset{\left(\ast\right)}{\lesssim} & \inf_{y\in\left[-1,1\right]}\left(Mf_{\lambda\left(I\right),-c/\lambda\left(I\right)}\right)\left(y\right)\\
 & = & \inf_{y\in\left[-1,1\right]}\left(Mf\right)\left(\lambda(I)(y+\frac{c}{\lambda\left(I\right)})\right)\leq \inf_{y\in I} Mf(y),
\end{eqnarray*}
where $\left(\ast\right)$ indicates an application of the result
for $I=\left[-1,1\right]$.
