average of maximal function is less than its infimum?

Question

Let M be the dyadic Hardy-Littlewood maximal operator. Prove the following: there is a constant $C$ such that for any $f$, $$ \inf_{x\in I}Mf(x)\le C 2^k\inf_{x\in J} Mf(x) $$ where $I$ and $J$ are dyadic intervals with $I\subset J$ and $2^k|I|=|J|$ for some positive integer $k$ (i.e. $I$ is the $k$-th generation of $J$).

I noticed that this theorem follows from $$\frac{1}{|J|}\int_JMf\le C\inf_{x\in J}Mf(x). $$ This is an interesting phenomenon, but I am unable to prove it.

Answer 1

You're unable to prove it because it's not true. Take $J=[0,1]$ and $f=\mathbb 1_{[0;2^{-k}]}$ with $k$ an integer. Then you have $\displaystyle\min_{x\in J} Mf(x)=Mf(1)=2^{-k}$. Now we also have

$Mf(x)=1$ if $x\in [0,2^{-k}[$

$Mf(x)=1/2$ if $x\in [2^{-k},2^{-k+1}[$

$Mf(x)=1/2^2$ if $x\in [2^{-k+1},2^{-k+2}[$

...

$Mf(x)=1/2^j$ if $x\in [2^{-k+j-1},2^{-k+j}[$

...

$Mf(x)=1/2^{-k}$ if $x\in [1/2,1[$

So $\displaystyle\int_{[0;1]} Mf(x)dx=2^{-k}+\frac{1}{2}2^{-k}+\frac{1}{2^2}2^{-k+1}+\frac{1}{2^3}2^{-k+2}+\ldots+\frac{1}{2^{-k}}2^{-1}=2^{-k}+k\cdot2^{-k-1}$

and so $\displaystyle\min_{x\in J} Mf(x)=o\left(\displaystyle\int_{[0;1]} Mf(x)dx\right)$ when $k$ goes to infinity, so your last inequality can't hold for every $f$.

Remember that even if $f$ is bounded with compact support $Mf$ is not integrable, so $\displaystyle\int_{-r}^r Mf(x)dx$ goes to infinity when $r$ goesto infinity while $\displaystyle\int_{-r}^r |f(x)|dx$ stays bounded so they're not "comparable" : the integral of $f$ is a little $o$ of the integral of $Mf$ when $r$ goest to infinity. "Rescale" everything with the change of variable $u=rx$ and you have another proof (without calculations) that your inequality can't hold. In fact the first proof is just a particular case of the second one.

This doesn't say anything about the first inequality (which is most likely true since it's in a book) but i'm not sure this was your question.

Answer 2

So, I don't know how to prove the estimate you ask for in your original question, but I believe I can prove the conclusion on the bottom of pg. 25 in the reference you gave me. The writing was a little confusing to me because I think Thiele is double using the index $k$, but here's what I think we want to show: Let $I$ be a generation $k$ dyadic subinterval of the maximal dyadic subinterval $J$ of the exceptional set $F$. For all $j=1,\ldots,n$,

$$\dfrac{|\langle{g_{j},\phi_{I,j}}\rangle|}{|I|^{1/2}}\lesssim 2^{k}$$

I am assuming that you are familiar with the notation used here, as you are evidently reading Thiele's lecture notes (there should be a Google Books preview here). For the benefit of other readers, $g_{j}:=f_{j}/|E_{j}|$, where $f_{j}$ is a measurable function such that $|f_{j}|\leq 1_{E_{j}}$ a.e. Since $\phi_{I,j}$ is a ($L^{2}$-normalized) bump function adapted to the interval $I$, we have that \begin{align*} \dfrac{|\langle{g_{j},\phi_{I,j}}\rangle|}{|I|^{1/2}}&\lesssim \dfrac{2^{k}}{|J|}\int \dfrac{1_{E_{j}}}{|E_{j}|}1_{J}+\dfrac{2^{k}}{|J|}\int_{|y|\geq|J|}(|y|/|J|)^{-100}dy\\ &\lesssim 2^{k}\max_{1\leq l\leq n}\dfrac{|E_{l}\cap J|}{|E_{l}||J|} + 2^{k}\int_{|y|\geq 1}|y|^{-100}dy\\ \end{align*} Evidently the second term is $\lesssim 2^{k}$. Recall that $J$ is a maximal dyadic interval such that $$\max_{1\leq l\leq n}\dfrac{|E_{l}\cap J|}{|E_{l}||J|}\geq C,$$ where $C\geq 1$ is a fixed large constant (see pg. 24). By maximality, we have that $$\max_{1\leq l\leq n}\dfrac{|E_{l}\cap 2J|}{|E_{l}||2J|}\leq C\Rightarrow \max_{1\leq l\leq n}\dfrac{|E_{l}\cap J|}{|E_{l}||J|}\leq 2C$$ Putting these estimates together, we conclude that $$\dfrac{|\langle{g_{j},\phi_{I,j}}\rangle|}{|I|^{1/2}}\lesssim 2^{k}C,\quad\forall j=1,\ldots,n$$