# average of maximal function is less than its infimum?

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.

• Is this is a conjecture or an exercise? Ultimately, it doesn't matter; I only ask because I don't have a lot of spare time to devote to something which isn't immediately obvious to me and may not be true. – Matt Rosenzweig Nov 23 '15 at 15:30
• @MattRosenzweig The statement is true, but the lemma I posed is my conjecture. – Tony B Nov 23 '15 at 17:19
• What is the text/paper from which the statement is taken? – Matt Rosenzweig Nov 23 '15 at 17:20
• @MattRosenzweig ams.org/bookstore?fn=20&arg1=cbmsseries&ikey=CBMS-105 bottom of page 25 – Tony B Dec 10 '15 at 21:31
• In the reference you gave, this result is not explicitly stated--plus, the intervals $I$ and $J$ considered have some structure. – Matt Rosenzweig Dec 18 '15 at 0:30

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.

• There are two typos: $Mf(x)=1/2^k$ if $x\in[1/2,1[$ and same mistake appears in the next line. Very nice construction. I have a conjecture for a modified statement: math.stackexchange.com/questions/1590031/… – Tony B Dec 26 '15 at 23:02

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 $$\dfrac{|\langle{g_{j},\phi_{I,j}}\rangle|}{|I|^{1/2}}\lesssim 2^{k}C,\quad\forall j=1,\ldots,n$$

• Thanks! Can you understand Thiele's argument? The display in the bottom of page 25 consists of three inequalities and I can understand all but the middle one, which is this question. – Tony B Dec 23 '15 at 19:38
• @Dony Thiele's argument showing what? The estimate for the sum over intervals contained in F? I didn't follow his reasoning on the bottom of pg 25, which is why I came up with my own argument to show the conclusion on the bottom of pg. 25. – Matt Rosenzweig Dec 23 '15 at 19:59
• Thiele wants to show exactly the same inequality you mentioned in your answer. He achieved this by three steps, each with an equality. Firstly, the LHS can be controlled by inf over $I$ of Maximal function. See math.stackexchange.com/questions/1066910/… for details. Secondly, enlarge the range of inf to ancestor $J$, but pay a price of $2^k$ (this is where I don't understand). Lastly, use the property of $J$ (you also used this property in your proof) to bound inf by a constant. – Tony B Dec 23 '15 at 20:31
• @Dony I don't know how to get the middle inequality. I told you that in my answer. Am I not understanding what you're asking? – Matt Rosenzweig Dec 23 '15 at 20:33
• Thank you anyway. I think you understand my question very well. Your argument is very straightforward and already satisfactory, whereas Thiele used some commonly used standard estimates, which can be hard to follow for beginners. Both of you make use of the property of $J$ in the last step. The difference is the initial setup. – Tony B Dec 23 '15 at 20:37