We know that the maximal operator is bounded on $L^{p}(\mathbb{R}^{n})$ where $n\geq 1$ and $1<p<\infty$ and the proof would be contained in many classical harmonic analysis books. Here I find a much more different method to verify this when $n=1$ and $p=2$ (This method would be found in the book "Some Topics in Dyadic Harmonic Analysis" by Michael T.Lacey, pp.13-14).

The general ideal is that since $L^{2}(\mathbb{R})$ is a Hilbert space, therefore, the bounded linear operator $T$ on $L^{2}(\mathbb{R})$ has the property $\left \| T \right \|^{2}\leq \left \| TT^{\star} \right \|$. Thus the essential thing is to construct a proper bounded linear operator $T$ on $L^{2}(\mathbb{R})$ such that $Mf(x)\leq CTf(x)$ where $C>0$ is a fixed constant and $f(x)$ is positive. This operator has been given by the form in Lacey's book: \begin{align*} Tf(x)=\sum_{I\in \mathfrak{D}}\frac{\chi_{E(I)}}{|I|}\int_{I}f(y)dy \end{align*} where $\mathfrak{D}$ denotes the set of dyadic intervals on $\mathbb{R}$ and for each $I\in \mathfrak{D}$ associate $E(I)\subseteq I$ so that the sets $\{E(I): I\in \mathfrak{D}\}$ are disjoint subsets. Lacey points out that $Mf(x)\leq 2Tf(x)$ in his book.(I could not verify this even if taking $C$ instead of $2$.)

To Bound the maximal operator, we calculate \begin{align*} TT^{\star}f(x)=\sum_{I\in \mathfrak{D}}\sum_{J\in \mathfrak{D}}\chi_{E(I)}\frac{<\chi_{I},\chi_{J}>}{|I|}\frac{<\chi_{E(J)},f>}{|J|}\leq Tf(x)+T^{\star}f(x). \end{align*} Therefore, from the inequalities \begin{align*} \left \| Tf \right \|^{2}_{2}\leq \left \| T \right \|^{2}\left \| f \right \|^{2}_{2}\leq \left \| TT^{\star} \right \|\left \| f \right \|^{2}_{2}\leq 2\left \| T \right \|\left \| f \right \|^{2}_{2} \end{align*} we have $\left \| T \right \|\leq 2$ and then $\left \| Mf \right \|_{2}\leq C\left \| Tf \right \|_{2}\leq 2C\left \| f \right \|_{2}$.

However, for the key step, the description of the choice of $E(I)$ in Lacey's book seems not too clear (maybe to me) . Hence, $\mathbf{my \ question}$ is:

How to choose a desirable set $\{E(I): I\in \mathfrak{D}\}$ such that the pointwise inequality $Mf(x)\leq CT(x)$ is valid?


1 Answer 1


I have just obtained an answer to this question but there would have some modifications that differ from Lacey's description. The detail is as follows (forgive me to pose it here):

First, we should proof a Lemma (it is just as an exercise in Lacey's book, p.4) :

$\textbf{Lemma.}$ $\mbox{}$For any interval $I$, there is an interval $J\in \mathfrak{D}\cup \mathfrak{D'}$ with $I\subseteq J$ and $|J|\leq 8|I|$, where $\mathfrak{D'}$ is another choice of dyadic intervals defined by $\mathfrak{D'}=\left \{ \left [j2^{k},(j+1)2^{k}\right)+(-1)^{k}\frac{2^{k}}{3}: j,k\in \mathbb{Z}\right \}$.

Proof of the Lemma: $\mbox{}$ Suppose the length of $I$ satisfies $2^{k}\leq |I|<2^{k+1}$ with $k\in \mathbb{Z}$. We should only separate it into two cases. Firstly, $I\subseteq J$ where $J\in\mathfrak{D}$ and $|J|=2^{k+1} $. Then it is easy to get $|J|\leq 8|I|$. Secondly, on the other hand, there must exists only one point $j_{0}2^{k+1}$ contained in $I$. We denote $J_{1}=\left[j_{0}2^{k+1},(j_{0}+1)2^{k+1}\right)$ and $J_{s}=J_{1}+(s-1)2^{k+1}$, $s\in \mathbb{N}$, in which $J_{1}$ is the right interval which has a nonempty intersection with $I$. Note that there exists only one interval $J'\in \mathfrak{D}$ such that $J'\supseteq J_{1}$ with $|J'|=2^{k+3}$. Then, if $J'\bigcap J_{4}=\varnothing$, we choose $J=J'\in \mathfrak{D}$ and the $J$ satisfies the Lemma. If $J'\bigcap J_{4}\neq \varnothing$ (hence $J_{4}\subseteq J'$), from the two inequalities $2^{k+1}<\frac{2^{k+3}}{3}$ and $\frac{2^{k+3}}{3}<2^{k+3}-2^{k+1}$, we have $J'-\frac{2^{k+3}}{3}\supseteq I$. Suppose that the interval $J''\in \mathfrak{D}$ is the left one which is adjacent to $J'$ with legth $|J''|=2^{k+3}$. Also we have $J''+\frac{2^{k+3}}{3}\supseteq I$. Hence, there must exists a $J\in \mathfrak{D'}$ with length $|J|=2^{k+3}$ such that $J\supseteq I$ and $|J|\leq 8|I|$. Therefore we complete the proof of the Lemma.

Now we can construct the set $\left \{E(I)\right \}$ and the operator $T$:

For any fixed $x\in \mathbb{R}$, suppose that $Mf(x)<\infty$, there exists an interval $Q\ni x$ such that $Mf(x)\leq \frac{2}{|Q|} \int_{Q}f(y)dy$, (recall that $f\geq 0$). By Lemma, there exists an $I_{x}\in \mathfrak{D}\bigcup \mathfrak{D'}$ such that $I_{x}\supseteq Q$ and $|I_{x}|\leq 8|Q|$. Hence $Mf(x)\leq \frac{16}{|I_{x}|} \int_{I_{x}}f(y)dy$. We relabel $\left \{ I_{x}:x\in\mathbb{R} \right \}$ as $\left \{I_{j}:j\in\mathbb{N} \right \}$. For each $j$, we set $$\tilde{E}(I_{j})=\left \{ x\in I_{j}:Mf(x)\leq \frac{16}{|I_{j}|}\int_{I_{j}}f(y)dy \right \}$$ and let $$E(I_{j})=\tilde{E}(I_{j})\backslash \bigcup_{k=1}^{j-1}\tilde{E}(I_{k})$$ for $j>1$, and $E(I_{1})=\tilde{E}(I_{1})$.

We can easily get $\bigcup_{k=1}^{\infty}E(I_{k})=\mathbb{R}$ and $E(I_{j_{1}})\bigcap E(I_{j_{2}})= \varnothing$, if $j_{1}\neq j_{2}$. Analogously, denote $$Tf(x)=\sum_{j=1}^{\infty}\frac{\chi_{E(I_{j})}(x)}{|I_{j}|}\int_{I_{j}}f(y)dy.$$ The most important thing is that the similar inequality $$Tf(x)\leq Mf(x)\leq 16Tf(x)$$ is now still valid for $x\in \mathbb{R}$.

$\textbf{Supplement}$ $\mbox{}$ The operator $T$ defined in the above is a little different from the one in Lacey's book. However, the rest proof for the boundedness of maximal operator $M$ on $L^{2}(\mathbb{R})$ is still followed by the method of Lacey's.

I have modified some mistakes in the proof of the Lemma. Now the proof will become more clear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.