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## January 29, 2011

### Harmonic Analysis Lecture Notes 9

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 12:17 am

Definition We define the Hardy-Littlewood Maximal function of ${f}$ to be

$\displaystyle [Mf](x) = \sup_{r > 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| \; dy$

Theorem Let ${\omega : {\mathbb R}^d \rightarrow (0,\infty)}$ be a “weight” and associate a measure via ${\omega(E) = \int_E \omega \; dx}$. For measure ${f : {\mathbb R}^d \rightarrow {\mathbb C}}$,

1. ${||Mf||_\infty \leq ||f||_\infty}$, true also for ${L^\infty (\omega dx)}$.
2. (Weak-type 1,1 bound) ${\omega ( \{ x : Mf > \lambda \} ) \leq \frac{4^d}{\lambda} \int |f(x)| [M\omega](x) \; dx}$
3. ${||Mf||_{L^p(\omega \; dx)} \lesssim_{p,d} ||f||_{L^p([M\omega] \; dx)}.}$

Remark

1. Assertion (1) is trivial and assertion (3) follows from (2) and Marcinkiewicz interpolation.
2. If ${\omega \equiv 1}$, then ${M \omega \equiv 1}$, and in this case we recover the result from a standard real analysis course.
3. ${(2) \equiv \int |Mf(x)|^p \; \omega(x)dx \lesssim \int |f(x)|^p \; [M\omega](x)dx}$
4. If ${M\omega \lesssim \omega}$, we say that ${\omega}$ obeys the ${A_1}$-condition and ${M : L^1(\omega dx) \rightarrow L^{1,\infty}(\omega dx)}$ as well as ${L^p(\omega dx) \rightarrow L^p(\omega dx)}$. The result ${M : L^1(\omega dx) \rightarrow L^{1,\infty}(\omega dx)}$ is only true when ${\omega \in A_1}$ (note that ${[M\delta_{x_0}](x) \approx |x-x_0|^{-d}}$, ${\{ M\delta_{x_0} > \lambda \} =\{ |x - x_0 \lesssim \lambda^{-1/d}\}}$ and so ${\omega(\{ |x-x_0| \lesssim \lambda^{-1/d}\}) = \omega\big(B(x_0, \lambda^{-1/d})\big) \lesssim \lambda^{-1} \omega(x_0)}$). A necessary and sufficient condition for ${M : L^p(\omega dx) \rightarrow L^p(\omega dx)}$ is known and is called the ${A_p}$ condition:

$\displaystyle \left[\frac{1}{|B|} \int_B \omega dx \right] \left[ \frac{1}{|B|} \int_B \omega^{-\frac{p'}{p}} dx \right]^{\frac{p}{p'}} \lesssim 1$

uniformly over all balls ${B}$. Note ${A_p \subseteq A_q}$ when ${p \leq q}$. It is know that if ${\omega A_p}$, then ${\omega \in A_{p - \epsilon}}$ for some ${\epsilon(\omega)> 0}$ (cf Reverse Hölder’s inequality).

5. Let ${(X,d,\mu)}$ be a metric measure space with a Borel measure ${\mu}$. The proof will show that when ${\omega \equiv 1}$, then

$\displaystyle M_\mu f := \sup_{r > 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f(y)| \; d\mu(y)$

maps ${L^1(d\mu) \rightarrow L^{1,\infty}(d\mu)}$ provided ${\mu}$ is doubling, i.e., ${\mu(B(x,2r)) \lesssim \mu(B(x,r))}$ uniformly in ${x}$ and ${r}$.

a.e convergence is equiv to boundedness of some sort of a maximal operator

Lemma (Vitali-type covering lemma) Given a finite collection of balls ${B_j}$, there is a subcollection ${S}$ that is disjoint and

$\displaystyle \bigcup_{\scriptsize\begin{matrix}\text{all}\\\text{balls}\end{matrix}} B \subseteq \bigcup_{S} 3B$

where ${B = B(x,r)}$ then ${3B = B(x,3r)}$.

Proof: Run the following algorithm

1. Take the largest (or equal) ball remaining and add that to ${S}$.
2. Discard (forever more) any ball that intersects our chosen one.
3. If balls remain, go back to step 1, else stop.

By construction, the balls in ${S}$ are disjoint. Also, all of the balls that meet the chosen one ${B}$ in a particular iteration must be smaller than ${B}$, and so ${3B}$ will be contain ${B}$ and all of its neighboring balls. $\Box$

Proof: (of Assertion 2) By inner regularity it suffices to control the measure of an arbitrary compact ${K \subseteq \{Mf> \lambda\}}$, the latter of which is an open set. Note that if ${x \in K}$ then there is an ${r_x}$ so that

$\displaystyle \int_{B(x,r_x)} |f| \; dy > \lambda |B(x,r_x)| \hbox{\hskip 68pt} (*)$

By compactness, we can cover ${K}$ by finitely many of these balls. We then apply the covering lemma to get a disjoint family ${B_j = B(x_j,r_j)}$ with ${K \subseteq \cup 3B_j}$.

$\displaystyle \begin{array}{rcl} \omega(3B_j) &=& \int_{3B_j} \omega(z) \; dz \\ &\leq& \displaystyle \int_{y,4r_j} \omega(z) \; dz \hbox{\hskip 18pt for all }y \in B_j \\ \\ &\leq& 4^d |B_j| [M\omega](y). \end{array}$

Now ${\omega(K) \leq \sum \omega(3B_j) \underbrace{\leq}_{\text{by }(*)} \sum \frac{\omega(3B_j)}{|B_j|} \lambda^{-1} \int_{B_j} |f| \leq 4^d \sum_j \lambda^{-1} \int_{B_j} |f| [M\omega] \; dy \leq \frac{4^d}{\lambda} \int_{{\mathbb R}^d} \; |f| [M\omega] }$ $\Box$