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)}.}


  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


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