January 31, 2011

Harmonic Analysis Lecture Notes 10

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:54 pm

The Vector Maximal Function

Definition Given {f : {\mathbb R}^d \rightarrow \ell^2({\mathbb Z})} with component functions {f_j}, define (the scalar-valued function)

\displaystyle  	[\bar M f] (x) = \sqrt{ \sum_j |Mf_j|^2}

Note that each function gets to choose its own maximizing radius at each point.


  1. {\big| \{ \bar M f > \lambda \} \big| \lesssim \frac{1}{\lambda} ||f||_{L^1}}, where {||f||_{L^1} = \int ||f||_{\ell^2({\mathbb Z})} \; dx}.
  2. {||Mf||_{L^p} \lesssim ||f||_{L^p(\ell^2)}} for {1 < p < \infty}.


  1. {\bar M} is not bounded on {L^\infty}. Indeed, consider {f_j : {\mathbb R} \rightarrow {\mathbb R}} defined by {f_j(x) = \chi_{[2^{j-1},2^{j})}(x)} so that {||f(x)||_{\ell^2} = 1} at each {x}. Now if {x \in [0,1]} and {j \geq 1}, then {[Mf_j](x) \geq \frac{1}{2^{j+1}} \cdot 2^{j-1} = \frac{1}{4}}. Thus {\bar M f = \infty } on {[0,1]}.
  2. For this operator it is the {L^2} bound that is easy:

    \displaystyle  		\int |\bar M f|^2 \; dx = \int \sum_j \big| [Mf_j] (x) \big|^2 \; dx 		\underbrace{\lesssim}_{L^2 \text{ bound for } M} \sum \int |f_j|^2 \; dx = \int ||f||^2_{\ell^2} \; dx

Before going on to the proof of the theorem, we first introduce some definitions,

Definition The family of dyadic cubes is

\displaystyle  \begin{array}{rcl}  	\displaystyle \Big\{ [ k_1 \cdot 2^n , (k_1 + 1) 2^n) \times [ k_2 \cdot 2^n , (k_2 + 1) 2^n) \times \cdots \times [ k_d \cdot 2^n , (k_d + 1) 2^n) \subseteq {\mathbb R}^d \\ 	\displaystyle : n \in {\mathbb Z}, (k_1, \ldots, k_d) \in {\mathbb Z}^d\Big\} 	\end{array}

Note that the cubes of each generation (i.e., of fixed side-length {2^n}) tile {{\mathbb R}^d}. The nice nesting properties of the dyadic cubes make them very convenient for covering and/or stopping time arguments (cf. if {Q_1 \cap Q_2 \neq \emptyset}, then {Q_1 \subseteq Q_2} or vice versa). Let

\displaystyle  	\mathcal{F}_n = \sigma ( \text{cubes of side length} = 2^{-n})

be the {\sigma}-algebra generated by cubes of side-length {2^{-n}}. It is equal to {\{\text{unions of these cubes}\}}. Note that {\mathcal{F}_{n} \subseteq \mathcal{F}_{n+1}}, which makes this a filtration. Note that {\mathop{\mathbb E}(f \mid \mathcal{F}_n)(x) = \frac{1}{|Q|} \int_Q f \; dx} where {Q \ni x} and {|Q| = 2^{-nd}} [{f_n = \mathop{\mathbb E}(f \mid \mathcal{F}_n)} is measurable with respect to {\mathcal{F}_n} and {\int fg = \int f_n g} whenever {g \in L^\infty(\mathcal{F}_n)}]. As {\mathop{\mathbb E}(f_{n+1} \mid \mathcal{F}_n) = f_n} we say {f_n} form a Martingale.


  1. Note that by Lebesgue differentiation {f_n \rightarrow f} Lebesgue a.e.
  2. Honest martingales are rare in analysis, but approximate martingales abound. For example, say {u} is harmonic on {\mathbb{D}}, and {u \equiv f} on {\partial \mathbb{D}} and consider {u(re^{i\theta}) \stackrel{r\nearrow 1}{\longrightarrow} f}.

Lemma (A Calderón–Zygmund-type decomposition) Suppose {f : {\mathbb R}^d \rightarrow \ell^2} is {L^1}. Then there is a family {\{Q_k\}} of disjoint dyadic cubes with {| \mathaccent\cdot\cup Q_k| \lesssim \frac{1}{\lambda} ||f||_{L^1}} so that

\displaystyle  \begin{array}{rcl}  	&&g = f \cdot \chi_{[\mathaccent\cdot\cup Q_k]^c} \text{ has } |g| \lesssim \lambda \\ 	&&b = f \cdot \chi_{\mathaccent\cdot\cup Q_k} \text{ has } \int_{Q_k} |b| \approx \lambda |Q_k| 	\end{array}

Proof: If {Q} is large enough then

\displaystyle  	\int_Q |f| < \lambda |Q|

e.g. {|Q| > \frac{1}{\lambda} \int_{{\mathbb R}^d} |f|}. Beginning with the tessellation of cubes of that size, we subdivide: if any cubes has

\displaystyle  	\int_Q |f| \; dx \geq \lambda |Q|

then we put {Q} into our collection {\{Q_k\}} and do not subdivide it any further. If a cube {Q_k} ended up in our collection, means that its parent did not:

\displaystyle  	\lambda |Q_k| \leq \int_{Q_k} |f| \leq \int_{parent} |f| < \lambda 2^d|Q_k|

Summing in {k}, we get {\sum |Q_k| \leq \frac{1}{\lambda} \int |f|}. By Lebesgue differentiation, {\displaystyle \lim_{\tiny \begin{matrix} |Q| \rightarrow 0 \\ Q \ni x\end{matrix}} \frac{1}{|Q|} \int |f| \; dx \rightarrow |f|} a.e. Off the cubes {\{Q_k\}} those averages are all less than {\lambda} and so {|f| < \lambda} a.e. there. \Box


  1. {g} is good: it lies in {L^1 \cap L^\infty}, so we can use some kind of “trivial bound.”
  2. The bad part has small support. Indeed a neighborhood of this set has small measure.
  3. Far from {\cup Q_k}, {b} looks good.

Proof of assertion (1): Do a Calderón–Zygmund decomposition at height {\lambda}

\displaystyle  	| \{ \bar M f > \lambda \} | \leq | \{ \bar M g > \lambda/2\}| + |\{ \bar M b > \lambda/2\}|


\displaystyle  \begin{array}{rcl}  	|\{ \bar M g > \frac{\lambda}{2}\}| &\lesssim& \displaystyle \lambda^{-2} ||\bar M g ||_{L^2}^2 \hbox{\hskip 38pt} \scriptsize \begin{bmatrix}L^2 \subseteq L^{2,\infty}\end{bmatrix} \\ 	&\lesssim& \lambda^{-2} ||g||_{L^2}^2 \hbox{\hskip 48pt} \scriptsize \begin{bmatrix}\bar M : L^2 \rightarrow L^2 \end{bmatrix} \\ 	&\lesssim& \lambda^{-2} ||g||_{L^1} ||g||_{L^\infty} \\ 	&\lesssim& \lambda^{-1} ||f||_{L^1} 	\end{array}

which is acceptable. As {\sum |Q_k| \lesssim \lambda^{-1} ||f||_{L^1}} so we do not need to worry about {\bar M b} on {\cup (2 Q_k)} (which is not a dyadic cube). We need to estimate

\displaystyle  	\Big| \{ x \not \in \cup(2Q_k) : \bar M b > \frac{\lambda}{2} \} \Big|

What would happen if we replaced {b} by {b^{av}} defined via

\displaystyle  	b_j^{av} = \sum_k \frac{\chi_{Q_k}}{|Q_k} \int_{Q_k} b_j \; dx

  1. For {x \in Q_k}, then

    \displaystyle  \begin{array}{rcl}  		|b^{av}(x) | &=& \displaystyle \sqrt{ \sum |b_J^{av}(x)|^2} \\ \\ 		&=& \displaystyle \sqrt{ \sum \big| \frac{1}{|Q_k|} \int_{Q_k} b_j(y) \; dy \big|^2} \\ \\ 	\scriptsize \begin{bmatrix} 	Jensen: \Phi : \vec b \mapsto || \vec b|| \text{convex} \\ \\ \mathop{\mathbb E}(\Phi(X)) \geq \Phi(\mathop{\mathbb E}(X)) 	\end{bmatrix}\:\:\:	&\leq& \frac{1}{|Q_k|} \int_{Q_k} \sqrt{ \sum_j |b_j(y)|^2} \; dy 		\end{array}


    \displaystyle  		||b^{av}||_{L^1} = \sum_k \int_{Q_k} |b^{av}(x)| \leq \sum_k |Q_k| \frac{1}{|Q_k|} \int_{Q_k} |b| \; dy \leq ||b||_{L^1}

  2. If {\bar M b(x) > \frac{\lambda}{2}} for some {x \not \in \cup 2 Q_k}, then is {\bar M b^{av} \gtrsim \lambda}? Well, {x \not \in \cup 2Q_k}, there exists {r_x} such that

    \displaystyle  \begin{array}{rcl}  		[Mb_j](x) &\lesssim& \displaystyle \frac{1}{|B(x,r_x)|} \int_{B(x,r_x)} |b_j| \; dy \\ \\ 		&\lesssim& \displaystyle \sum_k \frac{1}{|B(x,r_x)|} \int_{B(x,r_x) \cap Q_k} |b_j| \; dy \\ \\ 		&\lesssim& \displaystyle \sum_{Q_k \cap B(x,r_x) \neq \emptyset} \frac{1}{|B(x,r_x)|} \int_{Q_k} |b_j| \; dy \\ \\ 		&\lesssim& \displaystyle \sum_{Q_k \subseteq B(x,2r_x)} \frac{1}{|B(x,r_x)|} \int_{Q_k} |b^{av}| \; dy \\ \\ 		&\lesssim& \frac{2^d}{B(x,2r_x)} \int_{B(x,r2r_x)} |b^{av}| \; dy 		\end{array}

    (Misc notes: {x \not \in 2Q_k}, {Q_k \cap B(x,r_x) \neq \emptyset}, {r_x \gtrsim \text{side length}(Q_k)}) Therefore, {[Mb_j](x) \lesssim [Mb_j^{av}](x)}.



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