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## 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.

Theorem

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 ${1 < p < \infty}$.

Remark

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.

Remark

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. $\Box$

Morals

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 ${\lambda}$

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

and

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

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

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

Therefore

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

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, ${[Mb_j](x) \lesssim [Mb_j^{av}](x)}$.

${\Box}$

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