# Welcome.

## February 9, 2011

### Harmonic Analysis Lecture Notes 13

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:57 pm

Proposition Define ${K_\epsilon(x) = \chi_{|x| > \epsilon}K(x)}$ with ${K}$ being a Calderón–Zygmund kernel. Then

$\displaystyle ||K_\epsilon * f||_{L^2({\mathbb R}^d)} \lesssim ||f||_{L^2({\mathbb R}^d)}$

uniformly as ${\epsilon \downarrow 0}$. Consequently,

$\displaystyle K * f := \lim_{\epsilon \rightarrow 0} K_\epsilon * f$

extends from ${f \in \mathcal{S}({\mathbb R}^d)}$ to a bounded operator on ${L^2}$.

Proof: For the first part, we just need to show that ${|\hat K_\epsilon(\xi)| \lesssim 1}$ uniformly in ${\epsilon}$. Let’s compute (more honestly we ought to do some mollification near infinity)

$\displaystyle \begin{array}{rcl} \hat K_\epsilon(\xi) &=& \displaystyle \int_{|x| > \epsilon} e^{-2\pi i x \cdot \xi} K(x) \; dx \\ \\ &=& \displaystyle \int_{\epsilon < |x| < |\xi|^{-1}} [e^{-2\pi i x \cdot \xi} \underbrace{- 1]}_{\tiny\begin{matrix} K\text{ has mean}\\ \text{zero on spheres}\end{matrix}} K(x) \; dx+ \int_{|\xi|^{-1} < |x|} \left[\frac{1}{2} e^{-2\pi i x\cdot \xi} - \frac{1}{2} e^{-2\pi i (x + \frac{\xi}{2|\xi|^2}) \cdot \xi}\right]K(x) \; dx \end{array}$

So

$\displaystyle |\hat K_\epsilon(\xi)| \lesssim \int_{\epsilon < |x| < |\xi|^{-1}} |x| \cdot |\xi| \; |x|^{-d} \; dx + \bigg| \int_{s} \frac{1}{2} e^{-2\pi i x \cdot \xi} \left[ K(x) - K(x - \frac{\xi}{2|\xi|^2})\right] \; dx \bigg| + \bigg| \int_{B_1 \triangle B_2 \subseteq |x| \sim |\xi|^{-1}} |\xi|^{d} \; dx \bigg|$

where ${B_1 = B(0,|\xi|^{-1})}$ and ${B_2 = B(\frac{\xi}{2|\xi|^2}, 2|\xi|^{-1})}$ are the balls coming from two displays previous. Expanding each term in the right hand side of the previous display

$\displaystyle |\hat K_\epsilon(\xi)| \lesssim 1 + \underbrace{1}_{\tiny \begin{matrix}\text{property (3)}\\ \text{with }y= \frac{\xi}{2|x|^2} \end{matrix}} + 1$

If ${f \in \mathcal{S}}$ and ${0 < \epsilon_1 < \epsilon_2 < 1}$

$\displaystyle \begin{array}{rcl} |(K_{\epsilon_1} * f)(x) - (K_{\epsilon_2} * f) (x) | &=& \displaystyle \bigg| \int_{\epsilon_1 < |y| < \epsilon_2} K(y) f(x-y) \; dy \bigg| \\ \\ &=& \displaystyle \bigg| \int_{\epsilon_1 < |y| < \epsilon_2} K(y) [ f(x-y) - \underbrace{f(x)]}_{\tiny\begin{matrix} K\text{ has mean}\\ \text{zero on spheres}\end{matrix}} dy \bigg| \\ \\ &\lesssim& \displaystyle (1 + |x|^2)^{-13498273d} \int_{|y| \leq \epsilon_2} \frac{|y|}{|y|^d} \; dy \\ \\ &\lesssim& \epsilon_2 (1 + |x|^2)^{-13498273d} \end{array}$

and so ${\int |\cdots|^2 \; dx \lesssim \epsilon_2}$. Thus ${K_\epsilon * f}$ is Cauchy in ${L^2}$ and so it converges in ${L^2}$. To extend to general ${f}$ (i.e., not Schwartz), we approximate and use ${||K_\epsilon * (f-g)||_{L^2} \lesssim || f- g||_{L^2}}$ uniformly in ${\epsilon}$. $\Box$

Theorem If ${K}$ is a Calderón–Zygmund kernel, then

1. ${\displaystyle|\{ |K *f | > \lambda\}| \lesssim \frac{1}{\lambda} ||f||_{L^1}}$
2. ${||K*f||_{L^p} \lesssim ||f||_{L^p}}$ for ${1 < p < \infty}$.

Remark The fact that we are dealing with a convolution operator was essentially to prove ${L^2}$-boundedness. Once we know ${L^2}$-boundedness for a more general operator

$\displaystyle (Tf)(x) = \int K(x,y) f(y) \; dy,$

then the proof the Theorem goes through requiring only [the analog of property (3)]

$\displaystyle \int_{|x-y| \geq 2|y-y_0|} \big|K(x,y_0) - K(x,y)\big| \; dx \lesssim 1$

and

$\displaystyle \int_{|x-y| \geq 2|x-x_0|} \big|K(x,y) - K(x_0,y)\big| \; dy \lesssim 1$

Lemma (Calderón–Zygmund decomposition) Let ${f \in L^1({\mathbb R}^d)}$. Given ${\lambda > 0}$ we can decompose ${f = g+b}$ where ${b}$ is supported on disjoint dyadic cubes ${Q_k}$ with ${\sum |Q_k| \lesssim \frac{1}{\lambda} ||f||_{L^1}}$ and

1. ${|g(x)| \lesssim \lambda}$ a.e.
2. ${\int_{Q_k} b(y) \; dy = 0}$.

Moreover, ${||g||_{L^1} +||b||_{L^1} \lesssim ||f||_{L^1}}$.

Proof: Run a stopping time argument on the dyadic cubes with stopping rule

$\displaystyle \frac{1}{|Q|}\int_Q |f| \; dx > \lambda \implies \text{Stop!}$

This yields a collection of disjoint “stopping cubes” ${Q_k}$. Note

$\displaystyle ||f||_{L^1} \geq \sum \int_{Q_k} |f| \geq \sum \lambda |Q_k|$

Define

$\displaystyle b = \sum \chi_{Q_k}(x) \left[ f(x) - \frac{1}{|Q_k|}\int_{Q_k} f(y) \; dy\right]$

and ${g = f-b}$. Note that ${ \frac{1}{|Q_k|}\int_{Q_k} f(y) \; dy \leq \lambda 2^d}$, and the rest follows as before. $\Box$

Proof: (of Theorem) Let’s denote the operator ${T}$ by ${Tf(x) = \int K(x,y) f(y) \; dy}$. We need to bound ${|\{ |Tf | > \lambda \}|}$, for which we do a Calderón–Zygmund decomposition at height ${\lambda}$ and so bound ${|\{ |Tf | > \frac{\lambda}{2} \}|}$ and ${|\{ |Tb | > \frac{\lambda}{2} \}|}$. Now

$\displaystyle |\{ |Tf(x) | > \frac{\lambda}{2} \}| \lesssim \frac{||Tg||_{L^2}^2}{\lambda^2} \lesssim ||K||_{L^2 \rightarrow L^2}^2 \frac{||g||_{L^2}^2}{\lambda^2} \lesssim \frac{||g||_{L^1} ||g||_{L^\infty}}{\lambda^2} \lesssim \frac{||f||_{L^1}}{\lambda}$

Now let ${Q_k^* = 2\sqrt{d} Q_k}$ (draw a picture and inscribe some circles and their doubles) and so

$\displaystyle | \cup Q_k^*| \leq \sum |Q_k^*| \leq (2\sqrt{d})^d \sum |Q_k| \lesssim \frac{1}{\lambda} ||f||_{L^1}.$

It remains to estimate

$\displaystyle |\{ x \not \in \cup Q_k^* : |Tf(x)|> \lambda\}|$

Pick an ${x \not \in \cup Q_k^*}$, then

$\displaystyle \begin{array}{rcl} \displaystyle \int K(x,y) b(y) \; dy &=& \displaystyle \sum \int_{Q_k} K(x,y) b(y) \; dy \\ \\ \tiny \begin{bmatrix} \int_{Q_k} b = 0 \\ y_k = \text{center}(Q_k) \end{bmatrix} \hbox{\hskip 48pt} &=& \displaystyle \sum \int_{Q_k} [ K(x,y) - K(x,y_k)] b(y) \; dy. \end{array}$

Note that

$\displaystyle \begin{array}{rcl} \displaystyle \int_{{\mathbb R}^d \setminus Q_k^*} \int_{Q_k} |K(x,y) - K(x,y_k)| \: |b(y)| \; dy \; dx &\leq& \displaystyle \int_{Q_k} |b(y)| \int_{|x-y| \geq 2|y - y_k|} |K(x,y) - K(x,y_k)| \; dx \; dy \\ \\ &\lesssim& \displaystyle \int_{Q_k} |b(y) | \; dy. \end{array}$

Thus

$\displaystyle ||\int K(x,y) b(y) \; dy||_{L^1({\mathbb R}^d \setminus \cup Q_k^*)} \lesssim \sum_{k} \int_{Q_k} |b(y)| \lesssim ||b||_{L^1} \lesssim ||f||_{L^1}.$

Consequently,

$\displaystyle |\{ |Tb| > \lambda , x \not \in \cup Q_k^*\}| \lesssim \frac{||Tb||_{L^1}}{\lambda} \lesssim \frac{1}{\lambda} ||f||_{L^1}.$

Altogether, we have shown

$\displaystyle ||Tf||_{L^{1,\infty}} \lesssim ||f||_{L^1}$

but we also know that ${||Tf||_{L^2} \lesssim ||f||_{L^2}}$. Thus Marcinkiewicz implies boundedness on ${L^p}$ for ${1 < p \leq 2}$. For ${2 \leq p < \infty}$ we argue by duality,

$\displaystyle \begin{array}{rcl} \displaystyle \bigg|\iint \underbrace{\overline{g(x)}}_{\in L^{p'}} K(x,y) \underbrace{f(y)}_{\in L^p} \; dy \;dx \bigg| &=& \displaystyle \bigg| \int f(y) \overline{\int K^+(y,x) g(x)\; dx}\;dy \bigg| \\ &\leq& \displaystyle ||f||_{L^p} ||K^+||_{L^{p'} \rightarrow L^{p'}} ||g||_{L^{p'}} \\ \\ &\lesssim& ||f||_{L^p} ||g||_{L^{p'}}, \end{array}$

where the final inequality follows by applying the first part to ${K^+}$ instead of ${K}$. $\Box$

Remark Typical interesting Calderón–Zygmund operators are not bounded on ${L^1}$ or ${L^\infty}$. [In fact, if ${f \mapsto K * f}$ is bounded on ${L^1}$ or ${L^\infty}$ then ${K}$ is a finite measure.] Some examples include

$\displaystyle \frac{1}{z+i\epsilon}, \hbox{\hskip 48pt} \frac{\epsilon}{x^2 + \epsilon^2} \stackrel{*\frac{1}{\pi x}}{\longrightarrow} \frac{x}{x^2 + \epsilon^2}$

$\displaystyle Log(z), \hbox{\hskip 48pt} \pi \chi_{(-\infty,0)} \stackrel{*\frac{1}{\pi x}}{\longrightarrow} \log|x|$

If ${\xi > 0}$, then ${e^{2\pi i x \cdot \xi}}$ is bounded analytic on ${\text{Im }x \geq 0 \equiv {\mathbb C}^+}$. If ${\xi < 0}$, then ${e^{2\pi i x \cdot \xi}}$ is bounded analytic on ${\text{Im }x \leq 0 \equiv {\mathbb C}^-}$. So, splitting ${f \in L^p({\mathbb R})}$ into analytic in ${{\mathbb C}^+}$/${{\mathbb C}^-}$ parts.

$\displaystyle \hat f = \hat f \chi_{(0,\infty)} + \hat f \chi_{(-\infty, 0)}.$

For next time

Theorem (Mikhlin multiplier theorem, or possibly Marcinkiewicz, or Hörmander) If ${m : {\mathbb R}^d \rightarrow {\mathbb C}}$ obeys

$\displaystyle \Big|\frac{\partial^\alpha}{\partial \xi^\alpha} m(\xi) \Big| \lesssim |\xi|^{-|\alpha|} \hbox{ for } 0 \leq |\alpha| \leq \left\lceil \frac{d+1}{2} \right\rceil$

Then

$\displaystyle f \mapsto (m \hat f)\check{\:} = \check m * f$

is bounded on ${L^p}$ for ${1 < p < \infty}$.