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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}.

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