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February 7, 2011

Harmonic Analysis Lecture Notes 12

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:53 pm

Recall our previous computation

\displaystyle  	\pi^{-\frac{d-\alpha}{2}}\Gamma(\frac{d-\alpha}{2}) |x|^{-(d-\alpha)} \:\: \hat\longrightarrow \:\:\pi^{-\alpha/2} \Gamma(\frac{\alpha}{2}) |\xi|^{-\alpha}

Example (Newton) {d = 3, \alpha =2}, then

\displaystyle  	\frac{1}{4\pi |x|} \longleftrightarrow|2\pi \xi|^{-2}

Now,

\displaystyle  	\widehat{-\Delta u}(\xi) = -\sum_j (2\pi i \xi_j)^2 \hat u(\xi = |2\pi \xi|^2 \hat u (\xi)

Thus is {f \in L^p} then {-\Delta u = f} has solution

\displaystyle  	u = \frac{1}{4\pi |x|} * f \in L^r

by Hardy–Littlewood–Sobolev (note {|x|^{-1} \in L^{3,\infty}({\mathbb R}^3)}).

Definition Let {|\nabla|^s} be the linear transformation (defined initially for {f \in \mathcal{S}}) by

\displaystyle  	\widehat{|\nabla|^s f} (\xi) = |2\pi \xi|^s \hat f(\xi).

Here {s > -d}.

Theorem (Sobolev Embedding) For {f \in \mathcal{S}}, {0 \leq s < \frac{d}{p}}, and {1 < p < \infty}, then

\displaystyle  	||f||_{L^q} \lesssim \big|\big| |\nabla|^s f\big|\big|_{L^p}\hbox{\hskip 28pt where \hskip 28pt} \frac{d}{q} = \frac{d}{p} -s

Proof: By duality

\displaystyle  \begin{array}{rcl}  	\int \bar g f \; dx &=& \displaystyle \int \bar g (|\nabla|^{-s} |\nabla|^s f) \; dx \\ \\ 		&=& \displaystyle \int \overline{\big(|\nabla|^{-s} g\big)} \big( |\nabla|^s f\big) \; dy \\ \\ 		&\leq& \big|\big| |\nabla|^{-s} g\big|\big|_{L^{p'}} \big|\big| |\nabla|^s f \big|\big|_{L^p}. 	\end{array}

We’re done if we know {\big|\big| |\nabla|^{-s} g\big|\big|_{L^{p'}} \big|\big| \lesssim ||g||_{L^{q'}}}, but this is exactly what HLS says! \Box

If {s = 1}, it is natural to want to replace {\big|\big| |\nabla| f\big|\big|_{L^p}} by {\big|\big| \nabla f\big|\big|_{L^p}}. It would suffice to show that {m_j(\xi) = \frac{-i\xi_j}{|\xi|}} form bounded Fourier multipliers on {L^p} when {1 < p < \infty}:

\displaystyle  	|2\pi \xi| \hat f(\xi) = \sum_{j=1}^d m_j(\xi) \cdot 2\pi i \xi_j \hat f(\xi)

i.e.,

\displaystyle  	|\nabla| f = \sum_{j=1}^d \check m_j * \frac{\partial f}{\partial x_j}.

We will do this next. These convolution operators {R_j f = \check m_j * f = \pi^{-\frac{d+1}{2}} \Gamma(\frac{d+1}{2}) \frac{x_j}{|x|^{d+1}} *f} are called the Riesz transforms. E.g., {d = 1} gives {f \mapsto \frac{1}{\pi x} * f}, which is the Hilbert transform.

Additionally {L^p}-boundedness of {R_j} then

\displaystyle  	u = \frac{1}{4\pi |x|} * f \hbox{\hskip 28pt (is a solution to }-\Delta u = f)

obeys

\displaystyle  	\partial_j \partial_k u = \underbrace{\partial_j}_{2\pi i \xi_j} \underbrace{\partial_k}_{2\pi i \xi_k} \underbrace{(-\Delta)^{-1} f}_{|2\pi \xi|^{-2}} = R_j R_k f

and so

\displaystyle  	|| \partial_j \partial_k u||_{L^p} \lesssim ||f||_{L^p}.

Definition A Calderón–Zygmund (convolution) kernel {K} is a function {K : {\mathbb R}^d \setminus \{0\} \rightarrow {\mathbb C}} that obeys

  1. {|K(x)| \lesssim |x|^{-d}};
  2. {\int_{a < |x|<b} K(x) \; dx = 0} for all {0 < a< b< \infty};
  3. {\int_{|x| > 2|y|} |K(x) - K(x-y)| \; dx \lesssim 1} for all {y \in {\mathbb R}^d}.

The Riesz transforms obey these axioms. For property (3), it is easiest to use the following lemma.

Lemma If {K} obeys properties (1), (2), and

\displaystyle  	|\nabla K| \lesssim |x|^{-d-1},

then {K} is a Calderón–Zygmund kernel.

Wait do we really need (1) and (2)? Proof: The fundamental theorem of calculus gives

\displaystyle  \begin{array}{rcl}  	|K(x+y) - K(x)| &\leq& \displaystyle \Big| \int_0^1 y \cdot \nabla K(x + ty) \; dt \Big| \\ \\ 		&\lesssim& |y| \sup_{0 \leq t \leq 1} \frac{1}{|x + ty|^{d+1}} \\ 		&\lesssim& |y| |x|^{-(d+1)} \hbox{\hskip 18pt when } |x| \geq 2|y| 	\end{array}

Now

\displaystyle  	\int_{|x| \geq 2|y|} \frac{|y|}{|x|^{d+1}} \; dx \lesssim \frac{|y|}{2|y|} \lesssim 1.

\Box

Coming up in the next lecture{\ldots}

Proposition If {K} is a Calderón–Zygmund kernel, then “{\hat K \in L^\infty}”. Namely, {K_\epsilon = K \cdot \chi_{|x|> \epsilon}} obeys {|\hat K_\epsilon (\xi)| \lesssim 1} uniformly in {\epsilon} and {\xi}.

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