# Welcome.

## 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| 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 ${\xi}$.