# Welcome.

## February 25, 2011

### Harmonic Analysis Lecture Notes 15

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:51 pm

Now, recopying some handwritten notes:

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E} \{ | \sum c_n X_n|^p \} &=& \displaystyle \int_0^\infty p \lambda^{p-1} \mathbb{P} \{ |\sum c_n X_n | > \lambda \} \; d\lambda \\ \\ &\leq& \displaystyle 2p \int_0^\infty \lambda^{p-1} e^{-\frac{\lambda^2}{2 \sum |c_n|^2}} \; d\lambda \\ \\ &\leq& \displaystyle \left( \sum |c_n|^2 \right)^{p/2} \int_0^\infty 2p \mu^{p-1} e^{-u^2/2} \; du \\ \\ &\lesssim_p& \displaystyle \left( \sum |c_n|^2 \right)^{p/2} \end{array}$

For the other direction: when ${1

$\displaystyle \begin{array}{rcl} \sum |c_n|^2 &=& \displaystyle \mathop{\mathbb E} \{ |\sum c_n X_n|^2\} \\ \\ &\lesssim& \displaystyle \mathop{\mathbb E} \{ |\sum c_n X_n|^p\}^{1/p} \mathop{\mathbb E} \{ |\sum c_n X_n|^{p'}\}^{1/p'} \end{array}$

Corollary (of the Proof) For ${1 < p <\infty}$, then

1. For all ${s \in {\mathbb R}}$

$\displaystyle \big\| |\nabla|^s f\big\|_{L^p} \approx_{p,s} \left\| \sqrt{ \sum N^{2s} |f_N|^2} \right\|_{L^p},$

2. For ${s > 0}$

$\displaystyle \big\| |\nabla|^s f\big\|_{L^p} \approx_{p,s} \left\| \sqrt{ \sum N^{2s} |f_{\geq N}|^2} \right\|_{L^p}.$

Proof: We will prove assertion 1, first beginning with ${\gtrsim}$. Note that ${\theta(\xi) = |\xi|^{-s} \psi(\xi) \in C_c^\infty ({\mathbb R} \setminus \{0\})}$, so the first half of the regular square function proof shows that

$\displaystyle \left\| \sqrt{ \sum_{N \in 2^{\mathbb Z}} |\underbrace{N^d \check \theta (N \cdot)}_{(**)} * g|^2} \right\|_{L^p} \lesssim_{p,s} \| g\|_{L^p},$

noting that ${(**)}$ is the convolution operator associated to the multiplier ${\theta(\xi/N) = N^{s} |\xi|^{-s} \psi(\xi/N) = N^s \psi(\xi/N) |\xi|^{-s}}$. So the above says

$\displaystyle \left\| \sqrt{ \sum_{N \in 2^{\mathbb Z}} N^{2s}|P_N |\nabla|^{-s} g|^2} \right\|_{L^p} \lesssim_{p,s} \| g\|_{L^p}.$

Now substitute ${g = |\nabla|^s f}$. Note that the argument above also works for ${\tilde{\theta}(\xi)= |\xi|^s [ \psi(\xi/2) + \psi(\xi) + \psi(2\xi)]}$. Now, for ${g}$ a unit vector in ${L^{p'}}$, then

$\displaystyle \begin{array}{rcl} \int \bar g |\nabla|^s f &=& \displaystyle \int \overline{\sum_N |\nabla|^2 |\nabla|^{-s} \tilde P_N P_N g} |\nabla|^sf \\ \\ &=& \displaystyle \int \sum_N \underbrace{\overline{N^{-s} |\nabla|^s\tilde P_N g}}_{\text{this has multiplier }\tilde \theta(\xi/N)} N^s f_N \\ \\ &\leq& \displaystyle \int \sqrt{ \sum_N |N^d \check{\tilde{\theta}}(N \cdot) * g|^2} \sqrt{\sum_N N^{2s} |f_N|^2} \; dx \\ \\ &\leq& \displaystyle \underbrace{ \left\| \sqrt{ \sum_N |N^d \check{\tilde{\theta}}(N \cdot) * g|^2} \right\|_{L^{p'}}}_{ \tiny \begin{matrix}\text{boundedness of this in }L^{p'} \\ \text{and of } g \in L^{p'}\end{matrix} \lesssim 1} \left\| \sqrt{\sum_N N^{2s} |f_N|^2} \right\|_{L^p} \\ \\ &\lesssim& \left\| \sqrt{\sum_N N^{2s} |f_N|^2} \right\|_{L^p}. \end{array}$

Now choose ${g}$ to exhibit ${\big\| |\nabla|^s f \big\|_{L^p}}$ on the left hand side.

Now we turn to assertion 2, beginning with ${\gtrsim}$.

$\displaystyle \begin{array}{rcl} \displaystyle \sum N^{2s} |f_{\geq N}|^2 &=& \displaystyle \sum_{N \in 2^{\mathbb Z}} \; \sum_{\tiny\begin{matrix} M \geq N \\ M \in 2^{\mathbb Z}\end{matrix}} \; \sum_{\tiny\begin{matrix} K \geq N \\ K \in 2^{\mathbb Z}\end{matrix}} N^{2s} f_M \bar f_K \\ &\leq& \displaystyle \sum_{M,K \in 2^{\mathbb Z}} \; \left( \sum_{N \leq \min(M,K)} N^{2s} |f_K| \;|f_M|\right) \\ &\leq& \displaystyle 2 \sum_{M \leq K} \frac{M^{2s}}{1- 2^{-2s}} |f_K| \; |f_M| \\ &\leq& \displaystyle \sum_{M \leq K} \left( \frac{M}{K}\right)^s K^s |f_K| \cdot M^s |f_M| \\ \scriptsize\hbox{[Schur's test], } \scriptsize\begin{matrix}\sum_{k=0}^\infty 2^{-ks} \end{matrix} \hbox{\hskip 38pt} &\lesssim& \displaystyle \left( \sum_K K^{2s} |f_K|^2 \right)^{1/2} \left( \sum_M M^{2s} |f_M|^2 \right)^{1/2} \end{array}$

Therefore, the right hand side of assertion 2 is ${\lesssim}$ that the right hand side of assertion 1. Next we show the reverse (namely, replace “${\lesssim}$” from the previous sentence by “${\gtrsim}$”).

$\displaystyle \begin{array}{rcl} &\;&f_N = f_{\geq N} - f_{\geq 2N} \\ &\implies& |f_N|^2 \leq 2 |f_{\geq N}|^2 + 2 |f_{\geq 2N}|^2 \\ &\implies& \sum N^{2s} |f_N|^2 \leq 2 \sum N^{2s} |f_{\geq N}|^2 + 2^{1-2s} \sum (2N)^{2s} |f_{\geq 2N}|^2 \end{array}$

$\Box$

Now, Michael Bateman is lecturing.

Theorem (Fractional Product Rule, Christ–Weinstein 1991) For ${s > 0}$ and ${\frac{1}{p} = \frac{1}{p_1} + \frac{1}{p_2} = \frac{1}{p_3} + \frac{1}{p_4}}$ with ${p,p_j \in (1,\infty)}$, ${j = 1,2,3,4}$, then

$\displaystyle \big\| |\nabla|^s (fg) \big\|_p \leq \big\| |\nabla|^s f \big\|_{p_1} \|g\|_{p_2} + \|f\|_{p_3} + \big\| |\nabla|^s g \big\|_{p_4}.$

Proof: To prove this, we first use part 1 of the corollary to write the left hand side as a Littlewood–Paley sum:

$\displaystyle LHS \leq \left\| \sqrt{ \sum N^{2s} |(fg)_N|^2} \right\|_{p} \hbox{\hskip 38pt} (*)$

For each ${N \in 2^{\mathbb Z}}$, we have

$\displaystyle fg = f_{\geq \frac{N}{4}} g + f_{< \frac{N}{4}} g_{\geq \frac{N}{4}} + f_{<\frac{N}{4}} g_{<\frac{N}{4}}.$

This implies

$\displaystyle P_N(fg) = P_N(f_{\geq \frac{N}{4}} g) + P_N (f_{< \frac{N}{4}} g_{\geq \frac{N}{4}})$

Now we use the estimate ${P_Nh(x) \lesssim Mh(x)}$ to get

$\displaystyle N^{2s} |(fg)_N(x)|^2 \lesssim |M [N^s f_{\geq \frac{N}{4}} g(x)|^2 + | M[(Mf) N^s g_{\geq \frac{N}{4}}](x)|^2$

Thus,

$\displaystyle \begin{array}{rcl} (*) &\lesssim& \displaystyle \left\| \sqrt{\sum |M[N^sf_{\geq \frac{N}{4}} g](x)|^2} \right\|_p + \left\| \sqrt{| M[(Mf) N^s g_{\geq \frac{N}{4}}](x)|^2} \right\|_p \\ \\ &\lesssim& \displaystyle \left\| g \sqrt{ \sum_N N^{2s} |f_{\geq \frac{N}{4}}|^2} \right\|_p + \left\| (Mf) \sqrt{ N^{2s} | g_{\geq \frac{N}{4}}|^2} \right\|_p \end{array}$

Using Hölder and then assertion 2 of the corollary, then

$\displaystyle (*) \lesssim \|g\|_{p_1} \left\| \sqrt{ \sum_N N^{2s} |f_{\geq \frac{N}{4}}|^2} \right\| + \| f\|_{p^3} \left\| \sqrt{ \sum_N N^{2s} |g_{\geq \frac{N}{4}}|^2}\right\|$

$\Box$

Theorem (Fractional chain rule, Weinstein, 1991) Suppose ${F : {\mathbb C} \rightarrow {\mathbb C}}$ satisfies

$\displaystyle |F(u) - F(v)| \leq |u-v| [ a(u) + a(v)],$

where ${a : {\mathbb C} \rightarrow (0,\infty)}$. If ${s \in (0,1)}$, then

$\displaystyle \big\| |\nabla|^s F(u) \big\|_{p} \leq \big\| |\nabla|^s f \big\|_{p_1} \|a (u)\|_{p_2}$

whenever ${\frac{1}{p} = \frac{1}{p_1} + \frac{1}{p_2}}$.

Remark Recall the original chain rule: if ${F'}$ exists, then ${\| (F \circ u)'(x)\|_p \leq \| F'(u(x)) u'(x) \|_p \leq \|F'(u)\|_{p_1} \|u'\|_{p_2}}$.

Example ${F(u) = |u|^{p} u}$, then

$\displaystyle \begin{array}{rcl} F(u) - F(v) &=& \displaystyle \int_0^1 (u-v)\cdot \nabla F(v + t(u-v)) \; dt \\ \\ &\leq& |u-v| \sup_{t\in [0,1]} |\nabla F(v + t(u-v))| \end{array}$

Here ${a(u) = |u|^p}$, noting that if ${f(x) = x^p,}$ then ${f(x + (1-t)y) \leq C( |x|^p + |y|^p)}$.

Proof: Recalling that ${\psi(0) = 0 = \int \check \psi}$, then

$\displaystyle \begin{array}{rcl} |P_N(F \circ u)(x)| &=& \displaystyle \bigg| \int N^d \check \psi(Ny) F \circ u (x - y) \; dy \bigg| \\ \\ &=& \displaystyle \bigg| \int N^d \check \psi(Ny) [ F \circ u(x-y) - F \circ u(x)] \; dy \bigg| \\ \\ &\leq& \int |N^d \check \psi(Ny)| \: |u(x-y) - u(x)| [a \circ (x-y) + a \circ u(x)] \; dy \end{array}$

Let’s analyze ${|u(x-y) - u(x)|}$:

$\displaystyle |u(x-y) - u(x)| \leq |u_{> N}(x-y)| + |u_{> N} (x)| + \sum_{K \leq N} |u_K(x-y) - u_k(x)|$

and

$\displaystyle \begin{array}{rcl} u_K(x-y) - u_k(x) &=& \int \check{\tilde{\psi}}_K (x - y- z) u_K(z) \; dz - \int \check{\tilde{\psi}}_K(x -z) u_K(z) \; dz \\ \\ &=& \int [K^d \check{\tilde{\psi}}(K[z-y] - K^d \check{\tilde{\psi}}(Kz)) u_K (x-z) \; dz \hbox{\hskip 18pt} (**) \end{array}$

We claim that ${(**) \leq K|y|[ Mu_K(x-y) + Mu_K(x)]}$, the proof of which we break into two cases: ${K|y| \gtrsim 1}$ and ${K|y| \lesssim 1}$. For the first case, we change “${-}$” to “${+}$” and prove the estimate without ${K|y|}$. For the second case, we use the derivative of ${\check{\tilde{\psi}}}$:

$\displaystyle \begin{array}{rcl} | \check{\tilde{\psi}}(K[z-y]) - \check{\tilde{\psi}}(Kz)| &\leq& \displaystyle K|y| [ (\check{\tilde{\psi}})' (Kz) + (\check{\tilde{\psi}})' (K[z-y])] \\ \\ &\leq& K|y| \frac{1}{(1 + |z|K)^{100d}} \end{array}$

So,

$\displaystyle \begin{array}{rcl} (**) &\lesssim& \int K^d \frac{K|y|}{(1 + K|z|)^{100d}} u_K(x-z) \; dz \\ \\ &\lesssim& K|y| Mu_K(x), \end{array}$

which proves the claim. Now

$\displaystyle \begin{array}{rcl} |(F \circ u)_N (x) | &\leq& \displaystyle \int N^d |\check{\tilde{\psi}}(Ny)| \: |u(x-y) - u(x)| [a \circ u(x-y) - F \circ u(x) ] \; dy \\ \\ &=& \displaystyle \int N^d|\check{\tilde{\psi}}(Ny)| \: |u_{>N}(x-y)| \:[a \circ u(x-y) - F \circ u(x) ] \; dy \\ \\ &\;& + \displaystyle \int N^d|\check{\tilde{\psi}}(Ny)| \: |u_{>N}(x)| \: [a \circ u(x-y) - F \circ u(x) ] \; dy \\ \\ &\;& + \displaystyle \int N^d|\check{\tilde{\psi}}(Ny)| \: |u_{>N}(x-y)| \: [a \circ u(x-y) - F \circ u(x) ] \; dy \end{array}$

The first term is bounded above by ${M(u_{> N} a \circ u)(x) + a\circ u(x)M(u_{>N})(x)}$ and the second term can be handled similarly. The third term is bounded above by

$\displaystyle \begin{array}{rcl} \displaystyle\sum_{K \leq N} \int N^d \check \psi (Ny) K|y| [Mu_K(x-y) + Mu_K(x)] &=& \sum_{K \leq N} N^d \check{\psi}(Ny0 N|y| \frac{K}{N} [Mu_K(x-y) + Mu_K(x)] \\ \\ &\leq& \sum_{K \leq N} \frac{K}{N} [Mu_K(x-y) + Mu_K(x)] \end{array}$

(if ${\phi}$ is Schwartz, then ${\phi(x) |x|}$ is still Schwartz).

Now, ${\big\| |\nabla|^s (F \circ u) \big\|_p}$ consists of terms like

$\displaystyle \begin{array}{rcl} \left\| \sqrt{\sum_N N^{2s} |Mu_{> N} a\circ u|^2} \right\|_p &\leq& \big\| \sqrt{ \sum_N N^{2s} |u_{> N}(x)|^2} a \circ u(x) \big\|_{p} \\ \\ &\lesssim& \left\| \sqrt{ \sum_N N^{2s} |u_{> N}(x)|^2}\right\|_{p_1} \|a \circ u\|_{p_2} \\ \\ &\approx& \big\| |\nabla|^s u \big\|_{p_1} \|a \circ u\|_{p_2} \end{array}$

and terms like

$\displaystyle \left\| \sqrt{ \sum_N N^{2s} \Big| \sum_{K \leq N} \frac{K}{N} [Mu_K(x-y) + Mu_K(x)] \Big|^2}\right\|_p,$

and we claim ${\sum_N N^{2s} |\sum_{K \leq N} \frac{K}{N} c_K|^2 \lesssim \sum_N N^{2s} |c_N|^2}$. Given this claim, we are left with expressions like ${\left\| \sqrt{\sum_N N^{2s} |Mu_{> N} a\circ u|^2} \right\|_p}$. Now, using that ${s \in (0,1)}$,

$\displaystyle \begin{array}{rcl} \displaystyle \sum_N N^{2s} \sum_{K \leq N} \sum_{L \leq N} \frac{K}{N} \frac{L}{N} c_K c_L &=& \sum_K \sum_K \sum_{N \geq \max(K,L)} \frac{K}{N} \frac{L}{N} N^{2s} c_K c_L \\ \\ &\lesssim& \displaystyle \sum_K \sum_L \frac{KL}{\max(K,L)^{2-2s}} c_K c_L \\ \\ &\lesssim& \displaystyle \sum_{K \leq L} \left( \frac{K}{L} \right)^{1-s} K^s c_K L^s c_L \\ \\ &\leq& \sum_K |c_K|^2 K^{2s} \end{array}$

$\Box$