# Welcome.

## May 4, 2011

### Harmonic Analysis Lecture Notes 21

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:56 pm

In this discussion, we will prove the following theorem.

Theorem (Sharp Sobolev Embedding, Aubin & Talenti) Fix ${d \geq 3}$. The inequality

$\displaystyle \|f\|_{L^\frac{2d}{d-2}} \leq C_d \| \nabla f\|_{L^2}$

where the constant ${c_d}$ is determined by inserting ${W(x) = (1 + \frac{|x|^2}{d(d-2)})^{-\frac{d-1}{2}}}$. Moreover, equality occurs in the above if and only if ${f(x) = \alpha W(\frac{x-x_0}{\lambda})}$ for some ${\alpha \in {\mathbb C}}$, ${x_0 \in {\mathbb R}^d}$, and ${\lambda > 0}$.

The most difficult part of the proof is to show that there is indeed an optimizer (cf previous discussion with the sharp Gagliardo–Nirenberg inequality). Moreover, once we have the correct “bubble decomposition’, the existence of an optimizer is then a simple sub-additivity argument.

Theorem Let ${f_n}$ be a bounded sequence in ${\dot H^1({\mathbb R}^d)}$ with ${d \geq 3}$. Then there exist ${J_{\max} \in {\mathbb N} \cup \{ \infty\}}$, ${\{ \phi^j\}_{j=1}^{J_{\max}} \subseteq \dot H^1({\mathbb R}^d)}$, ${\{x_n^j\}_{j=1}^{J_{\max}}}$, ${\{\lambda_n^j\}_{j=1}^{J_{\max}} \in (0,\infty)}$ so that along a subsequence in ${n}$ we can write

$\displaystyle f_n(x) = \sum_{j=1}^J (\lambda_n^j)^\frac{2-d}{2}\phi\left(\frac{x-x_n^j}{\lambda_n^j}\right) + r_n^J$

for each finite ${J \geq J_{\max}}$, with the following properties

1. ${\phi^j = \text{w-lim}_{n\rightarrow \infty} (\lambda_n^j)^\frac{d-2}{2}\phi\left(\lambda_n^j[x-x_n^j]\right)}$ in ${\dot H^1}$.
2. ${\lim_{j \rightarrow J_{\max}} \lim_{n\rightarrow\infty} \| r_n^J\|_{L^\frac{2d}{d-2}} = 0}$.
3. For all ${J}$, ${\lim_{n\rightarrow\infty} \Big| \|f_n\|_{\dot H^1}^2 - \sum_{j=1}^J \|\phi^j\|_{\dot H^1}^2 - \|r_n^J\|_{\dot H^1}^2 \Big| = 0}$.
4. ${\lim_{J \rightarrow J_{\max}} \lim_{n\rightarrow\infty} \Big| \|f_n\|_{L^\frac{2d}{d-2}}^\frac{2d}{d-2} - \sum_{j=1}^J \|\phi_j\|_{L^\frac{2d}{d-2}}^\frac{2d}{d-2} \Big| = 0}$.
5. ${\frac{|x_n^j x_n^{j'}|}{\sqrt{\lambda_n^j\lambda_n^{j'}}} + \big| \log \left(\frac{\lambda_n^j}{\lambda_n^{j'}}\right) \big| \rightarrow \infty}$.

## April 27, 2011

### Harmonic Analysis Lecture Notes 20

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:53 pm

Concentration compactness approach to Gagliardo–Nirenberg in 1-dimensions

We have ${\|f_n\|_{L^2} = \| \nabla f \|_{L^2} = 1}$ with ${f_n}$ an optimizing sequence; in particular, ${J(f_n) = \|f_n\|_{L^p} \rightarrow J_{\max} > 0}$. Choose ${x_n}$ (Lebesgue points of ${f_n}$) so that

$\displaystyle |f(x_n)| \geq \frac{1}{2} \| f_n\|_{L^\infty}.$

Note that ${f_n \in L^\infty({\mathbb R})}$; we proved ${H^1 \hookrightarrow L^\infty}$. These functions are in fact continuous; indeed, they are Hölder ${\tfrac{1}{2}}$ continuous uniformly in ${n}$:

$\displaystyle |f_n(x) - f_n(y)| \leq \int_x^y |f_n'(t)| \; dt \leq \sqrt{\int_x^y 1 \; dt} \sqrt{\int_x^y |f_n'(t)|^2 \; dt} \leq |y-x|^\frac{1}{2}.$

Observe that ${\|f_n\|_{L^\infty}}$ cannot tend to ${0}$ as ${n \rightarrow \infty}$ since otherwise, by interpolation with ${L^2}$, we would have that ${\|f_n\|_{L^p} \rightarrow 0}$, contradicting the optimizing hypothesis. Henceforth we restrict ${n}$ to a subsequence so that ${\|f_n\|_{L^\infty} \rightarrow B > 0}$. By Alaoglu and Rellich–Kondrachov, we can pass to a further subsequence so that

$\displaystyle f(x + x_n) \rightarrow \phi$

weakly in ${H^1}$ (${\equiv}$ weakly in ${L^2}$ and in ${\dot H^1}$) and in the ${L^p}$ norm on any compact set.

We observe that ${\phi \not \equiv 0}$. As ${|f_n(x_n)| \gtrsim 1}$ and ${f_n}$ is Hölder ${\frac{1}{2}}$ continuous (uniformly in ${n}$), there is a small number ${\delta}$ so that

$\displaystyle |f_n(x + x_n)| \geq \delta \hbox{\hskip 18pt} \forall |x| < \delta.$

Thus, ${\|\phi\|_{L^p} \geq \left(\int_{-\delta}^\delta \delta^p \; dx \right)^\frac{1}{p} \gtrsim \delta^{1+\frac{1}{p}} > 0}$.

For the sake of discussion, suppose ${\|f_n\|_{L^p} \rightarrow \|\phi\|_{L^p}}$ (i.e., ${\phi}$ is capturing all the ${L^p}$ norm). Then, by Radon–Riesz, ${f_n \rightarrow \phi}$ in ${L^p}$ and we can finish the proof just as above. The remaining alternative is that ${0 < \|\phi\|_{L^p} < \lim_{n \rightarrow \infty} \|f_n\|_{L^p} = J_{\max}}$. In that case write

$\displaystyle f_n(x) = \phi(x - x_n) + \underbrace{r_n(x)}_{\text{remainder}}.$

Note that ${1 = \|f_n\|_{L^2} = \langle \phi, \phi \rangle + \langle r_n, r_n \rangle + 2 \text{ Re }\langle \phi(x) , f_n(x-x_n) - \phi(x) \rangle}$ and that ${\langle \phi(x) , f_n(x-x_n) - \phi(x) \rangle \rightarrow 0}$ as ${n \rightarrow \infty}$ because ${f_x(x-x_n) \rightharpoonup \phi}$. This shows that ${\phi(x-x_n)}$ and ${r_n}$ are “asymptotically orthogonal in the Pythagorean sense

$\displaystyle \lim \|r_n\|_{L^2}^2 + \|\phi\|_{L^2}^2 = 1 \hbox{\hskip 18pt but \hskip 18pt } 1= \|\phi(x-x_n) + r_n(x) \|_{L^2}.$

Similarly,

$\displaystyle J_{\max} = \lim \int |f_n|^p \; dx.$

Moreover,

$\displaystyle J_{\max} = \lim_{n \rightarrow \infty} \int|f_n|^p \; dx = \lim_{n \rightarrow \infty} |\phi(x) + r_n(x+x_n)|^p \; dx = \lim_{R \rightarrow \infty} \lim_{n \rightarrow \infty} \int_{|x| \leq R} + \int_{|x| \geq R} = \lim_{n \rightarrow \infty} \int |\phi|^p + \int |r_n(x+x_n)|^p \; dx,$

where the first term is due to ${r_n(x+x_n) \rightarrow 0}$ in ${L^p}$ con compact sets and the second term is due to ${\int_{|x| \geq R} |\phi|^2 \; dx}$ going to ${0}$ as ${R \rightarrow \infty}$. That is,

$\displaystyle \lim_{n \rightarrow \infty} \|\phi\|_{L^p}^p + \|r_n\|_{L^p}^p = \lim_n \|f_n\|_{L^p}^p = J_{\max}$

The remaining goal is to show that ${\|r_n\|_{L^p}}$, because then ${\|f_n\|_{L^p} \rightarrow \| \phi \|_{L^p}}$, which then implies ${\int |f_n(x+x_n) - \phi(x)|^p \rightarrow 0}$.

The key to showing this is sub-additivity:

$\displaystyle \begin{array}{rcl} J_{\max} &\longleftarrow& \displaystyle \Big(\|\phi\|_{L^p}^p + \|r_n\|_{L^p}^p\Big)^{1/p} \leq J_{\max} \Big( \|\phi\|_{L^2}^{p(1-\theta)} \|\nabla \phi\|_{L^2}^{p\theta} + \|r_n\|_{L^2}^{p(1-\theta)} \|\nabla r_n\|_{L^2}^{p\theta}\Big)^{1/p} \\ \\ &\qquad& \displaystyle \leq J_{\max} \Big( (1-\theta) \|\phi\|_{L^2}^p + \theta \|\nabla \phi\|_{L^2}^p + (1-\theta) \|r_n\|_{L^2}^p + \theta \|\nabla r_n\|_{L^2}^p\Big)^{1/p} \end{array}$

and so

$\displaystyle (1-\theta)\Big(\|\phi\|_{L^2}^p + \|r_n\|_{L^2}^p\Big) + \theta \Big( \|\nabla \phi\|_{L^2}^p + \|\nabla r_n\|_{L^2}^p \Big) \rightarrow 0.$

$\displaystyle \begin{array}{rcl} \|\phi\|_{L^2}^2 + \|r_n\|_{L^2}^2 &\rightarrow& 1 = \|f_n\|_{L^2} \\ \\ \|\nabla \phi\|_{L^2}^2 + \|\nabla r_n\|_{L^2}^2 &\rightarrow& = \|\nabla f_n\|_{L^2}. \end{array}$

Therefore, because ${p >2}$, the equalities

$\displaystyle a^2 + b^2 = 1, \hbox{\hskip 18pt and \hskip 18pt} a^p + b^p = 1$

must imply ${a = 0}$ or ${b = 0}$ (this can be seen by drawing the unit balls). Namely, ${a_n^2 + b_n^2 \rightarrow 1}$ and ${a_n^p + b_n^p \rightarrow 1}$ implies ${a_n \rightarrow 0}$ or ${b_n \rightarrow 0}$. As ${\|\phi\|_{L^2} \neq 0 }$ and ${\|\nabla \phi\|_{L^2} \neq 0}$, we must have ${\|r_n\|_{L^2} + \|\nabla r_n\|_{L^2} \rightarrow 0}$ and hence ${f_n(x+x_n) \rightarrow \phi}$ in ${H^1}$, and thus also in ${L^p}$. ${\square}$

In the original vocabulary of concentration compactness, there are three scenarios for sequences ${\|f_n\|_{L^2} = \|\nabla f_n\|_{L^2} = 1}$,

1. vanish, ${\|f_n\|_{L^p} \rightarrow 0}$.
2. compactness (co-compactness) [modulo translations ${f_n(x+x_n) \rightarrow \phi}$ in ${L^p}$.
3. dichotomy ${f_n \rightarrow \phi}$ strongly on compact sets bu ${0 < \|\phi\|_{L^p} < \limsup \|f_n\|_{L^p}}$.

at least after passing to a subsequence.

In the proof, we had a sequence ${f_n}$ and either ${\|f_n\|_{L^p} \rightarrow 0}$ or it contained (after passing to a subsequence and translating in space), a “bubble of concentration” ${\phi}$, leaving behind a remainder ${r_n = f_n(x) - \phi(x - x_n)}$. We can proceed inductively and look inside ${r_n}$ for a further bubble of concentration and so

Theorem Fix ${2 < p < \infty}$. Let ${f_n}$ be a bounded sequence in ${H^1({\mathbb R}^)}$. Then, after passing to a subsequence, we can decompose ${f_n}$ as

$\displaystyle f_n(x) = \sum_{j=1}^J \phi^j (x - x_n^j) + r_n^J(x)$

for each ${0 \leq J \leq M_\phi \in {\mathbb N} \cup \{\infty\}}$ where ${M_\phi}$ is the “total number of bubbles” such that

1. for all ${j}$ and ${\phi^j \not \equiv 0}$, ${f_n(x+x_n^j) \rightarrow \phi^j(x)}$ weakly in ${H^1}$ and strongly in ${L^p}$ on compact sets.
2. If ${j < j'}$, then ${|x_n - x_n^{j'}| \rightarrow \infty}$.
3. ${\lim_{J \rightarrow M_\phi} \lim_{n \rightarrow \infty} \| r_n^J\|_{L^p} = 0}$.
4. For all ${J}$, ${\|f_n\|_{L^2}^2 - \left( \sum_{j=1}^J \|\phi^j\|_{L^2}^2 + \|r_n^J\|_{L^2}^2\right) \rightarrow 0}$ as ${n \rightarrow \infty}$. Furthermore ${\|\nabla f_n\|_{L^2}^2 - \left( \sum_{j=1}^J \|\nabla \phi^j\|_{L^2}^2 + \|\nabla r_n^J\|_{L^2}^2\right) \rightarrow 0}$ and ${\lim_{J \rightarrow M_\phi} \limsup_{n \rightarrow \infty} \Big| \|f_n\|_{L^2}^p - \sum_{j=1}^J \|\phi^j\|_{L^2}^p\Big| = 0}$.

Remark For assertion 2, if ${(j,j')}$ was the minimal pair such that ${x_n^j - x_n^{j'}}$ does not diverge, then, along a subsequence, it converges, say to ${y}$ and so

$\displaystyle f_n(x + x_n^j) \rightarrow \phi^j \hbox{\hskip 18pt and \hskip 18pt} \phi^{j'} \longleftarrow f(x+x_n^{j'}) = f(x + x_n^{j} -(x_n^{j} - x_n^{j'})).$

Thus, ${\phi^{j'}(x) = \phi^j(x-y)}$, which is to say we are removing the same “bubble” twice, a troubling fact. In the proof, we find ${\phi^{j'}}$ by applying our “bubble-finding lemma” to ${r_n^{j'-1}}$, which converges to zero when translated by ${x_n^j}$. Any limit along that sequence was removed by ${\phi^j}$ and hence ${\phi^{j'}}$ would be identically ${0}$.

## April 18, 2011

### Harmonic Analysis Lecture Notes 19

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:51 pm

$\displaystyle \| f\|_{L^p} \lesssim \| f\|_{L^q}^{1-\theta} \|\nabla f\|_{L^2}^\theta$

provided

$\displaystyle \theta = \tfrac{2d(p-q)}{p\big(2d - (d-2) q\big)} \hbox{\hskip 18pt and \hskip 18pt} \left\{\begin{matrix} 1 \leq q< p < \tfrac{d}{d-2} &\hbox{\hskip 18pt} d \geq 3 \\ \\ 1 \leq q < p < \infty &\hbox{\hskip 18pt} d=2 \\ \\ 1 \leq q < p \leq \infty &\hbox{\hskip 18pt} d = 1 \end{matrix} \right. .$

Proof: From Bernstein’s inequality (because ${p > q}$)

$\displaystyle \| f_N \|_{L^p} \lesssim \underbrace{N^{d( \frac{1}{q} - \frac{1}{p})}}_{\text{positive power of }N} \| f\|_{L^q}.$

When ${p \geq 2}$

$\displaystyle \| f_N \|_{L^p} \lesssim N^{d( \frac{1}{2} - \frac{1}{p})} \|f_N \|_{L^2} \lesssim \underbrace{N^{d( \frac{1}{2} - \frac{1}{p})-1}}_{\text{negative power of }N} \|\nabla f \|_{L^2}.$

Moreover, by using the “wrong ${p}$” above (e.g. ${p = 2}$) and then interpolate with ${L^q}$ (actually, genuine Hölder by letting ${f \in L^q}$) we see that for ${q < p \leq 2}$,

$\displaystyle \| f_N\|_{L^p} \lesssim \underbrace{N^{-\phi}}_{\text{negative power of }N} \|f\|_{L^q}^{1-\phi} \|\nabla f\|_{L^2}^\phi$

with some ${\phi \in (0,1]}$ determined by ${\frac{1}{p} = \frac{1-\theta}{q} + \frac{\phi}{2}}$.

We treat the remainder of the proof in the ${p \geq 2}$ case—the other region is similar. Now,

$\displaystyle \| f \|_{L^p} \leq \sum_{N \in 2^{\mathbb Z}} \|f_N\|_{L^p} \lesssim \sum_{N \in 2^{\mathbb Z}} \min \left\{ \underbrace{N^{d(\frac{1}{q} - \frac{1}{p})}}_{\text{positive power of }N} \| f\|_{L^q} , \underbrace{N^{d(\frac{1}{2} - \frac{1}{p}) - 1}}_{\text{negative power of }N} \| \nabla f\|_{L^2} \right\}.$

To finish, we figure out when the minimum switches and then sum the geometric series. The details are left as an exercise. $\Box$

Question: If ${\int_{{\mathbb R}^2} |\psi|^2 = N}$, is

$\displaystyle E(\psi) := \int_{{\mathbb R}^2} \frac{1}{2} |\nabla \psi|^2 - \frac{1}{4} |\psi|^4$

positive, or even bounded from below? Think, stars.

Well, if ${N << 1}$, then we do have positivity, by Gagliardo–Nirenberg

$\displaystyle E = \frac{1}{2} \|\nabla \psi\|_{L^2}^2 - \frac{C_{GN}^4}{4} \underbrace{\| \psi\|_{L^2}^2}_{ = N} \| \nabla \psi\|_{L^2}^2$

(note that with ${p = 4, q = 2}$, ${\theta_{GN} = \frac{1}{2}}$). Thus if ${N}$ is small

$\displaystyle E \gtrsim \| \nabla \psi\|_{L^2}^2 \geq 0.$

Conversely, if we take ${\psi(x) = \lambda \psi_0(x)}$ with ${\psi_0 \in \mathcal{S}({\mathbb R}^d)}$, then

$\displaystyle E(\psi) = \frac{1}{2} \lambda^2 \| \nabla \psi_0 \|_{L^2}^2 - \frac{1}{4} \lambda^4 \|\psi_0\|_{L^4}^4$

which will be negative once ${\lambda}$ is large enough, though this makes ${N = \lambda^2 \int |\psi_0|^2}$ large.

Thus there is a transition size ${N}$ at which we switch from ${E > 0}$ always to ${E <0}$ sometimes. What happens if we rescale ${\psi}$ via ${\psi_\lambda(x) = \lambda \psi(\lambda x)}$ to keep ${\int |\psi|^2}$ constant? We obtain

$\displaystyle \begin{array}{rcl} \| \nabla \psi_\lambda\|_{L^2}^2 &=& \displaystyle \int_{{\mathbb R}^2} |\nabla \psi_\lambda|^2 \; dx \\ \\ \| \psi \|_{L^4}^4 &=& \displaystyle \int_{{\mathbb R}^2} \lambda^4 |\psi(\lambda x)|^4 \; dx = \lambda^2 \| \psi \|_{L^4}^4 \end{array}$

Therefore

$\displaystyle E(\psi_\lambda) = \lambda^2 E(\psi)$

and so as soon as ${\psi}$ has negative energy, a re-scaled version has arbitrarily negative energy.

Theorem (Rellich–Kondrachov) Let

$\displaystyle \begin{matrix} 1 \leq q< p < \tfrac{d}{d-2} &\hbox{\hskip 18pt} d \geq 3 \\ \\ 1 \leq q < p < \infty &\hbox{\hskip 18pt} d=2 \\ \\ 1 \leq q < p \leq \infty &\hbox{\hskip 18pt} d = 1 \end{matrix}.$

Then for all ${R > 0}$,

$\displaystyle \{ \chi_R f : \| f \|_{L^q} \leq 1 \hbox{ and } \| \nabla f \|_{L^2} \leq 1 \}$

is compact in ${L^p}$, where ${\chi_R = \chi_{B(0,R)}}$.

Remark Note that, without the ${\chi}$, this result would be false. For example, consider ${\{ \phi(x - x_0), \forall x_0 \in {\mathbb R}^d\}}$ with fixed ${\phi \in \mathcal{S}({\mathbb R}^d)}$.

Proof: From the previous proof, we know that

$\displaystyle \| f_N \|_{L^p} \lesssim N^{-\delta},\hbox{\hskip 18pt} \delta(p,q,d) > 0.$

and thus

$\displaystyle \| f_{\geq N} \|_{L^p} \lesssim \sum_{M \geq N} \|f_M\|_{L^p} \lesssim N^{-\delta}.$

Therefore, the question reduces to showing that

$\displaystyle \{ \chi_R f_{\leq N} : \| f \|_{L^q} \leq 1 \hbox{ and } \| \nabla f \|_{L^2} \leq 1 \}$

is compact for all ${N}$. There are many approaches to prove this.

The first approach is to recall that a ${\mathcal{F} \subseteq L^p}$ is pre-compact if and only if ${\mathcal{F}}$ is

1. Bounded: ${\exists C > 0}$ such that for all ${f \in \mathcal{F}}$, we have ${\|f\|_{L^p} \leq C}$.
2. Tight: ${\forall \epsilon > 0 \exists R >0}$ such that for all ${f \in \mathcal{F}}$, we have ${\|f\|_{L^p(|x| > R)} < \epsilon}$.
3. Equicontinuous: ${\forall \epsilon > 0 \exists \delta > 0}$ such that for all ${|y| < \delta}$ and for all ${f \in \mathcal{F}}$, we have ${\int |f(x+y) - f(x)|^o \; dx < \epsilon}$.

In our case, property 1 follows by the Gagliardo–Nirenberg inequality, property 2 follows by the analysis of ${\chi_R}$, and property 3 follows by the analysis of ${P_{\leq N} f = N^d \check \phi(N \cdot) * f}$.

The second approach is to note that, because we are on a compact/bounded region (namely, ${\overline{B(0,R)}}$), compactness of ${\{\chi_R f_{\leq N}\}}$ as a set of continuous functions in the supremum norm implies compactness in ${L^p}$. By Arzelà–Ascoli, we need only to check boundedness and equicontinuity:

$\displaystyle \|f_{\leq N}\|_{L^\infty} \lesssim N^{d/q}, \hbox{ and } \|f\|_{L^q} , \|\nabla f_{\leq N} \|_{L^\infty} \leq N^{1 + \frac{d}{q}} \|f\|_{L^q}.$

which follow by Bernstein’s inequalities. $\Box$

Theorem (Radial Gagliardo–Nirenberg) If ${f \in L^q \cap \dot H^1}$ and is spherically symmetric, then

$\displaystyle |x|^\frac{2(d-1)}{q+2} |f(x)| \lesssim \|f\|^{\frac{q}{q+2}}_{L^q} \| \nabla f \|_{L^2}^\frac{2}{q+2}$

Proof: Let us write ${f}$ as a function of radius ${r}$. By the fundamental theorem of calculus,

$\displaystyle (f(r))^{1+\frac{q}{2}} = -\int_r^\infty (1+\frac{q}{2}) |f(\rho)|^{\frac{q}{2} - 1} \bar f(\rho) f'(\rho) \; d\rho.$

Therefore,

$\displaystyle \begin{array}{rcl} |f(r)|^{1+\frac{q}{2}} &\underbrace{\lesssim}_{\rho > r}& \displaystyle r^{-(d-1)} \int_r^\infty \rho^\frac{d-1}{2} |f(\rho)|^\frac{q}{2} \rho^\frac{d-1}{2} |f'(\rho)| \; d\rho \\ \\ \scriptsize\hbox{[Cauchy--Schwarz] \hskip 18pt} &\lesssim& r^{-(d-1)} \sqrt{ \int_0^\infty |f|^q \rho^{d-1} \; d\rho} \sqrt{ \int_0^\infty |f'(\rho)|^2 \; \rho^{d-1} \; d\rho}, \end{array}$

namely,

$\displaystyle |f(x)|^\frac{2+q}{2} \lesssim |x|^{-(d-1)} \|f\|_{L^q}^\frac{q}{2} \|\nabla f \|_{L^2}.$

$\Box$

Corollary The set

$\displaystyle \{ f : \|f\|_{L^q} \leq 1, \|\nabla f\|_{L^2} \leq 1, \hbox{ and } f \hbox{ is spherically symmetric}\}$

is compact in ${L^p}$ when ${d \geq 2}$, where ${p}$ and ${d}$ are as above.

Remark This result is false when ${d = 1}$ (c.f. two bump functions who are symmetric with respect to each other about 0 and letting them move to infinity).

Proof: We just need to check tightness, so that we can apply Rellich–Kondrachov, which we can do via interpolation:

$\displaystyle \|f\|_{L^p(|x| > R)} \lesssim \|f\|_{L^q(|x| > R)}^\theta \underbrace{\|f\|_{L^\infty(|x| \geq R)}^{1-\theta}}_{\|f\|_{L^\infty(|x| > R)} \lesssim R^{-\delta} \|f\|_{L^q}^\phi \|\nabla f\|_{L^2}^{1-\phi}}.$

$\Box$

Theorem (Sharp Gagliardo–Nirenberg) Let ${d \geq 2}$ and ${2 < p < \frac{2d}{d-2}}$. The function

$\displaystyle J(f) = \frac{\|f\|_{L^p}}{\|f\|_{L^2}^{1-\theta} \|\nabla f\|_{L^2}^{\theta}}, \hbox{\hskip 18pt where \hskip 18pt} \theta = \frac{(p-2)d}{2p}$

achieves its maximum on ${H^1({\mathbb R}^d)\setminus\{0\} = \{ f \not \equiv 0 : f \in L^2, \nabla f \in L^2\}}$. Moreover, any such maximizer can be transformed via ${g(x) = \alpha f(\beta x)}$ to a solution of

$\displaystyle -\Delta g - |g|^{p-2} g = -g$

In particular, this partial differential equation has a solution.

Remark One can show (we will not do so here) that there is a unique radial non-negative solution ${Q}$ to this PDE in ${H^1}$. Moreover, one can deduce that every optimizer has the form ${f = \alpha Q(\beta(x - x_0))}$ and even that the Hessian at these points is positive definite perpendicular to the minimizing manifold.

Proof: By the Gagliardo–Nirenberg inequality, ${J}$ is bounded. Let ${f_n}$ be a sequence in ${H^1}$ so that ${J(f_n)}$ converges to the extremal value. Replacing ${f_n}$ by their radial symmetric decreasing rearrangements ${f_n^*}$ will only increase (or retain) the value of ${J}$, hence ${f_n^*}$ is also an optimizing sequence.

Next we consider replacing ${f_n^*(x)}$ by ${\alpha_n f_n^*(\beta_n x)}$. It is easy to check that this does not affect the value of ${J}$. We choose ${\alpha_n, \beta_n}$ so that ${\|\alpha_n f_n^*(\beta_n x)\|_{L^2} = \| \nabla (\alpha_n f_n^*(\beta_n x))\|_{L^2} = 1}$. Henceforth, we call ${\alpha_n f_n^*(\beta_n x) = f_n}$.

By Alaoglu, we can pass to a subsequence so that ${f_n \stackrel{\text{weak-}*}{\rightharpoonup} f}$ in ${L^2}$ and ${\nabla f_n \rightharpoonup F}$, also weak-${*}$ in ${L^2}$. Note that since ${\nabla f}$ is defined via testing against ${\phi \in C_c^\infty}$, we are guaranteed that ${F = \nabla f}$. Lower semi-continuity of norms under weak-${*}$ limits shows that ${\|f\|_{L^2} \leq \liminf \|f_n\|_{L^2} = 1}$ and, similarly, that ${\| \nabla f \|_{L^2} \leq 1}$. [Note that Alaoglu in ${L^p}$ leads nowhere.]

By Rellich–Kondrachov and tightness for radial functions, we know that a further subsequence converges strongly in ${L^p}$. [Actually, we do not need to pass to a further subsequence because testing with ${\phi \in C_c^\infty}$ identifies all subsequential limits as being ${f}$—the weak-${*}$ ${L^2}$ limit.] Note that as ${\|\nabla f_n\|_{L^2} = 1 = \|f_n\|_{L^2}}$, we thus have ${J(f_n) = \|f_n\|_{L^p} \rightarrow \|f\|_{L^p}}$, which is not zero as ${(\sup J) \neq 0}$. Noting that

$\displaystyle \sup J \geq \frac{\|f\|_{L^p}}{\|f\|_{L^2}^{1-\theta} \|\nabla f\|_{L^2}^\theta} \geq \|f\|_{L^p} = \sup J,$

therefore ${\|f_n\|_{L^2} \rightarrow \|f\|_{L^2}}$ and ${\|\nabla f_n\|_{L^2} \rightarrow \|\nabla f\|_{L^2}}$ (${f}$ cannot beat the optimal value). By the Radon–Riesz theorem, we see that ${f_n \rightarrow f}$ in ${H^1}$.

Now, let ${f}$ be an optimizer such that ${\|f\|_{L^2} = \| \nabla f \|_{L^2} = 1}$. Consider ${J(f+t\phi)}$ for some ${\phi \in C_c^\infty}$. This must have ${\frac{d}{dt} \big|_{t=0} J(f+t\phi) = 0}$. Furthermore

$\displaystyle \partial_t \big|_{t=0} \int |f + t\phi|^p \; dx = p \text{ Re} \int |f|^{p-2} f \bar \phi \; dx$

and

$\displaystyle \begin{array}{rcl} \displaystyle \partial_t |_{t=0} \int |\nabla (f + t\phi)|^2 \; dx &=& \partial_t \big|_{t=0} \int |\nabla f|^2 + 2t \text{ Re } \nabla f \cdot \nabla \bar \phi + t^2 |\nabla \phi|^2 \; dx \\ \\ &=& \displaystyle \int -2 \text{ Re } (\underbrace{\Delta f}_{\scriptsize\begin{matrix}\text{distributional} \\ \text{Laplacian}\end{matrix}} \phi) \; dx. \end{array}$

Therefore, using the fact that ${\|f\|_{L^2} = \| \nabla f \|_{L^2} = 1}$,

$\displaystyle \begin{array}{rcl} 0 &=& \displaystyle \frac{d}{dt} \big|_{t=0} J(f+t\phi) \\ &=& \displaystyle \| f \|_{L^p}^{1-p} \text{ Re } \int |f|^{p-2} f \bar \phi \; dx-\|f\|_{L^p} (1 - \theta) \text{ Re } \int f \bar \phi \; dx + \theta \|f\|_{L^p} \text{ Re } \int \Delta f \bar \phi \; dx, \end{array}$

valid for every ${\phi \in C_c^\infty}$. Choosing ${\phi}$ and ${i\phi}$ and canceling ${\| f\|_{L^p}}$, we obtain

$\displaystyle \int \big( \theta \Delta f - (1 - \theta) f + \|f\|_{L^p}^{-p} |f|^{p-2} f \big) \bar\phi \; dx = 0$

for all ${\phi \in C_c^\infty}$. Thus ${f}$ is a distributional solution to the PDE

$\displaystyle \theta \nabla f + (\max J)^p |f|^{p-2} f = (1-\theta) f$

which is called the Euler–Lagrange equation for our extremal problem. By rescaling ${g(x) = \tilde \alpha f(\tilde \beta x)}$, with ${\tilde \alpha, \tilde \beta}$ determined by ${\theta, \max J}$ alone, we get

$\displaystyle \Delta g + |g|^{p-2} g = g$

To recap, every optimizer will, after suitable rescaling, solve the above PDE. Moreover, there is at least one optimizer that has ${f = f^*}$. $\Box$

As a sketch of “uniqueness” of optimizers: given an optimizer ${f}$, note that ${f^*}$ is also an optimizer. Moreover, we must have ${\int |\nabla f|^2 = \int |\nabla f^*|^2}$. In general, this is not enough to guarantee that ${f(x) = f^*(x-x_0)}$ (see submarine remark above); however this conclusion does follow if we can show that the set ${\{f^*=\lambda\}}$ has zero measure for ${\lambda >0}$. Switching to polar coordinates, we see that radial optimizers obey an ODE (in a distributional sense)

$\displaystyle \partial_r^2 f^* + \frac{d-1}{r} \partial_r f^* + |f^*|^{p-2} f^* = f^*$

after rescaling. As such, we can see that ${f^*}$ is actually real analytic (at least away from ${r \equiv 0}$) and so has no positive measure level set — we can also realize this from the fact that ${f^*}$ radially decreasing means that a plateau would force our ODE to have a locally constant solution, but uniqueness for ODES implies this would be a globally constant solution ${1}$, which fails to be in ${H^1}$. Lastly, we need the uniqueness of non-negative ${H^1}$ solutions to our ODE.

The proof failed to cover the case ${d = 1}$ and ${2 < p < \infty}$ because ${\{ f \text{ even }: \|f\|_{H^1} \leq 1\}}$ is not tight in ${L^p}$ — for example, consider ${f = \phi(x - n) + \phi(x + n)}$. Nevertheless, ${\{ f^* : \|f\|_{H^1} \leq 1\}}$ is indeed tight in ${L^p}$, which allows us to adapt the preceding proof.

Proof of tightness: We compute

$\displaystyle 1 \geq \int |f^*(x)|^2 \; dx \geq \int_{-R}^R |f^*(x)|^2 \; dx \geq 2R \cdot |f^*(R)|^2$

because ${f^*}$ is decreasing in ${|x|}$. Thus

$\displaystyle \| f^* \|_{L^\infty(|x| \geq R)} = |f^*(R)| \leq \frac{\| f\|_{L^2}}{\sqrt{2R}}$

and so by Hölder’s inequality

$\displaystyle \|f^*\|_{L^p(|x| \geq R)} \leq \|f\|_{L^2}^\frac{2}{p} \|f\|_{L^\infty(|x| \geq R)}^{1-\frac{2}{p}} \leq \frac{\|f\|_{L^2}}{(2R)^\frac{p-2}{2p}}. \qed$

## April 13, 2011

### Harmonic Analysis Lecture Notes 18

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:53 pm

continuation of the Rearrangement Discussions

Theorem Fix ${1 \leq p, q < \infty}$. If ${f \in L^q}$ and ${|\nabla f| \in L^p}$, then ${\nabla |f| \in L^p}$. In fact, ${\big| \nabla |f| \big| \leq |\nabla f|}$ a.e. Similarly, if ${f,g \in L^p}$ with ${\nabla f, \nabla g \in L^p}$, then

$\displaystyle \begin{array}{rcl} |\nabla \max \{f,g\}| &\leq& \max \{ |\nabla f|, |\nabla g| \} \\ |\nabla \min \{f,g\}| &\leq& \max \{ |\nabla f|, |\nabla g| \} \\ \end{array}$

Proof: It is easy once we figure out what ${\nabla |f| \in L^p}$ means: There is an ${\vec F : {\mathbb R}^d \rightarrow {\mathbb C}^d}$ with ${|F| \in L^p}$ so that

$\displaystyle \int \nabla \phi |f| \; dx = - \int \phi \vec F \; dx$

for all ${\phi \in C_c^\infty}$.

Restart the proof: We wish to show that there is some ${\vec F : {\mathbb R}^d \rightarrow {\mathbb C}^d}$ with ${|F| \in L^p}$ so that

$\displaystyle \int \nabla \phi |f| \; dx = - \int \phi \vec F \; dx$

for all ${\phi \in C_c^\infty}$ with ${|\vec F| \leq |\nabla f|}$. In fact, we will see that

$\displaystyle \vec F = \left\{\begin{matrix} \tfrac{u \nabla u + v \nabla v}{\sqrt{u^2 + v^2}} & \hbox{ if } u^2 + v^2 \neq 0 \\ \\ 0 & \hbox{ if } u^2 + v^2 = 0 \end{matrix}\right.$

does the job; here ${f = u + iv}$.

Choose ${u_n, v_n \in C^\infty}$ converging to ${u}$ and ${v}$ in ${L^q}$ a.e., and so that ${\nabla u_n, \nabla v_n}$ converges to ${\nabla u}$ and ${\nabla v}$ a.e., and in ${L^p}$ (e.g., mollify ${u,v}$: convolve ${u}$ and ${v}$ with a smooth bump ${n^d \phi(nx)}$ with ${\int \phi = 1}$). We also smooth out the absolute-value function: ${(u+iv) \mapsto \sqrt{\epsilon^2 + u^2 + v^2}}$. As ${\sqrt{\epsilon^2 + u_n^2 + v_n^2}}$ is smooth,

$\displaystyle \int -\nabla \phi \sqrt{\epsilon^2 + u_n^2 + v_n^2} \; dx = \int \phi \frac{u_n \cdot \nabla u_n + v_n \cdot \nabla v_n}{\sqrt{\epsilon^2 + u_n^2 + v_n^2}}$

That ${(u+iv) \mapsto \sqrt{u^2 + v^2 +\epsilon^2}}$ is (uniformly) Lipschitz, so ${\sqrt{\epsilon^2 + u_n^2 + v_n^2} \rightarrow \sqrt{\epsilon^2 + u^2 + v^2}}$ in ${L^p}$. Furthermore, noting ${\nabla u_n = \nabla u + (\nabla u_n - \nabla u)}$, where we treat the first term by the dominated convergence theorem with ${|\phi| \; |\nabla u|}$ and treat the second term by ${L^p}$ convergence and ${L^{p'}}$ convergence of ${\phi \frac{u_n}{\sqrt{\epsilon^2 + u_n^2 + v_n^2}}}$ by dominated convergence theorem with ${|\phi}$. Sending letting ${n}$ tend to infinity on both sides of the above equality, we get

$\displaystyle \int - \nabla \phi \sqrt{\epsilon^2 + u^2 + v^2} \; dx = \int \phi \frac{u \nabla u + v \nabla v}{\sqrt{\epsilon^2 + u^2 + v^2}}.$

Sending ${\epsilon}$ to ${0}$ on the left side by DCT with ${|\nabla \phi| \sqrt{1 + u^2 + v^2}}$ and sending ${\epsilon}$ to ${0}$ on the right side by DCT with ${2|\phi| |\nabla f|}$, we get

$\displaystyle \int -\nabla \phi \sqrt{u^2 + v^2} = \int \phi \vec F.$

$\Box$

Theorem

$\displaystyle \int |\nabla f^*|^2 \; dx \leq \int |\nabla f|^2 \; dx$

Remark

1. This is also true when we replace the exponent ${2}$ above by some ${p}$ such that ${1 \leq p \leq \infty}$, but the proof is more complicated.
2. Equality can occur when ${f \neq f^*(x- x_0)}$, but only in special circumstances, exemplified perhaps by, say if we think of two bump functions on top of each other in the style of a submarine (viewed from the side).

Proof: Using Plancherel’s theorem

$\displaystyle \begin{array}{rcl} \text{RHS} &=& \displaystyle \int |\xi|^2 \; |\hat f (\xi)|^2 \; d\xi \\ \\ &=& \displaystyle \lim_{t \downarrow 0} \int \frac{1 - e^{-t|\xi|^2}}{t} |\hat f(\xi)|^2 \; d\xi \\ \\ &=& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\int |f|^2 \; dx - \langle f, e^{t \nabla} f \rangle \right] \\ \\ &=& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\int |f|^2 \; dx - \iint \overline{f(x)} \frac{e^{-|x-y|^2/(4t)}}{(4\pi t)^{d/2}} f(y) \; dy \; dx \right] \\ \\ &\geq& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\underbrace{\int |f^*|^2 \; dx}_{\text{equimeasurability}} - \underbrace{\iint \overline{f^*(x)} \frac{e^{-|x-y|^2/(4t)}}{(4\pi t)^{d/2}} f^*(y) \; dy \; dx}_{\text{Riesz rearrangement ineq.}} \right], \end{array}$

which is the left hand side of what we wanted.

Observed that this proof relied on ${f \in L^2}$. To see that this is enough, we treat non-negative ${f}$ via

$\displaystyle f_\epsilon = \left\{\begin{array}{ll} 0 &\qquad f \leq \epsilon \\ f - \epsilon &\qquad \epsilon < f < \epsilon + \tfrac{1}{\epsilon} \\ \tfrac{1}{\epsilon} &\qquad f \geq \epsilon +\tfrac{1}{\epsilon} \end{array}\right.$

which is defined via the minimum and maximum operations, noting ${\max\{a,0\} = \tfrac{1}{2}(|a| + a)}$, i.e., ${f_\epsilon = \min\{ \tfrac{1}{\epsilon} , \max \{ f- \epsilon, 0\} \}}$. This allows us to see that ${|\nabla f_\epsilon | \rightarrow |\nabla f|}$ monotonically, using the previous theorem and similarly ${|\nabla f_\epsilon^*| \rightarrow |\nabla f^*|}$ (note ${(f^*)_\epsilon = (f_\epsilon)^*}$). Sending ${\epsilon \downarrow 0}$, we get the result for non-negative for ${f}$ that are not necessarily in ${L^2}$. For general ${f}$ (possibly taking negative values), keep in mind that ${\big|\nabla |f| \big| \leq |\nabla f|}$. $\Box$

Theorem (Diamagnetic inequality) Let ${A : {\mathbb R}^d \rightarrow {\mathbb R}^d}$ then,

$\displaystyle \big| \nabla |f| \big| \leq |(\nabla + iA)f|, \hbox{ a.e.}$

Proof: From above, we get ${\partial_j |f| = \text{Re} \frac{\bar f \partial_j f}{|f|} = \text{Re} \frac{\bar f (\partial_j + i A_j) f}{|f|} \leq |(\partial_j + i A_j)f|}$. Summing them and taking the square-root, we obtain the result. $\Box$

The energy of a quantum mechanical particle in an electromagnetic field is given by

$\displaystyle E(\psi) = \int |(\nabla + i\vec{A})\psi|^2 + V |\psi|^2 \; dx$

where ${\psi}$ is the wave function, ${\vec{A}}$, ${V}$ are the electromagnetic potentials:

$\displaystyle \vec E = \nabla V, \hbox{\hskip 18pt} \vec B = \nabla \times \vec A$

Note that ${\vec E}$ does not determine ${V}$ uniquely, but only up to an additive constant, but that is alright. More dramatically, ${\nabla \times A_1 = \nabla \times A_2}$ just says ${A_1 - A_2 = \nabla \phi}$ for general ${\phi}$. Change ${A_1 \rightarrow A_2}$, we replace ${\psi}$ by ${e^{i\phi} \psi}$,

$\displaystyle \begin{array}{rcl} (\nabla + iA_2) e^{i\phi} \psi &=& e^{i\phi} \nabla \psi + i \nabla \phi e^{i\phi} \psi + iA_2 e^{i\phi} \psi \\ &=& e^{i\phi} (\nabla + iA_1) \psi \end{array}$

Equivalently, the operator ${\psi \mapsto e^{i\phi} \psi}$ is a unitary mapping that conjugates ${-(\nabla + i \vec A_2)^2 + V}$ to ${-(\nabla + i \vec A_1)^2 + V}$. By the diamagnetic inequality

$\displaystyle \begin{array}{rcl} \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} E_{A,V}(\psi) &\geq& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} \int \big| \nabla |\psi| \big|^2 + V|\psi|^2 \\ \\ &=& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} \int |\nabla \psi|^2 + V|\psi|^2 \\ \\ &=& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} E_{O,V}(\psi) \end{array}$

This shows that the lowest energy configuration has less (or equal) energy when the magnetic field is turned off. Naïvely, this says that it takes energy to push matter into a magnetic field. This property of a material is called diamagnetism (cf paramagnetism).

As an aside, ${\psi : {\mathbb R}^3 \rightarrow {\mathbb C}}$, actually not ${{\mathbb C}}$, but a one-dimensional vector space over ${{\mathbb C}}$, with an inner product. Note that this reduces to ${{\mathbb C}}$ once we choose who ${1}$ represents (a unit vector = orthonormal basis). Any pair of “frames” differs by a uni-modular complex number ${e^{i\phi}}$. Different “gauges” are just different choices of coordinates for such functions.

Now let us consider the idea that our notion of basis of our vector space depends upon our history. A “connection” is a means to perform “parallel transport”. The simplest version is: a function is “constant” if ${\nabla \psi = 0}$. More generally, we say that ${f}$ is “unchanging” if ${(\nabla + iA)\psi = 0}$. When ${\psi}$ traverses a circular path ${\gamma :[0,1] \rightarrow {\mathbb C}}$ with ${\gamma(0) = \gamma(1)}$, we have ${\psi \rightarrow e^{-i \int_0^1 A(\gamma(t)) \cdot \gamma'(t) \; dt} \psi}$. So, no monodromy for all paths is equivalent to ${\int_0^1 A(\gamma(t)) \cdot \gamma'(t) \; dt = 0}$ for all paths, which in turn is equivalent to ${\nabla \times A = 0}$ in a simply connected domain, i.e., no magnetic field. Indeed, the total phase change

$\displaystyle \oint_\gamma A \cdot d\vec r = \int_{\text{disk bounded by }\gamma} \nabla \times A \cdot d\vec S = \text{integral of curvature'' over this disk''}$

cf Foccault pendulum experiment.

## March 30, 2011

### Harmonic Analysis Lecture Notes 17

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:55 pm

Rearrangement Inequalities (cf Lieb–Loss, Ch 3)

Definition Given a Borel set ${A \subseteq {\mathbb R}^d}$ of finite measure, we define ${A^* = B(0,r)}$, an open ball in ${{\mathbb R}^d}$, with ${r}$ chosen so that ${|B(x,r)| = |A|}$. Similarly, we denote the (radially) symmetric decreasing rearrangement of a Borel function ${f : {\mathbb R}^d \rightarrow {\mathbb C}}$ as follows: Suppose ${\{x : |f| > \lambda\}}$ has finite measure for every ${\lambda > 0}$ and define

$\displaystyle f^*(x) = \int_0^\infty \chi_{\{x' : |f(x')| > \lambda\}^*}(x) \; d\lambda$

Remark

1. ${|f(x)| = \int_0^\infty \chi_{\{|f| > \lambda\}}(x) \; d\lambda}$.
2. As a positive linear combination of spherically symmetric functions that are decreasing in radius, so ${f^*}$ has both of these properties.
3. ${|f|}$ and ${f^*}$ are equi-measurable, that is,

$\displaystyle \big| \{|f| > t\} \big| = \big| \{ f^* > t\} \big|.$

To see this, note that

$\displaystyle \{ f^* > t \} = \{ |f| > t\}^*.$

So in particular

$\displaystyle | \{ x : f^* > t \}| = | \{ x : |f| > t\}|$

which is to say ${|f|}$ and ${f^*}$ are equi-measurable.

4. From the above,

$\displaystyle \int \underbrace{\phi(|f|)}_{\phi(0) = 0} \; dx = \int_0^\infty | \{ |f| > \lambda \}| \underbrace{\phi'(\lambda)}_{\tiny \begin{matrix} \text{want bounded variation} \\ \text{on any interval} \end{matrix}} = \int \phi(f^*) \; dx$

Proposition Fix ${f,g}$ measurable, then

$\displaystyle (1) \hbox{\hskip 28pt} \Big| \int fg \; dx \Big| \leq \int |fg| \; dx \leq \int f^* g^* \; dx.$

For any convex increasing function ${\phi [0, \infty) \rightarrow [0,\infty)}$ with ${\phi(0) = 0}$ (e.g., ${\phi(\lambda) = \lambda^p}$)

$\displaystyle (2) \hbox{\hskip 28pt} \int \phi(|f-g|) \; dx \geq \int \phi( \Big| |f| - |g| \Big| ) \; dx \geq \int \phi(f^* - g^*) \; dx.$

Moreover, if ${g = g^*}$ is strictly radially decreasing (that is, for all ${r}$ there is ${\lambda}$ such that ${\{ g > \lambda\} = B(0,r)}$ — note that ${g}$ cannot be compactly supported, nor have plateaus) and ${\phi}$ is strictly convex then equality occurs in either case if and only if ${f = f^*}$.

Example If ${g = \chi_{(-1,1)}}$, then both ${f_1 = \chi_{(-4/5,-3/5)} + \chi_{(-1/5,1/5)} + \chi_{(3/5, 4/5)}}$ and ${f_2 = \text{Gaussian-like"}}$ give the same answer as ${f_j^*}$ without ${f_j = f_j^*}$.

Proof: Suppose ${F : [0,\infty) \times [0,\infty) \rightarrow {\mathbb R}}$ obey

$\displaystyle (*) \hbox{\hskip 28pt} F(\lambda, \mu) - F(\lambda, 0) - F(0,\mu) + F(0,0) \geq 0$

and ${F(0,0) =0}$. This is implied by

$\displaystyle \frac{\partial^2 F}{\partial \lambda \partial \mu} \geq 0$

and integrating over ${[0,\lambda] \times [0,\mu]}$. For example, consider ${F(\lambda, \mu) = \lambda \mu}$ or ${F(\lambda, \mu) = -\phi(|\lambda - \mu|)}$. Note that in the second case we get

$\displaystyle \begin{array}{rcl} \displaystyle \frac{\partial F}{\partial \lambda} &=& - \phi'(|\lambda - \mu|) \text{signum}( \lambda - \mu) = \phi'(|\mu - \lambda|) \text{signum}( \lambda - \mu) \\ \displaystyle \frac{\partial^2 F}{\partial \mu \partial \lambda} = \phi''(|\lambda - \mu|) + \underbrace{\phi'(|\lambda - \mu|)}_{\geq 0 \text{, }\phi\text{ inc}} \delta(\lambda - \mu) \end{array}$

We will show that when ${F(\lambda, \mu) = F(\lambda,0) + F(0,\mu) + \int_0^\lambda \int_0^\mu \underbrace{\frac{\partial^2 F}{\partial \lambda \partial \mu}}_{\text{positive measure}}}$,

$\displaystyle \int F(|f|, |g|) \; dx \leq \int F(f^*, g^*) \; dx.$

Now

$\displaystyle \begin{array}{rcl} \text{LHS} &=& \displaystyle \int F(|f|, 0) \; dx + \int F(0, |g|) \; dx+ \iiint \frac{\partial^2 F}{\partial \lambda \partial \mu}(\lambda, \mu) \chi_{\{|f| > \lambda\}} (x) \chi_{\{ |g| > \mu \}}(x) \; d\lambda \; d\mu \; dx \\ \\ &=& \displaystyle \int F(f^*, 0) + F(0, g^*) \; dx + \iint \underbrace{| \{|f| > \lambda\} \cap \{ |g| > \mu\} |}_{\scriptsize \begin{matrix} \leq \hbox{min of meas} \\ = |\{ |f| > \lambda\}^* \cap \{ |g| > \mu \}^*| \\ = |\{ f^* > \lambda \} \cap \{ g^* > \mu \}| \end{matrix}} \frac{\partial^2 F}{\partial \lambda \partial \mu} \; d\lambda \; d\mu \\ \\ &\leq& \displaystyle \int F(f^*, g^*) \; dx = \text{RHS} \end{array}$

Note that for equality we must have ${ |\{ |f| > \lambda\}^* \cap \{ |g| > \mu \}^*| = |\{ f^* > \lambda \} \cap \{ g^* > \mu \}| }$ for ${\frac{\partial^2 F}{\partial \lambda \partial \mu} \; d\lambda \; d\mu}$ almost everywhere and so in our example for a.e. ${\lambda, \mu}$. $\Box$

Theorem (Riesz Rearrangement Inequality) For measurable ${f,g,h}$ non-negative

$\displaystyle \iint_{{\mathbb R}^d \times {\mathbb R}^d} f(x) g(x-y) h(y) \; dy \; dx \leq \iint f^*(x) g^*(x-y) h^*(y) \; dx \; dy.$

Moreover, if ${g = g^*}$ is strictly radially decreasing and equality eoccurs, then ${f(x) = f^*(x-x_0)}$ and ${h(x) = h^*(x-x_0)}$ for some ${x_0 \in {\mathbb R}^d}$.

Remark By linear changes of variables, one can essentially permute ${f,g,}$ and ${h}$ modulo possible additional reflections.

Application (Non-rotating) stars are spherically symmetric. Minimize

$\displaystyle F = \int \underbrace{\Phi}_{\tiny \begin{matrix} \text{complicated thermodynamic} \\ \text{energy contribution} \end{matrix}}\big(\underbrace{\rho}_{\text{density}}(x) \big) \; dx - \iint \frac{\rho(x) \rho(y)}{|x-y|} \; dx \; dy.$

Passing from ${\rho}$ to ${\rho^*}$ preserves ${\int \Phi\big(\rho(x)\big) \; dx}$ but will decrease ${F}$ (because of the second term, cf theorem above) unless ${\rho}$ is already spherically symmetric about some point ${x_0 \in {\mathbb R}^3}$.

Proof: (of the 1D case, without cases of equality) First write ${f(x) = \int_0^\infty \chi_{\{ |f| > \lambda\}}(x) \; d\lambda}$ and similarly for ${g,h}$ to see that it suffices to treat the case where ${f,g,}$ and ${h}$ are characteristic functions of measurable sets.

By standard approximation arguments, we can pretend that our our open sets are finite collections of open intervals, e.g., ${f(x) = \sum_{j=1}^J \chi_{(-\frac{l_j}{2}, \frac{l_j}{2})} (x - a_j)}$ with mutually disjoint closures.

Now add a new time’ parameter

$\displaystyle f_t(x) = \sum_{j=1}^J \chi_{(-\frac{l_j}{2}, \frac{l_j}{2})} (x - ta_j)$

and similarly for ${g_t,h_t}$. Note that when ${t = 1}$, then ${f_t = f, g_t = g, h_t = h}$. We claim that ${\int f_t(x) g_t(x-y) h_t(y) \; dx \; dy}$ is a decreasing function of ${t}$, at least up until the first collision of a pair of intervals, for which it suffices to treat one summand from each of ${f,g,}$ and ${h}$. In this way, we are claiming that

$\displaystyle \begin{array}{rcl} 0 &\geq& \displaystyle \frac{d}{dt} \iint \chi_{(-\frac{l}{2}, \frac{l}{2})} (x - ta) \chi_{(-\frac{w}{2}, \frac{w}{2})} (x - y - tb) \chi_{(-\frac{l'}{2}, \frac{l'}{2})}(y - tc) \\ \\ &=& \frac{d}{dt} \int_{-\frac{l}{2}}^\frac{l}{2} \int_{-\frac{l'}{2}}^\frac{l'}{2} \chi_{(-\frac{w}{2}, \frac{w}{2})} \big(x' - y' + t(a-b-c) \big) \; dx' \; dy' \\ \\ &=& \frac{d}{dt} (pic) \end{array}$

Combining this with the natural amalgamation on collision algorithm yields the claim.

Before going to the multi-dimensional case, we first concretize a notion. $\Box$

Definition (Steiner symmetrization) Given ${f : {\mathbb R}^d \mapsto [0,\infty)}$ and a direction ${e \in S^{d-1} \subseteq {\mathbb R}^d}$ we define, writing ${x = x'' e + x^\perp}$ with ${x^\perp \perp e}$,

$\displaystyle f^{*e}(x) = [t \mapsto f(te + x^\perp)]^* (x'')$

Note that this is actually measurable! Oops I missed a picture. By Fubini, the ${(d-1)}$-dimensional measure of ${F \cap \{{\mathbb R} e + x^\perp\}}$ is a measurable function of ${x^\perp}$, and this measurable function completely describes the set ${F^{*e}}$.

Proof: (of the 2D case, as a model for general dimension, without cases of equality) Let ${X}$ and ${Y}$ denote the Steiner symmetrization in the ${e_1}$ and ${e_2}$ directions, respectively: ${(Xf)(x) = f^{*e_1}(x)}$.

Let ${R_\alpha}$ denote rotation by some ${\alpha \not \in {\mathbb Q} \pi}$. As in the 1D cases, it suffices to treat the case where ${f,g}$, and ${h}$ are characteristic functions of bounded measurable sets, say ${F, G,H}$. Now define ${F_{k+1} = XYR_\alpha F_k}$ with ${F_0 = F}$ and similarly for ${G_k}$ and ${H_k}$. By the 1D Riesz inequality (and the fact that ${R_\alpha}$ is linear and measure preserving)

$\displaystyle \iint \chi_{F_{k+1}}(x) \chi_{G_{k+1}}(x-y) \chi_{H_{k+1}}(y) \; dx \; dy \geq \iint \chi_{F_k}(x) \chi_{G_k}(x-y) \chi_{H_k}(y) \; dx \; dy.$

We just need to show that ${F_k \stackrel{subseq}{\longrightarrow} F^*}$ say a.e. (and DCT applies because all of the ${F_k}$‘s live in the same big ball ${B(0,R)}$ that contain ${F}$) or ${L^2}$ or others, and same for ${G_k, H_k}$.

Oops there is another picture here. Now, letting ${K}$ be some measurable subset of ${{\mathbb R}^2}$, we have ${XY K = \{ |y| < w(|x|)\}}$ with ${w}$ decreasing (lower semicontinuous).

We have sets ${F_k = \{ |y| < w_k(|x|)\}}$ with ${w_k}$ decreasing and uniformly bounded (by the radius ${R}$ above) with support contained in a common, compact set (e.g., ${[0,R]}$). By the Cantor diagonal slash trick we can find a subsequence of our ${w_k}$ that converge at each rational ${x}$. Because our functions are monotone, we obtain pointwise convergence everywhere except a countable set where the limit function has jumps (there are at most countably many of them). Thus, along this sequence, ${\chi_{F_k}}$ converge almost everywhere (and so also in ${L^2}$) to ${\chi_{\tilde F}}$ for some ${\tilde F = \{ |y| < w(|x|)\}}$ with ${w}$ decreasing. We need to show ${\tilde F}$ is actually ${F^*}$; by the Dominated Convergence Theorem, we obtain ${| \tilde F| = |F| = |F^*|}$. So, it suffices to show ${\tilde F}$ is actually a ball. Fix ${\gamma(x) = e^{-|x|^2}}$, which is strictly radially decreasing (in the technical sense that every ${B(0,r)}$ appears as a super-level set). By the ${L^2}$-contracting property of rearrangements (which incidentally is equivalent to the enhancement-of-dot-products property, noting ${\| f - g\|^2 = \|f\|^2 + \|g\|^2 - 2 \text{Re}\langle f,g \rangle}$), therefore ${\| \gamma - (XYR_\alpha)^k\chi_F\|_{L^2}}$ is decreasing in ${k}$ and so converges (even without passing to our subsequence). In particular,

$\displaystyle \|\gamma - (XY R_\alpha) \chi_{\tilde F}\|_{L^2} = \lim_{\text{subseq}} \| \gamma - (XYR_\alpha)^{k+1} \chi_F\| = \lim_{\text{subseq}} \|\gamma - (XYR_\alpha)^k F\|_{L^2} = \|\gamma - \chi_{\tilde F}\|_{L^2}.$

From the case of equality for our first arrangement inequalities,

$\displaystyle \big\| X[ YR_\alpha \chi_{\tilde F} - \gamma] \big\|_{L^2} = \|YR_\alpha \chi_{\tilde F} - \gamma \|_{L^2}$

By our first rearrangement inequalities, we have

$\displaystyle \| XYR_\alpha \chi_{\tilde F} - \gamma\|_{L^2} = \|XYR_\alpha \chi_{\tilde F} - \underbrace{X\gamma}_{= \gamma}\| \leq \|YR_\alpha \chi_{\tilde F} - \gamma\|_{L^2} = \|YR_{\alpha} \chi_{\tilde F} - \underbrace{Y\gamma}_{= \gamma}\|_{L^2} \leq \|R_\alpha \chi_{\tilde F} - \gamma \|_{L^2} = \|\chi_{\tilde F} - \gamma\|_{L^2}$

By our previous computation, therefore equality holds.

So now let us recall when equality holds in the 1D case: as ${\gamma}$ is already symmetrical, so must the other functions be symmetrical. Therefore, ${|YR_\alpha \tilde F \triangle R_\alpha \tilde F| = 0}$, and hence ${R_\alpha \tilde F}$ is symmetrical across the ${x}$-axis. Undoing the rotation, we see that ${\tilde F}$ has two a.e. reflection axes that are at an angle ${\alpha \not \in \pi{\mathbb Q}}$. Thus, the group generated by these two reflections, which is dense in ${O(2)}$, leaves ${|R\tilde F \triangle \tilde F|= 0}$ for all ${R}$ in this group. Note that ${R \mapsto |R \tilde F \triangle \tilde F|}$ acts on ${O(2)}$, indeed ${|R \tilde F \triangle \tilde F| = \|\chi_{\tilde F} \circ R^{_1}(\cdot) - \chi_{\tilde F}\|_{L^2}}$ and consequently ${|R\tilde F \triangle \tilde F| = 0}$ for all ${R \in O(2)}$ and so ${\tilde F}$ is almost every a ball.

Recall that if ${g = g^*}$ is strictly radially decreasing, then equality requires ${f(x) = f^*(x - x_0)}$ and ${h(x) = h^*(x-x_0)}$. So, for the proof of cases of equality, we use induction on dimension. Moreover we just need to consider the case when ${f = \chi_F}$ and ${h = \chi_H}$ for ${F}$ and ${H}$ being sets of finite measure. Similarly, writing ${g = g^*}$ as an integral over the characteristic functions of its super-level sets, our problem reduces to the following: if for a.e. ${r>0}$,

$\displaystyle (*)\hbox{\hskip 28pt}\iint \chi_F \chi_{B(0,r)}(x-y) \chi_H(y) \; dx \; dy = \iint \chi_{F^*}(x) \chi_{B(0,r)}(x-y) \chi_{H^*}(y) \; dx \; dy,$

then ${\int |\chi_F(x) - \chi_{F^*}(x-x_0)| + |\chi_H(x) - \chi_{H^*}(x-x_0)| \; dx = 0}$.

In the one-dimensional case, if ${r > \tfrac{1}{2}|F^*| + \tfrac{1}{2}|H^*|= \tfrac{1}{2}|F| + \tfrac{1}{2}|H|}$, then ${|x - y| < r}$ for almost ${x \in F^*}$ and ${y \in H^*}$ and so ${\text{RHS}(*) = |F| \cdot |H|}$ and thus ${\text{LHS}(*) = |F| \cdot |H|}$ for a.e. ${r > \tfrac{1}{2}|F| + \tfrac{1}{2}|H|}$. Sending ${r \downarrow \tfrac{1}{2}|F| + \tfrac{1}{2}|H|}$ within this set, we see that every pair of Lebesgue points of ${F}$ and ${H}$ are separated by no more than ${\tfrac{1}{2} |F| + \tfrac{1}{2}|H|}$. From this and elementary geometry, we see that ${F}$ and ${H}$ are a.e. intervals with a common center.

For the higher dimensional case,

$\displaystyle \text{LHS}(*) = \iint \left[\iint \chi_F(x_1, x') \chi_{B(0,\sqrt{r^2 - |x_1 - y_1|^2})}(x'-y') \chi_H(y_1,y') \; dx' \; dy' \right] \; dx_1 \; dy_1$

and similarly for the right hand side. By Riesz, we have inequality of inner integrals for all ${x_1, y_1}$, so equality of ${\text{LHS}(*) = \text{RHS}(*)}$ requires equality for a.e. ${x_1}$ and ${y_1}$. Thus for a.e. ${x_1}$ and ${y_1}$,

$\displaystyle F|_{x_1} := \{ x' : (x_1, x') \in F\} = \big[(F|_{x_1})^* \text{ translated by some } x_0'(x_1, y_1) \big]$

and similarly for the slices of ${H}$, with the same translation parameter ${x_0'}$. Because ${F|_{x_1}}$ does not depend on ${y_1}$ nor ${H|_{y_1}}$ on ${x_1}$, we see that ${x_0'}$ is a.e. constant. Thus, there is an axis parallel to ${e_1}$ and passing through ${(0,x_0')}$ so that ${F}$ and ${H}$ both are ${O(d-1)}$ symmetric about this axis. We can make the same argument for a second axis, say one with an angle ${\alpha \not \in \pi {\mathbb Q}}$, and so see that ${F}$ and ${H}$ are a.e. invariant for a dense set of rotations about the intersection of the axes (If the intersections do not meet, we could use successive rotations to march our set to infinite, contradicting the fact that ${F}$ is invariant and of finite measure.) To finish, we use the fact that ${g\mapsto |gF \triangle F|}$ is a continuous function on the group of Euclidean motions (affine group, or ${O(d) \rtimes {\mathbb R}^d}$). $\Box$

## March 4, 2011

### Harmonic Analysis Lecture Notes 16

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:54 pm

Method of Stationary Phase

Proposition (non-stationary phase) Let ${a \in C^\infty_c}$ and ${\phi \in C^\infty}$ (and real-valued) with ${\phi \geq 1}$ on ${supp(a)}$, then

$\displaystyle \bigg| \int e^{\pm i \lambda \phi(x)} a(x) \; dx \bigg| \lesssim_N \lambda^{-N}$

for every integer ${N \geq 0}$ and ${\lambda > 0}$.

Proof: Let us take the ${e^{+ i \lambda \phi}}$ case,

$\displaystyle I(\lambda) := \int e^{i\lambda \phi(x)} a(x) \; dx.$

In the spirit of integration by parts,

$\displaystyle \begin{array}{rcl} \displaystyle I(\lambda) &=& \displaystyle \int \left[ \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} e^{i\lambda \phi(x)}\right] a(x) \; dx \\ \\ &=& \displaystyle \int e^{i \lambda \phi(x)} \frac{d}{dx} \left( \frac{i}{\lambda \phi'(x)} a(x) \right) \; dx. \end{array}$

Thus, by the quotient rule,

$\displaystyle |I(\lambda)| \leq \frac{1}{\lambda} \left\{ \left\| \frac{a'}{\phi'}\right\|_{L^1} + \left\| \frac{a\phi''}{(\phi')^2}\right\|_{L^1} \right\}$

More generally

$\displaystyle e^{i\lambda \phi} = \left( \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} \right)^N e^{i\lambda \phi}$

and so

$\displaystyle I(\lambda) = \int e^{i\lambda \phi} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right)^N a(x) \; dx.$

Thus

$\displaystyle | I(\lambda) | \lesssim \lambda^{-N} \sum_{k=0}^N \;\;\; \sum_{\tiny \begin{matrix} \alpha_1 + \cdots + \alpha_k + \beta = N \\ \alpha_j \geq 1 , \beta \geq 0 \end{matrix} } \left\| \frac{(\partial^\beta a)(\partial^{\alpha_1} \phi') \cdots (\partial^{\alpha_k} \phi')}{(\phi')^{N+k}} \right\|_{L^1}$

$\Box$

The ideas we use work for sums as well. Traditionally, dealing with sums are much harder.

Theorem (van der Corput lemma) Suppose ${\phi \in C^\infty}$ and ${\phi^{(k)}(x) \geq 1}$ throughout the interval ${[a,b]}$ for some ${k \geq 1}$. If ${k =1}$, then we additionally assume that ${\phi'}$ is monotone. Then

$\displaystyle \bigg| \int_a^b e^{\pm i \lambda \phi(x) } \; dx \bigg| \lesssim_k \lambda^{-\frac{1}{k}}$

independent of ${a,b,}$ and ${\phi}$.

Remark

1. If ${k = 1}$ then the phase is non-stationary; however, we cannot do better because of the end-point contributions from integration by parts (alternatively, the “amplitude function” is ${\chi_{[a,b]}}$, which is not smooth). If ${\phi(x) = x}$, we can compute exactly and see that we cannot do better.
2. To get a bound which is independent of the length of ${[a,b]}$, we need to assume ${\phi'}$ is monotone (or at least some additional hypothesis). Recall

$\displaystyle \text{Im}\; \int_a^b e^{i \phi(x)} \; dx = \int_a^b \sin(\phi(x)) \; dx.$

We define ${\phi}$ heuristically, slow down when ${\sin(x)}$ is positive and speed up when ${\sin(x)}$ is negative. Doing this appropriately, we may make the right side arbitrarily large.

Proof: (by induction in ${k}$). For ${k = 1}$ we argue as before

$\displaystyle \begin{array}{rcl} I(\lambda) &=& \displaystyle \int_a^b e^{i\lambda\phi(x)} \; dx \\ \\ &=& \displaystyle \int_a^{b} \frac{1}{i \lambda \phi'(x)} \left( \frac{d}{dx} e^{i\lambda \phi(x)} \right) \; dx \\ \\ &=& \displaystyle \left[ \frac{e^{i\lambda \phi(x)}}{i \lambda \phi'(x) } \right]^b_a + \int_a^b e^{i\lambda \phi(x)} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right) \; dx. \end{array}$

Thus

$\displaystyle \begin{array}{rcl} I(\lambda) &\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \int_a^b \underbrace{\frac{d}{dx} \frac{1}{\phi'(x)}}_{\tiny \begin{matrix}\text{sign definite as} \\ \phi' \text{ is monotone} \end{matrix}} \; dx \bigg| \\ \\ &\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)}\bigg| \leq \frac{3}{\lambda}. \end{array}$

Now, let us do the induction step (assume the result is true for ${k}$, now show for ${k+1}$). By hypothesis, ${\phi^{(k+1)} \geq 1}$. Let us look at ${\phi^{(k)}}$. This vanishes at most at one point ${c \in [a,b]}$; if it does not vanish on ${[a,b]}$, then choose ${c}$ to be the end point where ${|\phi^{(k)}|}$ is least. Consider the partition of ${[a,b]}$ into interval ${(c- \delta, c+ \delta) \cap [a,b]}$ and its complement, which we will write as ${J}$. On ${J}$, we have ${|\phi^{(k)}| \geq \delta}$ and ${\phi^{(k)}}$ is monotone. Using the inductive hypothesis,

$\displaystyle |I(\lambda)| \lesssim_k \delta + (\lambda \delta)^{-\frac{1}{k}}.$

Choosing ${\delta \approx \lambda^{-\frac{1}{k+1}},}$ we get ${| I(\lambda)| \lesssim_k \lambda^{-\frac{1}{k+1}}}$, as desired. $\Box$

Remark A computation using ${\phi(x) = x^k}$ shows that these bounds are the best that we may expect.

Corollary Let ${\phi}$ obey the hypothesis from the van der Corput lemma and let ${\psi}$ be of bounded variation (or at least ${C^1}$). Consider

$\displaystyle I(\lambda) = \int_a^b e^{\pm i \lambda \phi(x)} \psi(x) \; dx$

then

$\displaystyle | I(\lambda) | \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)$

Proof:

$\displaystyle \begin{array}{rcl} I(\lambda) &=& \displaystyle \int_a^b \psi(x) \frac{d}{dx} \int_a^x e^{i\lambda \phi(y)} \; dy \; dx \\ \\ &=& \displaystyle \Big[ \psi(x) \int_a^x e^{i \lambda \phi(y)} \; dy \Big]_a^b - \int_a^b \psi'(x) \int_a^x e^{i \lambda \phi(y)} \; dy \; dx. \end{array}$

Using the van der Corput lemma twice,

$\displaystyle | I(\lambda) | \lesssim_k |\psi(b)| \lambda^{-\frac{1}{k}} + \|\psi'\|_{L^1} \sup_{a \leq x \leq b} \Big| \int_a^x e^{i\lambda\phi(y)} \; dy \Big| \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)$

$\Box$

An application of this lemma

Definition A sequence ${\{x_k\}_{k=1}^\infty \subseteq {\mathbb R}^z / {\mathbb Z}^n}$ is equi-distributed if

$\displaystyle \frac{1}{K} \sum_{k=1}^K f(x_k) \stackrel{K\rightarrow \infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y) \; dy$

for all continuous functions ${f}$. Analogously, a curve ${x : [0,\infty) \rightarrow {\mathbb R}^n / {\mathbb Z}^n}$ is equi-distributed if

$\displaystyle \frac{1}{T} \int_0^T f(x(t)) \; dt \stackrel{T\rightarrow\infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y)$

for all continuous functions ${f}$.

Remark

1. The notion is unchanged if we replace ${f \in C({\mathbb R}^n/{\mathbb Z}^n)}$ by ${\chi_E}$ or ${\chi_{\mathcal{O}}}$ for arbitrary closed/open sets (Urysohn’s lemma).
2. This replacement does not work for ${F_\sigma}$‘s or ${G_\delta}$‘s (heuristically because they mimic the sequences too well).
3. That ${e^{2\pi i t^2}}$ is not uniformly continuous is implied by the fact that the curve ${(t^2, (t+h)^2)}$ is equi-distributed in ${{\mathbb R}^2 / {\mathbb Z}^2}$ when ${h \neq 0}$, because

Lemma If ${x(t)}$ is equi-distributed then it is dense.

Proof: Choose ${f}$ to be a bump in the open set that ${x(t)}$ misses. $\Box$

Theorem (Weyl’s criterion) The curve ${x(t)}$ is equi-distributed in ${{\mathbb R}^n/{\mathbb Z}^n}$ if and only if

$\displaystyle \frac{1}{T}\int_0^T e^{2\pi i \vec m \cdot \vec x(t)} \; dt \rightarrow 0$

for all ${\vec m \in {\mathbb Z}^n\setminus\{0\}}$.

Proof: The forward implication is a fortiori clear. For the reverse implication, trigonometric polynomials are dense. $\Box$

Theorem Let ${p_1(t), \ldots, p_n(t)}$ be rationally independent “of constants”: that is, if ${\vec m \cdot \vec p(t) := \sum m_j p_j(t) \equiv const}$ with ${\vec m \in {\mathbb Z}^n}$, then ${\vec m \equiv 0}$. Then the curve ${\vec p (t)}$ is equi-distributed in ${{\mathbb R}^n / {\mathbb Z}^n}$.

Remark Conversely, if ${\vec m \cdot \vec p \equiv c}$ where ${\vec m \in {\mathbb Z}^n \setminus\{ 0\}}$, then the curve is no equi-distributed: it lives in a co-dimension one sub-manifold.

Proof: By Weyl’s criterion, we need to show that ${\frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \stackrel{T \rightarrow \infty}{\longrightarrow} 0}$ for all ${\vec m \in {\mathbb Z}^n \setminus\{0\}}$. For fixed ${\vec m}$, we know that ${\vec m \cdot \vec p(t)}$ is a non-constant polynomial of degree ${k \geq 1}$, so ${\vec m \cdot \vec p(t) = a_k t^k + a_{k-1}t^{k-1} + \cdots + a_0}$ with ${a_k \neq 0}$. Changing variables ${t = uT}$ gives

$\displaystyle \frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt = \int_0^1 e^{2\pi iT^k \phi_T(u)} \; du$

where

$\displaystyle \phi_T(u) = T^{-k} \vec m \cdot \vec p(Tu) = a_ku^k + a_{k-1} T^{-1} u^{k-1} + \cdots + a_0T^{-k}.$

Now, note that ${\Big| \frac{d^k \phi_T}{du^k} \Big| = |a_k| > 0}$ uniformly in ${T}$. By Van der Corput’s lemma,

$\displaystyle \Big| \frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \Big| \lesssim |a_k T^k|^{-1/k}$

uniformly in ${T}$. This goes to zero as ${T \rightarrow \infty}$. $\Box$

Proposition (Multi-dimensional non-stationary phase) Fix ${a \in C_c^\infty({\mathbb R}^d)}$ and ${\phi \in C^\infty({\mathbb R}^d)}$ with ${\nabla \phi \neq 0}$ on ${\mathrm{supp}(a)}$. Then

$\displaystyle \Big| \int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx \Big| \lesssim_{N,a,\phi} \lambda^{-N}$

for any integer ${N \geq 0}$.

Proof: By introducing a partition of unity, we can assume that ${| \frac{\partial \phi}{\partial x_j} | \geq \frac{1}{2d} \min_{\mathrm{supp}(a)} |\nabla \phi| > 0}$ on the support of ${a}$ for some fixed ${1 \leq j \leq d}$. Now writing, with ${x'=(x_1, \ldots, \hat{x_j}, \ldots, x_d)}$,

$\displaystyle \int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx = \int_{\text{compact set}} \left( \int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j \right)\; dx'.$

Now, we apply the one dimensional result to ${\int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j }$ to get ${O(\lambda^{-N})}$ with a constant that depends continuously on ${x'}$. $\Box$

An alternative proof is to use ${\frac{\nabla \phi}{i \lambda |\nabla \phi|^2} \cdot \nabla e^{i \lambda \phi} = e^{i\lambda \phi}}$ and integrate by parts sufficiently many times. For example

$\displaystyle \int e^{i \lambda \phi(x)} a(x) \; dx = \int e^{i\lambda \phi} \nabla \cdot \left( \frac{ia\nabla\phi}{\lambda |\nabla \phi|^2}\right) \; dx.$

Proposition (Morse Lemma) Let ${\phi \in C^\infty({\mathbb R}^d)}$ have a non-degenerate critical point at ${0 \in {\mathbb R}^d}$; that is,

$\displaystyle \phi(0) = 0 \hbox{\hskip 28pt but \hskip 28pt} \det \left( \frac{\partial^2 \phi}{\partial x_i \partial x_j}\right) \neq 0.$

Then there is a local diffeomorphism ${x \mapsto y}$ in a neighborhood of ${x =0}$ with ${0 \mapsto }$,

$\displaystyle \frac{\partial y_j}{\partial x_i}(0) = \partial_{ij}$

and

$\displaystyle \phi(x) = \phi(0) + \frac{1}{2} \sum_{i,j} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0) y_i(x) y_j(x)$

Proof: By Taylor’s formula (with integral remainder)

$\displaystyle \begin{array}{rcl} \phi(x) &=& \displaystyle \phi(0) + x \cdot \nabla \phi(0) + \int_0^1 (1-t) \left( \frac{d^2}{dt^2} \phi(t,x)\right) \; dt \\ \\ &=& \phi(0) + \int_0^1 (1-t) \sum_{i,j} x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt \end{array}$

Note that, for ${|x|}$ sufficiently small, we can write

$\displaystyle \int_0^1 (1-t) x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt = \frac{1}{2}\big(T(x)^t \phi''(0) T(x) \big)_{i,j} = \frac{1}{2} \sum_{k,l} T_{kj}(x) \frac{\partial^2 \phi}{\partial x_k \partial x_l}(0) T_{li}(x)$

with ${T(x)}$ a smooth, non-singular matrix with ${T(0) = \mathrm{Id}}$. (Eg. first make a linear change of variables to reduce ${\phi''(0)}$ to ${\mathrm{diag}(\pm 1, \cdots, \pm 1)}$, then explicitly do the completing the square argument).

Now, set ${y_i(x) = \sum_k T_{ik}(x) x_k}$. Note that ${y(0) = 0}$ and ${\frac{\partial y_i}{\partial x_j}(0) = \sum_{k} T_{ik}(0) \delta_{kj} = \delta{ij}}$ $\Box$

Theorem Let ${a \in C_c^\infty}$, and ${\phi \in C^\infty}$ with a non-degenerate critical point at ${x =0}$ but otherwise, ${\nabla \phi}$ does not vanish on ${\mathrm{supp}(a)}$. Then

$\displaystyle \int_{{\mathbb R}^d} e^{i\lambda \phi(x)} a(x) \; dx = e^{i\lambda \phi(0)} a(0) \lambda^{-\frac{d}{2}} \det \left( \frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)\right)^{-\frac{1}{2}} + O_{a,\phi}(\lambda^{-\frac{d+2}{2}})$

as ${\lambda \rightarrow \infty}$.

Note that the ${\frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)}$ (wait… I forgot what I wanted to say here)

Remark One can use a partition of unity (and spatial translations) to generalize to multiple stationary phase points.

Proof: By the Morse lemma, we can find some small neighborhood of ${x = 0}$ where we can change variables ${x = \Phi(y)}$ to simplify ${\phi}$. Now use a smooth partition of unity to write ${a(x) = a_1(x) + a_2(x)}$ where ${a_1}$ is supported in the neighborhood of ${x = 0}$ where the change of variables is defined, while ${a_2(x}$ vanishes in a neighborhood of ${x=0}$. (${a_1}$, ${a_2 \in C_c^\infty}$ with ${\mathrm{supp}(a_j) \subseteq \mathrm{supp}(a))}$. By the non-stationary phase position (and ${\nabla \phi \neq 0}$ on ${\text{supp}(a_2)}$),

$\displaystyle \int e^{i \lambda \phi(x)} a_2(x) \; dx = O(\lambda^{-N})$

for ${N}$ as large as we wish. This leaves

$\displaystyle \int e^{i\lambda \phi(x)} a_1(x) \;dx = \int e^{i \lambda ( \phi(0) + \frac{1}{2} y^t \phi''(0) y)} a_1 \circ \phi(y) \det(\Phi' x) |.$

Lets call ${a_1 \circ \Phi(y) \det(\Phi'(y)) = \tilde a(y)}$; it is ${C_c^\infty}$ and ${\tilde a(0) = a_1 \Phi \circ \phi(0) = a(0)}$.

Remembering that ${\int fg = \int \hat f \check g}$, we see that

$\displaystyle \mathrm{above} = e^{i\lambda \phi(0)} \det \left( \lambda \frac{\phi''}{2\pi i}\right)^{-\frac{1}{2}} \iint e^{-2\pi^2 i\lambda^{-1} \xi^t (\phi'')^{-1} \xi} \check{\tilde{a}}(\xi) \; d\xi$

which has the correct asymptotics. $\Box$

Application of the theorem

For the following discussion, we shall denote the Fourier transform (and its inverse) by

$\displaystyle \begin{array}{rcl} \hat f(x) &=& \displaystyle (2\pi)^{-d/2} \int e^{-ix \cdot \xi} f(x) \; dx \\ \\ f(x) &=& (2\pi)^{-d/2} \int e^{+ix \cdot \xi} f(x) \; dx \end{array}$

noting that ${\| \hat f\|_{L^2} = \|f\|_{L^2}}$.

Example (Schrödinger’s equation) Consider the long-time behavior of the solutions to

$\displaystyle i \partial_t \psi = -\frac{1}{2} \Delta \psi$

for ${\psi: {\mathbb R}_t \times {\mathbb R}_x^d \rightarrow {\mathbb C}}$ with ${\psi(0,x) = \psi_0(x)}$.

As the equation is space-translation invariant (it is also time-translation invariant, but we will not use that here), we take the spatial Fourier transform to find

$\displaystyle i \partial_t \hat \psi (t,\xi) = \frac{1}{2} |\xi|^2 \hat \psi (t,\xi)$

with ${\hat \psi(0,\xi) = \hat \psi_0(\xi)}$, which we take (for now) to be Schwartz ${\mathcal{S}({\mathbb R}^d)}$. Therefore ${\hat \psi(t,\xi) = e^{-t|\xi|^2/2}\hat \psi_0(\xi)}$, i.e. ${\psi(t,x) = e^{-it|\nabla|^2/2} \psi_0 = e^{it\Delta/2}\psi_0}$. Taking inverse Fourier transforms, we get

$\displaystyle \psi(t,x) = (2\pi)^{-d/2} \int e^{i x \cdot \xi - \frac{1}{2} i t|\xi|^2} \hat \psi_0(\xi) \; d\xi$

and so we may apply the stationary phase theorem. Looking for stationary phase points,

$\displaystyle 0 = \nabla \phi = x - t\xi \iff \xi = \frac{x}{t}.$

Also,

$\displaystyle \mathrm{Hessian}(\phi) = \partial_{\xi_i} \partial_{x_j} \left[ -\tfrac{1}{2} |\xi|^2 + \frac{x}{t} \cdot \xi \right] = -\delta_{ij}$

So the theorem says

$\displaystyle \psi(t,x) = [e^{it\Delta/2}\psi_0](x) = (2\pi)^{-d/2} \cdot e^{\frac{i}{2} |x|^2/t} \cdot \hat \psi_0\left(\tfrac{x}{t}\right) \cdot e^{-i\frac{d\pi}{4}} \cdot \left(\tfrac{1}{2\pi}\right)^{-d/2} t^{-d/2} + O(t^{-\frac{d+2}{2}})$

with the big-${O}$ notation for ${\frac{x}{t}}$ constant and ${t \rightarrow +\infty}$ (i.e., this describes the decay if we travel along a world-line). Simplifying the above, we get

$\displaystyle \psi(t,x) = e^{-i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{smaller"}$

with the interpretation that ${\xi \equiv \mathrm{momentum} \equiv \mathrm{velocity}}$ and so ${\mathrm{mass} = 1}$. Note that for ${t \rightarrow -\infty}$, the asymptotics pick up a “discrete” change

$\displaystyle \psi(t,x) = e^{+i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{smaller"}.$

Proposition We have, uniformly in ${t}$ and ${\psi_0}$, the dispersive estimate,

$\displaystyle \|e^{it\Delta/2}\psi_0 \|_{L_x^\infty} \lesssim |t|^{-d/2} \|\psi_0\|_{L_x^1}$

and the “Fraunhofer formula”

$\displaystyle \left\|e^{it\Delta/2} \psi_0(x) - e^{-i\frac{d\pi}{4} + i \frac{|x|^2}{2t}} t^{-d/2} \hat \psi_0\left(\tfrac{x}{t}\right) \right\|_{L_x^2} \stackrel{t\rightarrow \infty}{\longrightarrow} 0$

for every ${\psi_0 \in L^2({\mathbb R}^d)}$.

Remark The motivation for the dispersive estimate comes from the combination of our stationary phase computation and the a priori estimate ${\| \hat \psi_0\|_{L^\infty} \leq (2\pi)^{-d/2} \|\psi_0\|_{L^1}}$.

Proof: As a Fourier multiplier, ${e^{it\Delta/2}}$ is also just a convolution operator. To compute the convolution kernel, take a ${\psi_0 \in \mathcal{S}({\mathbb R}^d)}$ and find

$\displaystyle \begin{array}{rcl} \psi(t,x) &=& \displaystyle (2\pi)^{-d/2} \int e^{i \xi \cdot x - i t|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ &=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int e^{i \xi \cdot x - (\epsilon+i t)|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ &=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int \int e^{i \xi \cdot (x-y) - (\epsilon+i t)|\xi|^2/2} \psi_0(y) \; dy \; d\xi \\ \\ &\vdots& \scriptsize\hbox{using Fubini and Gaussian integrals} \\ &=& e^{-i\pi d/4}(2\pi t)^{-d/2} \int e^{+ i|x-y|^2/(2t)} \psi_0(y) \; dy \end{array}$

In particular, the above computation gives the dispersive estimate

$\displaystyle \|e^{it\Delta/2} \psi_0\|_{L_x^\infty} \leq (2\pi t)^{-d/2} \| \psi_0\|_{L_x^1}.$

Furthermore, we get

$\displaystyle [e^{it\Delta} \psi_0] (x) = (2\pi t)^{-d/2} e^{-i\pi d/2 + i|x|^2/(2t)} \int e^{i x\cdot y/t} \underbrace{e^{i y^2/(2t)}}_{\scriptsize \begin{matrix} \text{the only factor making}\\ \text{Fraunhofer non-exact} \end{matrix}} \psi_0(y) \; dy.$

Thus,

$\displaystyle \begin{array}{rcl} \text{LHS (Fraunhofer)} &=& \displaystyle \Big\| (2\pi t)^{-d/2} \int (1 - e^{iy^2/(2t)}) e^{i x\cdot y/t} \psi_0(y) \; dy \Big\|_{L_x^2} \\ \\ \tiny \begin{bmatrix} \text{Plancherel and changing}\\ \text{variables to } x' = \frac{x}{t} \end{bmatrix} &=& \| (1-e^{i|y|^2/(2t)}) \psi_0(y)\|_{L_x^2}, \end{array}$

which, by the dominated convergence theorem with dominating function ${4|\psi_0(y)|^2}$, goes to ${0}$. $\Box$

As an aside, we discuss the result in the setting of optics. If we shine a laser through a screen with an aperture, let us denote ${\psi_0(y_1,y_2) = (\text{amplitude }) e^{i \text{phase}}}$ with ${y_1, y_2}$ being the horizontal and vertical axes, respectively. At the far screen we have

$\displaystyle \psi(x_1, x_2) = \int_{{\mathbb R}^2} e^{i \sqrt{L^2 + |x-y|^2}} \; \psi_0(y) \; dy$

where ${\sqrt{L^2 + |x-y|^2}}$ is the distance from the source point to the observation point. Because ${\sqrt{L^2 + |x-y|^2} \approx \sqrt{L^2 + x^2} - \frac{x \cdot y}{\sqrt{L^2 + x^2}} + (\text{smaller})}$ and that ${\sqrt{L^2 + x^2} \approx L + \frac{x^2}{2L} + \cdots}$, the dominant term is essentially the Fourier transform.

Also

$\displaystyle \begin{array}{rcl} 0 &=& (\partial_t^2 - \Delta_{x,y,z}) ( e^{i\omega (t-z)} \psi(x,y,z)) \\ &=& \big( 2i\omega \psi_z + \underbrace{\psi_{zz}}_{\text{`negligible''}} - \Delta_{x,y} \psi \big) e^{i\omega(t-z)} \end{array}$

cf Paraxial approximations.

## 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 ${s \in (0,1)}$,

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

$\Box$

## February 14, 2011

### Harmonic Analysis Lecture Notes 14

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:52 pm

Definition (Construction of the dyadic partition of unity) Fix a ${C^\infty}$ mapping ${\phi : [0,\infty) \rightarrow [0,1]}$ with

$\displaystyle \phi(r) = \left\{\begin{matrix} 1 & r < 1.40 \\ \\ 0 & r > 1.42 \end{matrix}\right. \hbox{\hskip 28pt (note } 1.40 < \sqrt{2} < 1.42)$

which we also regard as a ${C^\infty}$ map ${\phi : {\mathbb R}^d \rightarrow [0,1]}$ via ${\phi(\xi) = \phi(|\xi|)}$. We further define ${\psi: {\mathbb R}^d \rightarrow [0,1]}$ by

$\displaystyle \psi(\xi) := \phi(\xi) - \phi(2\xi)$

Note that

$\displaystyle 1 = \phi(\xi) + \sum_{N\in 2^{\mathbb Z}, N \geq 2} \psi( \frac{\xi}{N}) \quad\underbrace{=}_{a.e}\quad \sum_{N \in 2^{\mathbb Z}} \psi(\xi/N).$

Now, from previous time for this time:

Theorem (Mikhlin multiplier theorem, or possibly Marcinkiewicz, or Hörmander) If ${m : {\mathbb R}^d \setminus\{0\} \rightarrow {\mathbb C}}$ 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

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

maps ${L^p \rightarrow L^p}$ (in a bounded manner) for ${1 < p < \infty}$.

Proof: We will just prove the theorem for ${m}$ obeying “symbol estimates” for multi-indices ${0 \leq \alpha \leq d+2}$. (For the full case, see [Stein, Harmonic Analysis, VI.4.4]) Clearly ${m}$ is bounded on ${L^2}$ (just requires ${m \in L^\infty(d\xi)}$, i.e., ${\alpha = 0}$). To prove it is bounded on ${L^p}$ we just need to check that ${\check m}$ obeys the Calderón–Zygmund cancellation condition:

$\displaystyle \int_{|x| \geq 2|y|} |\check m (x-y) - \check m(x) | \; dx \lesssim 1.$

This is implied by

$\displaystyle | \nabla \check m (x)| \lesssim |x|^{-d-1}$

which is what we will show. Decompose ${m(\xi) = \sum_{N \in 2^{\mathbb Z}} m_N(\xi)}$, where ${m_N(\xi) = \psi(\frac{\xi}{N}) m(\xi)}$. Now

$\displaystyle \begin{array}{rcl} \displaystyle \frac{\partial^\gamma}{\partial \xi^\gamma} [\xi m_N(\xi)] &=& \displaystyle \frac{\partial^\gamma}{\partial \xi^\gamma} [ \psi(\frac{\xi}{N}) \xi m(\xi)] \\ \\ &=& \sum_{\alpha + \beta = \gamma} c^{\alpha\beta}_{\gamma} N^{-|\beta|} (\partial^\beta \psi)(\frac{\psi}{N}) [\frac{\partial^\alpha}{\partial \xi^\alpha} (\xi m(\xi))] \end{array}$

where ${c^{\alpha\beta}_\gamma}$ are some combinatorial coefficients. So,

$\displaystyle \begin{array}{rcl} \displaystyle \left\| \frac{\partial^\gamma}{\partial \xi^\gamma} [\xi m_N(\xi)] \right\|_{L^1(d\xi)} &\lesssim_\gamma& \displaystyle \sum_{\alpha + \beta = \gamma} N^{-|\beta|} \int_{|\xi| \sim N} |\xi|^{1-|\alpha|} d\xi \\ \\ &\lesssim_\gamma& N^{d+1-\gamma} \end{array}$

Hence, because F.T${:L^1 \rightarrow L^\infty}$,

$\displaystyle |x^\gamma (\nabla \check m_N)(x) | \lesssim_\gamma N^{d+1 -\gamma}$

and so, choosing various ${\gamma}$‘s,

$\displaystyle |(\nabla \check m_N)(x) | \lesssim \min( \;\underbrace{N^{d+1}}_{\gamma = 0} \; ,\; \underbrace{N^{-1}|x|^{-(d+2)}}_{|\gamma| = d+2} \;).$

Now we sum

$\displaystyle |\nabla \check m (x) | \lesssim \displaystyle \sum_{N \leq |x|^{-1}} N^{d+1} + \sum_{N \geq |x|^{-1}} N^{-1}|x|^{-(d+2)} \lesssim |x|^{-d-1}.$

$\Box$

Definition (Littlewood–Paley projections) Let ${\psi}$ be as above and ${N \in 2^{\mathbb Z}}$, then

$\displaystyle f_N := P_N f := ( \psi(\frac{\xi}{N} \hat f ))\check\; = \int N^d \check\psi(Ny) f (x - y) \; dy$

and similarly

$\displaystyle f_{\leq N} := P_{\leq N} f := (\phi(\frac{\xi}{N}) \hat f(\xi))\check\; = \int N^d \check \phi(Ny) f(x-y) \; dy.$

These are not honest projections ${P_N^2 \neq P_N}$ (namely, ${\phi^2 \neq \phi}$).

Proposition

1. If ${f \in L^p}$, ${1 \leq p < \infty}$, then

$\displaystyle \sum_{N \in 2^{\mathbb Z}} P_Nf = P_{\leq 1} f + \sum_{N \in 2^{\mathbb Z}, N \geq 2} P_N f = f$

both a.e. and in ${L^p}$.

2. We have

$\displaystyle ||P_N f||_{L^p} + ||P_{\leq N} f||_{L^p} \lesssim ||f||_{L^p}$

for all ${1 \leq p \leq \infty}$.

3. ${\big| [P_N f](x) \big| + \big| [P_{\leq N} f] (x) \big| \lesssim [Mf](x)}$.
4. For ${1 \leq p \leq q \leq \infty}$,

$\displaystyle ||P_N f||_{L^q} + ||P_{\leq N} f||_{L^q} \lesssim N^{d \left( \frac{1}{p} - \frac{1}{q}\right)} ||f||_{L^p} \hbox{\hskip 28pt (S. Bernstein's inequalities)}$

5. ${\| |\nabla|^s f_N\|_{L^p} \approx N^s \| f_N \|_{L^p}}$ for all ${1 \leq p \leq \infty}$ and for all ${s \in {\mathbb R}}$.
6. ${\| |\nabla|^s f_{\leq N} \|_{L^p} \lesssim N^s \| f \|_{L^p}}$ for all ${1 \leq p \leq \infty}$ and for all ${s > 0}$.

Remark

1. ${L^\infty}$ norm convergence fails because smooth functions are not dense in ${L^\infty}$. A.e. convergence holds for the second decomposition in ${L^\infty}$ but not for the first, consider ${f \equiv 1}$.
2. Note that ${P_{\leq N} = P_{\leq 2N} P_{\leq N}}$ and so ${\|P_{\leq N} f\|_{L^q} \lesssim N^{\frac{d}{p} - \frac{d}{q}} \| P_{\leq N} f\|}$. Similarly for ${P_N = (P_{N/2} + P_{N} + P_{2N}) P_N}$. Bernstein’s name is also attached to inequalities of the form ${\| P'\|_{L^\infty([0,1])} \lesssim (\text{deg of }P) \| P\|_{L^\infty([0,1])}}$ for polynomials ${P}$.

Proof: For assertion 2, this is Minkowski’s inequality:

$\displaystyle \| P_{\leq N} f\|_{L^p} = \| N^d \check \phi (N \cdot) * f\|_{L^p} \leq \underbrace{\| N^d \check \phi(Nx)\|_{L^1}}_{= \|\check \phi\|_{L^1} < \infty, \text{ because } \phi \in \mathcal{S}} \| f \|_{L^p}$

and the reasoning for ${\|P_N\|}$ is similar.

For assertion 3,

$\displaystyle \big| [P_{\leq N} f[ (x) \big| \leq \int | N^d \check \phi(Ny)| \; |f(x-y)| \; dy \leq \int \frac{N^d}{(1 + |Ny|)^{100(d+1)}} |f(x-y)| \; dy \leq [Mf](x).$

To realize the final inequality, observe

$\displaystyle g(r) = \int_r^\infty - g'(\rho) \; d\rho = \int_0^\infty -\frac{\chi_{B(0,\rho)}(r)}{\rho^d} g'(\rho) \rho^d, \; d\rho$

i.e., ${g}$ is a positive linear combination of characteristic functions of balls\footnote{${|B(0,1)| \int_0^\infty \rho^d [-g'(\rho) d\rho] = \underbrace{d|B(0,1)|}_{(d-1)\hbox{-volume of }\partial B(0,1)} \int_0^\infty g(\rho) \rho^{d-1} d\rho = \|g\|_{L^1({\mathbb R}^d)}}$} and so

$\displaystyle \int g(y) |f(x-y)| \; dy = \int_0^\infty \int \frac{1}{\rho^d} \chi_{B(0,\rho)}(y) |f(x-y)| \; |g'(\rho)| \rho^d \; dy \lesssim \int_0^\infty [Mf](x) |g'(\rho)| \rho^d \; d\rho$

For assertion 1, when ${f \in \mathcal{S}({\mathbb R}^d)}$, ${\| P_N f\|}$ converges to ${f}$ when ${N \rightarrow \infty}$, to ${0}$ when ${N \rightarrow 0}$, both a.e. and in ${L^p}$. This extends to ${f \in L^p}$ via assertions 3 and 2, respectively.

For assertion 4,

$\displaystyle \begin{array}{rcl} \|P_{\leq N} f\|_{L^q} &=& \|N^d \check \phi(N \cdot) * f\|_{L^q}\\ \scriptsize \begin{bmatrix} \text{Young's} \end{bmatrix} &\leq& \displaystyle \underbrace{\| N^d \check \phi(N x) \|_{L^r}}_{\lesssim N^{d-\frac{d}{r}}} \| f \|_{L^p}, \hbox{\hskip 18pt} \frac{d}{q} + d = \frac{d}{p} + \frac{d}{r} \end{array}$

For assertion 5, let ${\theta_s(\xi) := |\xi|^s (\psi(\xi/2) + \psi(\xi) + \psi(2\xi))}$ and so ${\theta_s(\xi/N) \psi(\xi/N) = N^{-s} |\xi|^s \psi(\xi/N)}$. Thus, by Minkowski’s inequality

$\displaystyle \| N^{-s} |\nabla|^{s} P_N f\|_{L^p} = \| N^d \check \theta_s (N \cdot) * f_N \|_{L^p} \lesssim \| \check \theta_s\|_{L^1} \| f_N \|_{L^p} \lesssim_s \|f_N \|_{L^p}.$

For the converse inequality note that

$\displaystyle \theta_{-s}(\xi/N) \theta_{s} (\xi/N) \psi(\xi/N) = \psi(\xi/N)$

and so

$\displaystyle \| f_N \|_{L^p} = \| \underbrace{N^d \check \theta_{-s} (N \cdot )}_{\theta_{-s}(i \nabla)} * N^d \check \theta_s(N \cdot) * f_N \|_{L^p} \lesssim \underbrace{\|\check \theta_{-s} \|_{L^1}}_{\lesssim 1} \|N^{-s} |\nabla|^s f_N\|_{L^p}.$

For assertion 6,

$\displaystyle \begin{array}{rcl} \| |\nabla|^s f_{\leq N}\| &\leq& \displaystyle \sum_{M \leq N} \| |\nabla|^s f_M \|_{L^p} \\ \\ &\lesssim& \sum_{M \leq N} M^s \| f_M \|_{L^p} \\ \\ &\leq& \underbrace{(\sum_{M \leq N} M^s)}_{\lesssim N^s} \|f \|_{L^p} \end{array}$

$\Box$

Theorem (Littlewood–Paley Square function estimate) Given ${f \in L^p}$ when ${1 < p < \infty}$, define

$\displaystyle S(f)(x) = \sqrt{\sum_{N \in 2^{\mathbb Z}} |f_N(x)|^2}$

Then

$\displaystyle \| S(f) \|_{L^p} \approx_p \| f \|_{L^p}.$

Remark

1. We can view ${S}$ as a sub-linear operator, or as a vestige of the linear operator ${f \mapsto (f_N) \in \ell^2}$ and then the square function estimate is about its mapping properties from ${L^p \rightarrow L^p({\mathbb R}^d \rightarrow\ell^2)}$
2. The Littlewood–Paley operators decompose ${f}$ into its constituent frequencies/wave-lengths/length-scales. The square function says that we do not need to fear subtle cancellations between these different components—note the absolute value signs.
3. Compare ${\|f\|_{L^p} \leq \sum \|f_N\|_{L^p}}$, the triangle inequality, but this is not reversible.

Lemma (Khinchin) Let ${X_n}$ be (statistically) independent identically distributed random variables, Bernoulli ${X_n = \pm 1}$ with equal probability. For complex numbers ${c_n}$,

$\displaystyle \mathop{\mathbb E}( |\sum c_n x_n|^p )^{1/p} \approx_p \left( \sum |c_n|^2\right)^{1/2}$

for each ${0 < p < \infty}$.

Proof: We will just treat ${c_n}$ as real numbers. Note that for all ${t > 0}$

$\displaystyle \mathbb{P}( \sum c_n X_n > \lambda ) \underbrace{\leq}_{\text{Markov}} e^{-\lambda t} \mathop{\mathbb E} \{ e^{t\sum c_n X_n}\} \leq e^{-\lambda t} \prod_n \cosh(t c_n) \leq^* e^{-\lambda t} \prod_n e^{t^2 c_n^2/2} \leq e^{-\lambda t} \exp\{ \frac{t^2}{2} \sum |c_n|^2\}$

and optimizing in ${t}$,

$\displaystyle \mathbb{P} (\sum c_n X_n > \lambda)\leq \exp \left\{ -\frac{1}{2} \frac{\lambda^2}{\sum |c_n|^2} \right\}$

where the ${*}$‘ed inequality follows from noting ${\cosh(x) = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\leq \exp(x^2/2) = \sum_{k=0}^\infty \frac{x^{2k}}{2^k k!}}$. We will finish this proof next time. $\Box$

## 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 ${1 < p < \infty}$.

## 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 ${\xi}$.

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