Posted by: Shuanglin Shao | June 14, 2010

## On localization of the Schrodinger maximal function

I was interested in the almost everywhere pointwise convergence problem of the Schrödinger solution $e^{it\Delta}f$ as time goes to zero since I was a graduate student. The question was raised by Carleson, and remains open in high dimensions. It is closely related to the boundedness of the Schrodinger maximal operator (local or global); in turn it is closely related to the interesting oscillatory integrals in harmonic analysis and Strichartz estimates in PDE. Recently I understood some of the problem and wrote down a note to re-construct the two dimensional proof. These results contained in this note are not new; in the note I explore them from a slightly different perspective.

Posted by: Shuanglin Shao | December 6, 2009

## Kato-smoothing effect

“Kato smoothing” “is ” important in dispersive PDE.“Smoothing”, as it stands, sounds like a very good word. But why is it good? How will it be used remains as vague questions to me.

In the last week, I began to report a paper by Alazard-Burq-Zuily, “On the water waves questions with surface tension”, which contains a “local-smoothing” result: roughly speaking, solutions to 2-D water waves with surface tension in $C^0_tH^s_x$ will get $\frac 14$ regularity upgrade to $L^2_tH^{s+\frac 14}_x$, to the price of locally in space and being averaged in time. This result was first proven by Christianson-Hur-Staffilani.

A question came to me, why $1/4$? Where can I quickly see it? Soon I found out it is determined by the structure of the water-wave equation. To oversimplify the major result in the paper on the derivation of water-wave equations by means of paralinearization, it can be written as a type of dispersive equations,
$\partial_t u+iD^{3/2} u=0, \,D:=|\nabla|, \, (*)$

(I have dropped a lot of terms, for instance, only looking at a linear equation, and no flow-terms in (*); it is because I am only looking at main terms which I think reflect dispersion). (*) suggests a dispersive relation $\tau=|\xi|^{3/2}$. If one look for
$\int \int_{|x|\le 1} |D^\alpha u|^2 dxdt \lesssim \|u_0\|_{L^2}$, the maximum value of $\alpha$ turns out to be $1/4$. ($1/4$ was proved for 2D, does this suggest that it was the case in all dimensions?)

To understand this necessary condition on $\alpha$, I would like to draw an analogy with the “1/2-local smoothing” for the Schrodinger equations: for any $\epsilon >0$,

$\int\int \langle x \rangle^{-1-\epsilon} |D^{1/2} e^{it\Delta} f|^2 dx dt \le C \|f\|_2^2.$

Why $1/2$ there? and how it was proved? Then the clarifications of these matters provides a model (to me) that $1/4$ for water-wave sounds reasonable.

Let us first motivate “local-smoothing” for Schr\”odinger? It is well known that the solutions $e^{it\Delta} f$ to a free Schr\”odinger equation

$i\partial_t u+\Delta u=0, u(0,x)=f(x)$

obeys the conservation law of mass, i.e.,

$\| e^{it\Delta} f \|_{L^{\infty}_tL^2_x} =\|f\|_{L^2_x}. \,(1)$

Note that we can not add any $D^\alpha$ with $\alpha>0$ to the left hand side of (1) due to the Galilean transform (or simply by creating a bump at high frequency). One may argue that, is this no-gain-of-derivative due to we asking for too much in time by requiring a $L^\infty_t$? For instance, on a time interval, $[0,1]$, $L^\infty$-norm is stronger than any $L^q$ norm with $1\le q<\infty$? So is the following true,

$\|D^\alpha e^{it\Delta} f\|_{L^q_tL^2_x}\le C\|f\|_2,\,(2)$

for some $\alpha>0$ and $1\le q<\infty$. Unfortunately (2) does not hold for any $\alpha>0$ due to the same reason as above (one can take the same examples.).

Is there any hope that some variant of (2) holds true? By using the heuristic that solutions to a dispersive equations at high frequency travels much faster than low frequency, after waiting for a long time, only low-frequencies are left behind around spatial origin and they do not hurt positive derivatives. So if we were asking for an estimate locally in space, was it possible? The answer turns out to be Yes. We have the following estimate,

$\int \int_{|x|\le 1} |D^{1/2} e^{it\Delta}|^2 dxdt \lesssim \|f\|^2_2.\, (3)$

This is referred to as “Kato smoothing estimate" for Schr\"odinger in literature. Moreover $1/2$ is the most one can expect due to the obstruction (counterexample) we mentioned above. (3) will lead to a more general estimate, for any $\epsilon>0$

$\int \int_{\mathbb{R}^d} (1+|x|)^{-1-\epsilon} |D^{1/2} e^{it\Delta}|^2 dxdt \lesssim \|f\|^2_2.\, (4)$

The implication of (4) from (3) is easier by partitioning $\{|x|\ge 1\}$ and rescaling, and using the information that $\epsilon>0$.

We focus on proving (3). It will follow from a Plancherel argument.
Writing

$D^{1/2} e^{it\Delta} f(x)=\int_{\xi^\prime;, \tau} e^{ix^\prime\cdot \xi^\prime+it\tau}\int_{\xi_d} e^{ix_d\xi_d} \delta (\tau-|\xi|^2)|\xi|^{1/2} \widehat{f}(\xi)d\xi_d d\xi'd\tau,$

where $\delta$ is the Dirac mass, and $\xi=(\xi^\prime,\xi_d)$. We set

$F(\xi^\prime,\tau)=\int_{\xi_d} e^{ix_d\xi_d} \delta (\tau-|\xi|^2) |\xi|^{1/2}\widehat{f}(\xi)d\xi_d.$

To prove (3), it suffices to prove

$\int_{|x_d|\le 1} \int_{\mathbb{R}^d } |\widehat{F}(x^\prime,t)|^2 dx^\prime dt dx_d \le C\|f\|_2^2, \, (5)$

where $x=(x^\prime,x_d)$. Obviously by the Plancherel theorem, the left hand side of (5) is bounded by

$\int_{|x_d|\le 1} \int_{(\xi^\prime, \tau)} |F|^2 d\xi^\prime d\tau dx_d.$

Fixing $(\xi^\prime \tau)$, Cauchy-Schwarz yields

$|F|^2\le \int_{\xi_d} |\widehat{f}|^2 \delta(\tau-|\xi|^2 )d\xi_d \int_{\xi_d} |\xi|\delta(\tau-|\xi|^2) d\xi_d. \, (6)$

The second factor on the right hand side of (6) is bounded by an absolute if we restricting $\xi$ to to set $\{\xi: |\xi|\le \sqrt{d} |\xi_d|\}$; it is not hard to do so if at the beginning we aim to prove (3) under this restriction; then the general estimate follows from the triangle inequality and partition the frequency space into $laex d$ pieces.

So plugging (6) in (5) and interchanging the integration order, and using $|x_d|\le 1$, we find out the left hand side of (5) is bounded by $\|f\|_2^2$. This is exactly what we need. So we finish proving (3).

I like the previous type of argument very much; but I did not remember where it was the first place/time I saw it. So I recorded it for my own benefits.

Posted by: Shuanglin Shao | November 14, 2009

## Reading: GWP for critical 2D dissipative SQG

Last weekend I attended a wonderful conference SCAPDE at UC Irvine, where I learned an interesting theorem from Kiselev: solutions to critical 2D dissipative quasi-geostrophic equations (SQG) are globally wellposed

$\theta_t=u\cdot \nabla \theta-(-\Delta)^\alpha\theta, u=(u_1,u_2)=(-R_2\theta, R_1\theta),$

where $\theta:\mathbb{R}^2\to \mathbb{R}$ is a scalar function, $R_1, R_2$ are the usual Riesz transform in $\mathbb{R}^2$ defined via $\widehat{R_i(f)}(\xi)=\frac {i\xi_i}{|\xi|}\widehat{f}(\xi)$ and $\alpha\geq 0$.

This is his joint work with Nazarov and Volberg. I took a look at this paper in the past few days. The proof makes good use of classical tools in Fourier analysis such as singular integrals and modulus of continuity, which are familiar topics in Stein’s book, Singular integrals and differentiability properties of functions. It is a place where I do not see Strichartz estimates or Littlewood Paley decompositions and see the power of classical Fourier analysis.

The monotonicity formula they have is

$\|\nabla \theta\|_\infty\le C\|\nabla \theta_0\|_\infty \exp \exp\{C\|\theta_0\|_\infty\}, (1)$

for periodic smooth initial data $\theta_0$. I am new to this field of fluid dynamics but I would still like to say a few words on this estimate. The proof is to find an “upper bound “, modulus of continuity $\omega$, for solutions $\theta$; then they show that a family of modulus of continuity is preserved under evolution of the equation, which is strong enough to control $\|\nabla \theta\|_\infty$. Quoted from the paper, the idea is to show that the critical SQG possesses a stronger “nonlocal” maximum principle than $L^\infty$ control.

Recall a modulus of continuity $\omega: [0,\infty)\mapsto [0,\infty)$ is just an arbitrarily increasing continuous and concave function such that $\omega(0)=0$. A function $f: \mathbb{R}^n\mapsto \mathbb{R}^m$ has modulus of continuity $\omega$ if $|f(x)-f(y)|\le \omega (|x-y|)$ for all $x,y\in \mathbb{R}^n$.

We choose not to report the crucial/essential part of finding the continuity of modulus (maybe later). Instead we assume that $f$ has modolus of continuity $\omega$, which is unbounded and $\omega^\prime (0)<\infty$ and $\lim_{\xi\to 0+}\omega^{\prime\prime}=-\infty$. Then there holds

$\|\nabla f\|_\infty< \omega^\prime (0), (2)$.

The proof is actually very simple. The explicit form of $\omega$ will take care of the implication of (1) from (2).

Assume that $\|\nabla f\|_\infty=|\nabla f(x)|$ for some $x$. We consider the point $y=x+\xi e$ for $e=\frac {\nabla f}{|\nabla f|}$. On the one hand, we have

$f(y)-f(x)\le \omega (\xi), (3)$

for all $\xi\geq 0$. On the other hand, the left hand side of (3) is at least $|\nabla f(x)|\xi -C\xi^2$ where $C=\frac 12 \|\nabla^2 f\|_\infty$ while its right hand side can be represented as $\omega^\prime (0)-\rho(\xi)\xi^2$ with $\rho(\xi)\to\infty$ as $\xi\to 0+$.

Then

$|\nabla f(x)| \le \omega^\prime (0)-(\rho(\xi)-C)\xi$

for all sufficiently small $\xi>0$, and it remains to choose some $\xi>0$ satisfying $\rho(\xi)>C$.

Posted by: Shuanglin Shao | May 1, 2009

## The Cotlar-Stein lemma

The Coltar-Stein lemma is a powerful tool to deal with $L^2$ boundedness of some translation-invariant operators such as convolution operators, which can be expressed as

$Tf(x)=\int K(x-y) f(y)dy.$

I recently began to understand how powerful it might be via the $TT^*$ method. I would like to reproduce its proof following Fefferman’s presentation.

$\mathbf{Statement.}$

Suppose $T:=\sum_{k=1}^M T_k$ is a sum of operators on Hilbert spaces (say the usual $L^2$). Assume that
$\|T^*_jT_k\|\le a(j-k)$ and $\|T_jT^*_k\|\le a(j-k)$. Then $\|T\|\le \sum_{|j|\le M }\sqrt{a(j)}$.
Here $a(\cdot)$ is a non-negative even function.

$\mathbf{Proof.}$ The argument follows an idea of iteration.

Step 1. $\|Tf\|^2=\langle Tf, Tf \rangle =\sum_{j=1}^M\sum_{k=1}^M \langle T_j f, T_k f\rangle$. We estimate $\langle T_j f, T_k f\rangle$ into two ways. Firstly it is easy to see that
$\langle T_j f, T_k f\rangle \le (\max \|T_j\|)^2 \|f\|^2$; secondly it is also trival that
$\langle T_j f, T_k f\rangle = \langle T^*_kT_j f, f\rangle \le a(j-k) \|f\|^2$. Hence by taking the geometric means of these two bounds, we have

$\langle T_j f, T_k f\rangle \le \max \|T_j\| \sqrt {a(j-k)} \|f\|^2$.

Hence

$\|T\|^2\le \sum_{j=1}^M\sum_{k=1}^M \max \|T_j\| \sqrt {a(j-k)} \le M \max \|T_j\| \sum_{|j|\le M} \sqrt{a(j)}$.

This gives that

$\|T\|\le (M\max\|T_j\|)^{1/2} (\sum_{|j|\le M} \sqrt{a(j)})^{1/2}.$

Step 2. We investigate one more iteration of $TT^*$. We write

$\langle TT^* f, TT^* f\rangle= \sum_{1\le k_1,k_2,j_1,j_2\le M} \langle T_{k_1}T^*_{j_1}f, T_{k_2}T^*_{j_2}f\rangle$
$\qquad = \sum_{1\le k_1,k_2,j_1,j_2\le M} \langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle.$

Also we estimate $\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle$ by organizing the operators in two ways.
Firstly since $T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}=\bigl(T_{j_2}T^*_{k_2}\bigr) \bigl(T_{k_1}T^*_{j_1}\bigr)$, we see that

$\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle \le a(j_2-k_2)a(j_1-k_1) \|f\|^2;$

secondly since $T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}=T_{j_2}\bigl(T^*_{k_2}T_{k_1}\bigr)T^*_{j_1}$,

$\langle T_{j_2}T^*_{k_2}T_{k_1}T^*_{j_1}f, f\rangle \le ( \max\|T_j\|)^2 a(k_2-k_1) \|f\|^2;$

By taking geometric means, we see that

$\|T\|^4 \le \sum_{1\le k_1,k_2,j_1,j_2\le M} \max\|T_j\| \sqrt{a(j_2-k_2)}\sqrt{a(k_2-k_1)}\sqrt{a(k_1-j_1)}$
$\qquad \le M\max(\sum_{|j|\le M}\sqrt{a(j)})^3 .$

Hence
$\|T\|\le (M\max\|T_j\|)^{1/4} (\sum_{|j|\le M}\sqrt{a(j)})^{3/4}$.

Step 3. by induction, we see that for any $n\ge 1$, we have

$\|T\|\le (M\max\|T_j\|)^{\frac 1{2n}} (\sum_{|j|\le M}\sqrt{a(j)})^{\frac {2n-1}{2n}}$.

Let $n\to\infty$, we see that

$\|T\|\le \sum_{|j|\le M} \sqrt{a(j)}$. The proof of this lemma is complete.

We remark that in the argument both bounds $\|T_jT^*_k\|$ and $\|T^*_jT_k\|$ are used.

$\mathbf{Example.}$

Let $P:=\sum_{k\in \mathcal{Z}}P_k$ where $P_k$ be the Littlerwood-Paley projection operator. It is easy to see that $P_k$ is a self-adjoint operator. Also $a(0)\le 1$ and $a(j)=0$ for $j\neq 0$.
Then Stein-Cotlar gives,

$\|P\|\le 1$,

which matches that $\|P\|=1$.

Posted by: Shuanglin Shao | February 28, 2009

## A misleading argument for the adjoint Fourier restriction for the sphere in 3 dimension

Today I thought that I had a new proof of the adjoint Fourier transform for the sphere in three dimensions; but it turns out it was wrong. I would like to record it here, which I hope is useful at some point.

First let us define the notion of Fourier transform. Let
$\widehat{f}(\xi) :=(2\pi)^{-3/2}\int_{\mathbf R^3} e^{-ix\xi} f(x)dx.$
$\|\widehat{f}\|_{L^2(\mathbf R^3)}=\|f\|_{L^2(\mathbf R^3)}.$
From this definition the convolution of two functions behaves under the Fourier transform like
$\widehat{ f\ast g}= (2\pi)^{3/2} \widehat{f}\widehat{g}.$

Now let us begin the argument. By Plancherel,
$\|\widehat{fd\sigma}\|^2_{L^2(\mathbf{R}^3)} =(2\pi)^{-3/2} \|fd\sigma \ast fd\sigma \|_{L^2(\mathbf{R}^3)}.$
In view of this, we may assume that $f\ge 0$.
We write the integrand out,
$fd\sigma \ast fd\sigma (x) =\int_{S^2} [f\sigma](x-y)[fd\sigma](y).$

(1) By Cauchy-Schwarz inequality, we have
$f(x-y)f(y)\le \dfrac {|f(x-y)|^2+|f(y)|^2}{2}.$
We observe that
$\int_{S^2}f^2(x-y)d\sigma(x-y)d\sigma(y)=\int_{S^2}d\sigma(x-y)f^2(y)d\sigma(y).$
Then
$\|fd\sigma \ast fd\sigma \|_{L^2(\mathbf{R}^3)}$
$\le\|\int_{S^2}f^2(y)d\sigma(x-y)d\sigma(y) \|_{L^2(\mathbf{R}^3)}.$

(2)Then by Minkowsik inequality,
$\|\int_{S^2}d\sigma(x-y)f^2(y)d\sigma(y) \|_{L^2(\mathbf{R}^3)}$
$\le \|d\sigma\|_{L^2} \int_{S^2} f^2(y)d\sigma=(4\pi)^{1/2} \|f\|^2_{L^2(S^2)}.$

Note that the best constant for Young’s inequality $L^1\times L^2 \to L^2$ is $1$.

(I orginially thought that $\|d\sigma\|_{L^2(\mathbf R^3)} =(\int_{S^2} d\sigma)^{1/2}=(4\pi)^{1/2}$. However, it is wrong since $d\sigma\neq d\sigma^2$ or $d\sigma\notin L^2(\mathbf R^3)$, which can be seen from the asympotics $\widehat{d\sigma}(\xi)\sim (1+|\xi|)^{-1}$ for large $\xi$ in $\mathbf R^3$. Incidentally, today I found that Terry remarked on his blog that the Dirac mass $\delta^2 \notin L^1$ by a trick of epsilon regularization (another way is to use Pancherel theorem and observe that $\widehat{\delta} =1$).)

At this point, there is no need to read on since the following is based on Step 2. I worte it down before I realized the mistake; so I backed it up as my notes.

{Then we can conclude that
$\|\widehat{fd\sigma}\|^2_{L^2 (\mathbf{R}^3)}\le (2\pi)^{-3/2}(4\pi)^{1/2}\|f\|^2_{L^2(S^2)}=2^{-1/2}\pi^{-1/2}\|f\|^2_{L^2(S^2)}.$
}

Posted by: Shuanglin Shao | February 18, 2009

## Pohazav argument: no moving-to-left “solitions” for the critical gKdV

I would like to show a short but beautiful Pohazav argument to exclude “moving-to-left” solitons in the form of $u(t,x):=u(x-ct)$ for the focusing generalized Korteweg de Vries equation
$u_t+u_{xxx}+ (u^5)_x=0.$
This is to show that $c>0$.

Firstly we assume that $u(x,t)=u(x-ct)$ is a solution for the equation above and also assume that $u$ decays to $0$ in the infinity (to justify the arguments). Then
$-cu_x+u_{xxx}+(u^5)_x=0.$
Then an integration yields $-cu+u_{xx}+u^5=0. \qquad (1)$.

We multiply (1) by $u$ and integrate,
$-c\int u^2-\int u_x^2 + \int u^6=0. \qquad (2)$

We multiply (1) by $xu$ and integrate,
$-c\int xuu_x +\int xu u_{xxx}+\lambda \int xu (u^5)_x=0,$
which is simiplified to
$\frac c2 \int u^2 +\frac 32 \int u_x^2-\int u^6=0. \qquad (3)$

Now we see that $(2)\times \frac 32+(3)$ yields that
$-c\int u^2 +\frac {4}3 \int \lambda^6=0$,
This forces $c>0$.

In general, for the defocusing critical gKdV, to exclude solutions in the “solitions” sense is really a hard topic.
A decay estimate on the right is desired.

Posted by: Shuanglin Shao | February 17, 2009

## A continuation in reading: heat flow monotonicity of Strichartz norms

It has been a long time that I haven’t updated this blog. I promised to post my readings on the Bennett-Bez-Carbery-Hundertmark’s short but beautiful paper,  “heat flow monotonicity of Strichartz norm”. Here it goes!

In this paper, the authors applied the method of heat-flow in the setting of  Strichartz inequality for the Schr\”odinger equation and amusingly obtained, among other things, the Strichartz norm

$\|e^{is\Delta}f\|_{ L^{2+\frac 4d}_{s,x}(\mathbf{R}\times \mathbf{R}^d)}$

is nondecreasing as the initial datum $f$ evolvs under a under quadratic flow when $d=1, 2$. This immediately yields that the Gaussians are extremisers to the classical Strichartz inequality,
$\|e^{is\Delta}f\|_{ L^{2+\frac 4d}_{s,x}(\mathbf{R}\times \mathbf{R}^d)} \le S_d\|f\|_{L^2(\bf{R}^d)}, \text{ with } S_d:=\frac {\|e^{is\Delta}f\|_{ L^{2+\frac 4d}_{s,x}(\mathbf{R}\times \mathbf{R}^d)}}{\|f\|_{ L^2(\bf{R}^d)}}.$
The paper also consider the monotonicity given by some other flows such as Mehler-flow; it also considers the higher dimensions’s analogues by embedding the usual Strichartz norm into one-parameter family of norms $\||\cdot\||_p$. In this post I will focus on the monotonicity induced by the heat-flow.

The method of heat-flow deformation is interesting in itself. It is expected to be applied to related problems such as the existence of extremisers to the adjoint Fourier restriction operators to the sphere; but it is a different story; so far I can see an immediate difficulty in the latter setting, i.e., an representation formula for $\|\widehat{fd\mu}\|_{L^{2+\frac 4d}_x(\bf{R}^{d+1})},$ where $d\mu$ is the standard surface measure for the unit sphere in $\bf R^{d+1}$.

Fortunately the representation formula of this kind is available in the paraboloid setting thanks to Hundertmark-Zharnitsky’s work; I already reported it in my previous post but I would like to record it here again basically because an “equation” is rare in analysis: for nonnegative $f\in L^2(\bf R)$,
$\|e^{is\Delta}f\|^6_{L^6_{s,x}(\bf R\times \bf R)}=\frac {1}{\sqrt{12}}\int_{\bf R^3} (f\bigotimes f\bigotimes f) (X)P_1( f\bigotimes f\bigotimes f)(X) dX,$
where $P_1: L^2(\bf{R}^3)\to L^2(\bf{R}^3)$ is the projection operator onto the subspace of functions on $\bf R^3$ which are invariant under the isometries which fix the direction $(1,1,1)$. ( We have a similiar representation formula when $d=2$. )

The paper innovatively combines this formula with monotonicity given by some heat-flow version of Cauchy-Schwartz inequality together to prove the following main theorem.

Let $f\in L^2(\bf{R}^d).$ If $(p,q,d)$ is Schr\”odinger admissible and $q$ is an even integer which divides $p$ then the one-parameter family of functions

$Q_{p,q}(t):= \|e^{is\Delta}(e^{t\Delta}|f|^2)^{1/2}\|_{ L^p_sL^q_x(\mathbf{R}\times \mathbf{R}^d)}$

is nondecreasing for all $t>0$; i.e., $Q_{p,q}(t)$ is nondecreasing in the case $(1,6,6), (1,8,4)$ and $(2,4,4)$. Next we show how to prove this theorem in the case $(1,6,6)$.

${\bf 1. \text{ Monotonicity of } \Lambda(t)}$
Given $n\in \mathcal{N}$ and nonnegative integrable functions $f_1, f_2$ on $\bf R^n$, we define $\Lambda(t):=\int_{\bf R^n} (e^{t\Delta}f_1)^{1/2}(e^{t\Delta}f_2)^{1/2}$. Then $\Lambda(t)$ is nondecreasing for all $t>0$. Instead of repeating the proof in the paper of using the convolution with the heat-kernel, we will try working out how it is deduced from the divergence theorem. Our goal is to prove the following explicit formula
$\Lambda'(t)=\frac 14 \int_{\bf R^n} |\nabla (\log e^{t\Delta}f_1)-\nabla (\log e^{t\Delta}f_2)|^2 (e^{t\Delta}f_1)^{1/2}(e^{t\Delta}f_2)^{1/2}.$

Let $U_j:=e^{t\Delta }f_j=\frac {1}{(4\pi t)^{n/2}} \int_{\bf R^n} e^{-\frac {|x-y|^2}{4t}}f_j(y)dy$ for $j=1,2$. Then
$\partial_t U_j=-\frac {n}{2t} U_j +\frac {1}{(4\pi t)^{n/2}}\int_{\bf R^n}e^{-\frac {|x-y|^2}{4t}}\frac {|x-y|^2}{4t^2}f_j(y)dy.$

We write $U_j= \frac {1}{(4\pi t)^{n/2}} \int_{\bf R^n} e^{-\frac {|y|^2}{4t}}f_j(x-y)dy$. Then
$\frac {d^2}{d^2x_i}U_j=\frac {1}{(4\pi t)^{n/2}} \int_{\bf R^n} (\frac {y_i^2}{4t^2}-\frac {1}{2t})e^{-\frac {|y|^2}{4t}}f_j(x-y)dy$. Then
$\Delta U_j= \frac {1}{(4\pi t)^{n/2}} \int_{\bf R^n}(\frac {|y|^2}{4t^2} -\frac {n}{2t})e^{-\frac {|y|^2}{4t}}f_j(x-y)dy$.
Hence
$\partial_t U_j = \Delta U_j$. This is not surprising because $U_j$ formally solves the heat equation.

Let $V_j:= \nabla(\log U_j )$, then
$\partial_t (\log U_j) =\frac {(U_j)_t}{U_j}=\frac {\Delta U_j}{U_j}=div (V_j)+|V_j|^2$.

Hence we see
$\Lambda'(t) =\frac 12 \int_{\bf R^n} [\partial_t (\log V_1)+\partial_t \log V_2] \sqrt{U_1U_2}$
$\qquad =\frac 12 \int_{\bf R^n} [div (V_1)+div (V_2)+|V_1|^2+|V_2|^2] \sqrt{U_1U_2}$
$\qquad =\frac 14 \int_{\bf R^n } |V_1-V_2|^2 \sqrt{U_1U_2}.$
Here we have used,
$div(\sqrt(U_1U_2) V_1)=\sum_{i=1}^n (V_1)_i \sqrt{U_1U_2}+ (\frac {V_1}{2}+\frac {V_2}{2})\sqrt{U_1U_2}V_1$, which yields
$0=\int_{\bf R^n} div(\sqrt{U_1U_2} V_1)+div(\sqrt{U_1U_2} V_2)$
$\quad \int_{\bf R^n } \left(div (V_1)+div (V_2)+\frac {|V_1|^2}{2}+\frac {|V_2|^2}{2}+V_1V_2 \right) \sqrt{U_1U_2}$,
i.e.,
$\int_{\bf R^n } \left(div (V_1)+div (V_2)\right) \sqrt{U_1U_2}=-\int_{\bf R^n}\left(\frac {|V_1|^2}{2}+\frac {|V_2|^2}{2}+V_1V_2\right)\sqrt{U_1U_2}.$

${\bf 2. \text{ Monotonicity of } Q_{6,6}(t) }$

We start with a general representation formula for the projection map in $L^2$. For functions in $G\in L^2(\bf R^3)$ we write
$P_1 G(X) =\int_O G(\rho X) d\mathcal{H}(\rho),$
where $O$ is the group of isometries on $\bf R^3$ which fix the direction $(1,1,1)$, and $d\mathcal{H}(\rho)$ denotes the right-invariant Haar probability measure on $O$.

Let $F:=f\bigotimes f\bigotimes f$ and $F_\rho:=F(\rho \cdot)$. Then by using the fact the heat-flow operator commutes with tensor product and Hundertmark-Zharnistsky representation formula, we have
$Q_{6,6}^6=\frac {1}{\sqrt{12}} \int_O \int_{\bf R^3} (e^{t\Delta}|F|^2)^{1/2}(X) (e^{t\Delta}|F_\rho|^2)^{1/2}(X)dXd\mathcal{H}(\rho).$
This is in form of $\Lambda(t)$. Then the monotonicity of $Q_{6,6}$ follows from that of $\Lambda(t)$ and the non-negativity of the measure $d\mathcal{H}$.

${\bf 3. \text{ Gaussians are extremisers }}$

For $f\in L^2$, define $u(t,x)=H_t*|f|^2(x)$ with $H_t:=\frac {1}{(4\pi t)^{d/2}}e^{-\frac {|x|^2}{4t}}$, and the rescaled
$\tilde{u}(t,x) =t^{-d} u(t^{-2},t^{-1}x)=\frac {1}{(4\pi)^{d/2}}\int_{\bf R^d} e^{-\frac 14 |x-tv|^2}|f(v)|^2dv.$
Then
$Q_{6,6}(t^{-2}) =\|e^{is\Delta} (\tilde{u}(t,\cdot))^{1/2}\|_{L^6_{s,x}}.$
$\lim_{t\to \infty} Q_{6,6}(t)=\|e^{is\Delta}(H_1^{1/2})\|_{L^6_{s,x}} \|f\|_{L^2}.$
On the other hand,
$\lim_{t\to 0} Q_{6,6}(t)=\|e^{is\Delta}|f|\|_{L^6_{s,x}}.$
We observe that, by Plancherel theorem in both variables,
$\|e^{is\Delta}f\|^3_{L^6_{t,x}}=\|(fd\sigma)^\vee(fd\sigma)^\vee(fd\sigma)^\vee\|_{L^2_{s,x}}$
$\qquad = \|(fd\sigma)*(fd\sigma)*(fd\sigma)\|_{L^2_{\tau,\xi}}$
$\qquad \le \|(|f|d\sigma)*(|f|d\sigma)*(|f|d\sigma)\|_{L^2_{\tau,\xi}}=Q^3_{6,6},$
where $d\sigma$ denotes the surface measure of the paraboloid in $\bf R^2$. Hence we obtain
$\|e^{is\Delta}f\|_{L^6_{t,x}}\le \|e^{is\Delta}(H_1^{1/2})\|_{L^6_{s,x}} \|f\|_{L^2}.$
This shows that $H_1^{1/2}$ is an extremiser. So we are done.

${\bf 4. \text{ A final word}}$
You can see that the heat-flow deformation method is a very nice method; it has been applied by the authors in several settings such as the Hausdorff-Young inequality, and Young’s inequality (but I haven’t carefully read their previous works). It is also proved effective in treating $d$-linear analogues of the Strichartz estimate (Bennett-Carbery-Tao’s work) and in the setting of multilinear Brascamp_Lieb inequalities (Bennett-Carbery-Christ-Tao) (I admitted that I haven’t read them carefully either). The “disadvantage” is that this method doesn’t charaterize the set of extremisers. Of course, we can’t hope it is a tool of catch-all. The question of charaterization is a completely different problem.

Posted by: Shuanglin Shao | October 1, 2008

## Readings on sharp Strichartz inequalities for the Schrodinger equation

Since last weekend, I have read several papers due to Bennett-Bez-Carbery-Hundertmark, Carneiro, Foschi, Hundertmark-Zharnitsky and  on the sharp Strichartz inequality

$\|e^{it\Delta}f\|_{L_{t,x}^{2+4/d}(\Bbb R\times \Bbb R^d)}\le C\|f\|_{L^2_x(\Bbb R^d)}$

where $e^{it\triangle}f$ is the solution to the following free Schrodinger equation

$iu_t+\triangle u=0$

with initial data $u(0,x)=f(x)$, where $u(t,x):\Bbb R\times {\Bbb R}^d\to \Bbb C$.

I am happy that the latest two papers cited mine on the existence of a maximiser for the nonendpoint Strichartz inequality.  For my own benefits, I would like to record here the main idea of their proofs respectively.

I choose to start with Hundertmark-Zharnitsky’s paper, “on sharp Strichartz inequalities in low dimensions”(IMRN) since this paper introduces a beatiful argument which seems to be of general interest (at least essential to papers by Carneiro and Bennett-Bez-Carbery-Hundertmark).  In this paper, Hundertmark-Zharnitsky obtained the sharp value for

$C_{p,q}:=\sup_{f\neq 0} \frac {\|e^{it\Delta}f\|_{L^q_sL^r_x(\Bbb R\times {\Bbb R}^d)}}{\|\|_{L^2_x({\Bbb R}^d)}}$

when $q=r=2+4/d$ in lower dimensions $d=1,2$.  They first built an interesting representation formula for the Strichartz norm $\|e^{it\Delta}f\|^{2+4/d}_{L^{2+4/d}(\Bbb R\times {\Bbb R}^d)}$ as follows,

If $d=1$, then

$\int\int_{\Bbb R} |e^{it\Delta}f|^6dxdt=\frac 1{\sqrt{12}}\langle f\bigotimes f\bigotimes f, P_1(f\bigotimes f\bigotimes f)\rangle_{L^2_x({\Bbb R}^3)},$

where $f\bigotimes g$ denotes the usual tensor product and $P_1:L^2_x({\Bbb R}^3)\to L^2_x({\Bbb R}^3)$ denotes the orthogonal projection to the closed linear subspace which consists of functions invariant under rotations of ${\Bbb R}^3$ which keep the $(1,1,1)$ direction fixed.

If $d=2$,

$\int\int_{{\Bbb R}^2} |e^{it\Delta}f|^4dxdt=\frac 14\langle f\bigotimes f, P_2(f\bigotimes f)\rangle_{L^2_x({\Bbb R}^4)},$

$P_2:L^2_x({\Bbb R}^4)\to L^2_x({\Bbb R}^4)$ denotes the orthogonal projection to the closed linear subspace which consists of functions invariant under rotations of ${\Bbb R}^4$ which keep the $(1,0,1,0)$ and $(0,1,0,1)$ directions fixed.

They proved them in a simple way.  we only look at the $d=1$ case and its explicit maximizers. By using the definition of the delta function $\delta(\xi)=(2\pi)^{-1}\int e^{-ix\xi}dx,$ they expressed the left side $\int\int_{\Bbb R} |e^{it\Delta}f|^6dxdt$ as

$\frac 1{2\pi}\int \int_{{\Bbb R}^3}\delta((1,1,1)\cdot(\eta-\zeta))\delta(|\eta|^2-|\zeta|^2)\overline{f\bigotimes f\bigotimes f(\eta)}f\bigotimes f\bigotimes f(\zeta)d\eta d\zeta.$

This leads to define the following symmetric linear operator: for $G\in \mathcal{C}_0^\infty({\Bbb R}^3)$,

$A_1(G)(\eta):=\frac 1{2\pi}\int_{{\Bbb R}^3}G(\zeta)\delta((1,1,1)\cdot(\eta-\zeta))\delta(|\eta|^2-|\zeta|^2)d\zeta$

Then they showed that $A_1$ is a bounded operator with operator bound $\frac 1{\sqrt{12}}$ on $\mathcal{C}_0^\infty({\Bbb R}^3)$  by showing the measure $m_{\eta} (d\zeta):=\frac {\sqrt{3}}{\pi}\delta((1,1,1)\cdot(\eta-\zeta))\delta(|\eta|^2-|\zeta|^2)$ is a probability measure on ${\Bbb R}^3$ for almost every $\eta\in {\Bbb R}^3$.  Finally  they extended $A_1$ onto the whole $L^2_x({\Bbb R}^3)$ and showed that $A_1$ is a multiple of the orthogonal projection $P_1$.  Now they obtain the representation formula for the Strichartz norm.

Once having representation formula, they obtained that

$\|e^{it\Delta}f\|^6_{L^6_{t,x}(\Bbb R\times {\Bbb R}^3)}\le \frac 1{\sqrt{12}}\langle f\bigotimes f\bigotimes f, P_1(f\bigotimes f\bigotimes f)\rangle_{L^2_x({\Bbb R}^3)},$

which gives $C_{6,6}\le \frac 1{\sqrt{12}}$. Also they saw that in order to have equality, the function $f\bigotimes f\bigotimes f$ must lie in the rangle of $P_1$, that is to say, it is invariant under rotations of ${\Bbb R}^3$ which fix $(1,1,1)$. Obviously, all functions of the form

$Ae^{(-\lambda+i\mu )x^2+cx}$

with $\lambda >0, \mu\in \Bbb R, A\in \Bbb C$, i.e., Gausssians,  are maximisres.

In the second part of their paper, they managed to show that the Gausssians turn out to be the only maximizers in the following three steps,

“1”. Assume $f$ is a maximizer for the Strichartz inequality (then $f\bigotimes f\bigotimes f$ is in the range of $P_1$). Then $Q_t(f)$ never vanishes for all $t>0$. Here $Q_t(f)$ is the convolution of $f$ with the approximation to identity $Q_t(f):=\frac {1}{(2\pi t)^{1/2}}\int e^{-\frac{|x-y|^2}{2t} }f(y)dy$.

“2”. $Q_t(f)$ never vanishes and  is differentiable, then $Q_t(f)$ is a Gausssian. hence as a limit, $f$ is a Gausssian too.

The fact “2” above is actually the following general theme: if $f$ is differentialable and never vanishes and $f\bigotimes f\bigotimes f$ is in the range of $P_1$, then $f$ is a Gausssian, which is proven by working out the explicit forms of the rotations $M(\theta)$ in ${\Bbb R}^3$  which keep $(1,1,1)$ invariant and use $h(\eta)=h(M(\theta)\eta)$ for all differentiable functions invariant under rotations.  At this step, we note that $f\bigotimes f\bigotimes f$ is a product of one dimensional function, which we need in the one dimensional argument. In higher dimensions, for $f\in L^2_x({\Bbb R}^d)$ and $f\bigotimes \cdots\bigotimes f$ invariant under rotations which keep $(e_i,\cdots, e_i)_{1\le i\le d}$ with standard basis $e_i\in {\Bbb R}^d$, we will have to prove that $f$ is a product of  one dimensional functions.

Recently, Carneiro generaized this argument to a prove a sharp form of  “Strichartz inequalities” as follows, for $k\in \mathcal{Z}, k\ge 2$ and $(d,k)\neq (1,2)$,

$\|u(t,x)\|_{L^{2k}_tL^{2k}_x(\Bbb{R}\times \Bbb{R}^d)}\le \left(C_{d,k}\int_{\Bbb{R}^{dk}} |\hat{F}(\eta)|^2K(\eta)^{\frac {d(k-1)-2}{2}}\right)^{1/2k}d\eta$

with

$C_{d,k}=[2^{d(k-1)-1}k^{d/2}\pi^{(d(k-1)-2)/2}\Gamma(\frac {d(k-1)}{2})]^{-1}$

and $F(\eta)=f(\eta_1)\cdots f(\eta_d)$ with $\eta_i\in \Bbb{R}^d$ for $1\le i\le d$ and the kernel $K(\eta)=\frac{1}{k}\sum_{1\le i.

This inequality is sharp if and only if $f$ is a Gaussian.

How to find $K$ is mysterious for me here.

(I leave the discussion on Foschi’s and Bennett-Bez-Carbery-Hundertmark’s papers to a later time)