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)