In a paper, I see the following inequality:

$$\vert e^{it\Delta} g(x)- e^{it'\Delta} g(x') \vert\leqslant C (\vert t-t'\vert + \vert x-x'\vert),$$ where $C$ depends on $d$ and $\Vert g \Vert_{L^{\infty}}$ ($g\in\mathcal{S}(\mathbb{R}^d)$). Here the $\Delta$ means the Laplacian. What's more, $e^{it\Delta}g(x)=\mathcal{F}^{-1}(e^{-t\vert\eta\vert^2}\mathcal{F}(g)(\eta))(x),$ where $\mathcal{F}$ means the Fourier transform in $\mathbb{R}^d$.

I do not know how to prove this? Can anybody give me some suggestions?


1 Answer 1


$e^{it\Delta}g(x) - e^{it'\Delta}g(x') = e^{i t \Delta}(g(x) - g(x') + g(x')(e^{it\Delta} - e^{it'\Delta})$.

Combining with $\left|\,g(x) - g(x')\right| \leq d \max_{k,y} \left|\; \partial g/\partial x_k(y)\right|\; \left|\,x - x'\right|$ gives

$\left|\,e^{it\Delta}g(x) - e^{it'\Delta}g(x')\right| \leq d \| g'\|_{L^\infty} \; \left|\,x - x'\right| + \| g\|_{L^\infty} \Delta \left|\,t - t'\right|$

This depends on $\| g'\|_{L^\infty}$ as well as $\| g\|_{L^\infty}$ but it must be that way. Take $d=1$, $t=t'$ and $g(x) = Mx\phi(x)$ for $ 0 \leq x \leq 1/M$ where $\phi$ is $C^\infty$ and identically $1$ on $ 0 \leq x \leq 1/M$.

Similarly, you can't get rid of the $\Delta$ in $\Delta \left|\,t - t'\right|$. Take $x = x'$, $t=0$ and $t' = \pi/\Delta$.

  • $\begingroup$ I made a mistake, Here the $\Delta$ means the Laplacian, not a number. What's more, $e^{it\Delta}g=\mathcal{F}^{-1}(e^{-t\vert\eta\vert^2}\mathcal{F}(g)(\eta)),$ where $\mathcal{F}$ means the Fourier transform. $\endgroup$
    – x tang
    Oct 6, 2013 at 3:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .