Let $f$ be a Schwartz function. Prove that, whenever $2\le r < \infty,$

$$\| e^{it \Delta} f\|_{L^{3r}(\mathbb{R}^2_{xt})} \le c \| \widehat{f}\|_{r'},$$

Where $1/r + 1/r' = 1.$

My Attempt

My main idea was to use some Plancherel-like Theorem, either for one or for two variables. More specifically, my main idea was proving the statement for a specific class of exponents, such as integers or even integers, and after that finding a way to interpolate.

For these exponents, the idea should be to write the norm above as the $L^2-$norm of a function. For even exponents, this works just fine, so we can express the desired quantity as we want.

As also the Schrödinger Kernel can be expressed as the Fourier Transform of a finite measure, we would like then to use Plancherel's Theorem to express the norm above as a convolution product of measures with $L^2-$norms taken.

This all is pretty justifiable, but the last part is the hardest one: passing from this $L^2$ norm of a convolution to a $L^{r'}$ norm of a Fourier Transform. I think I am close enough, but I cannot conclude. Maybe Young's inequality will come handy, but I cannot see how at this moment.

Any help would be appreciated.

  • $\begingroup$ Can you clarify the notation for the norm on the LHS? Is it the $L^{3r}$ norm of the function $g(x,t):=(e^{it\Delta}f)(x)$ defined on $\mathbb{R}^{2}$? $\endgroup$ – Matt Rosenzweig Feb 11 '16 at 19:16

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