Let $\Delta$ be the Laplacian operator and denote $e^{i\Delta}$ the Schrödinger semi-group. It is well known that $e^{i\Delta} \in \mathcal{L}(L^2(\mathbb{R}^d)).$ Indeed, we can define $e^{i\Delta}$ as follows; for $f\in L^2(\mathbb{R}^d),$ $$e^{i\Delta}f = \mathcal{F}^{-1} (e^{-i|\xi|^2} \hat{f}),$$ where $\mathcal{F}^{-1}$ is the invers Fourier transform. Moreover, using Parseval's identity, we have $$\|e^{i\Delta}f\|_{L^2} = \|f\|_{L^2}.$$ I wonder whether $e^{i\Delta} \in \mathcal{L}(L^p(\mathbb{R}^d),L^q(\mathbb{R}^d)) $ for $1\leq p\leq2$ and for some $q.$
For $1\leq p\leq2$ and $q = p/(p-1) $ the Fourier transform is bounded from $L^p(\mathbb{R}^d)$ to $L^q(\mathbb{R}^d).$ Then for $f\in L^p$ the function $\xi \mapsto e^{-i|\xi|^2} \hat{f}(\xi) \in L^q.$
In which sense could the inverse Fourier transform be defined and do we have a similar estimate as $\|e^{i\Delta}f\|_{L^2} = \|f\|_{L^2}$ ?
Thank you for any hint.
YES. $e^{i\Delta} \in \mathcal{L}(L^p(\mathbb{R}^d),L^q(\mathbb{R}^d)) $ for $1\leq p\leq2$ and for $q= p/(p-1)$.
In fact, we have the so called the dispersive inequality for Schrödinger operator
$$\|e^{it\Delta} f\|_{p'} \lesssim |t|^{-d(\frac{1}{p} - \frac{1}{2})}\|f\|_p,$$
where $f \in L^p(\mathbb R^d)$ and $p' = p/(p-1)$.
For the proof and more information please see Tao, Terence. Nonlinear dispersive equations. American Mathematical Society, Providence, RI, 2006, especially, Section2.3 and Exercise 2.35 in this book. I am sorry to avoid tips on writing great answers, ;-)