Put $\langle x \rangle ^s = (1 + |x|)^{1/2},$ and $|\nabla |^s $ denotes the Fourier multiplier with symbol $|\xi|^s$, that is, $\widehat{|\nabla |^s f} = |\xi|^s \hat{f}.$
Put $ \langle \nabla \rangle = (1- \Delta)^{1/2}$ (And how should I define this? any motivation?) and
we define $\|f\|_{H^{s}} = \|\langle \nabla \rangle ^{s} f \|_{L^2}$
Questions:
(1) How I should define $(- \Delta )^{1/2}$?
How to justify $(i\partial_t)^{\frac{1}{2}} e^{it\Delta} f = (-\Delta )^{\frac{1}{2}} e^{it\Delta} f$, $f\in \mathcal{S}(\mathbb R \times \mathbb R^d)$ (Schwartz space) (where $e^{it\Delta}$ is a Schrodinger propagator ) (2) Let $K$ be any compact set in $\mathbb R \times \mathbb R^d.$ Can we show
$\|e^{it\Delta } f \|_{H^{1/2}_{t,x}(K)} \leq C \| \langle - \Delta \rangle^{1/2} e^{it\Delta} f \|_{L^{2}(K)} \leq C \|\nabla f \|_{L^2}$
(Do I need to use the H\"olders inequality and (1) to prove (2))
Motivation: I am trying to understand the proof of Inverse Strichatz estimates Proposition 3.2, p.242 (Chapter 3). The above fact has been used in it (see p.242).
Edit: I just saw here, p.12, we may say $|\nabla|^s = (-\Delta)^{\frac{s}{2}}$ . And so $(-\Delta )^{\frac{1}{2}} e^{it\Delta} f(\xi) = |\xi| \widehat{e^{it\Delta}f}(\xi)= |\xi| e^{-it |\xi|^2}\hat{f}(\xi) $, But now how should I handle $(i\partial_t)^{\frac{1}{2}} e^{it\Delta} f$?
We may write $\widehat {\langle \nabla \rangle^s} f(\xi) = \langle \xi \rangle ^s \hat{f}(\xi).$ How should I define $\langle -\Delta \rangle ^{s}$?