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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 $\langle -\Delta \rangle ^{s}$?

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    $\begingroup$ It might help if you define $(i\partial_t)^{1/2}$. $\endgroup$ Feb 22, 2017 at 5:33
  • $\begingroup$ (1) via the spectral theorem? $\endgroup$ Feb 22, 2017 at 5:46
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    $\begingroup$ @MichaelRenardy $i\partial_t$ with domain $H^1(\mathbb R)$ is a self-adjoint operator on $L^2(\mathbb R)$: can't one use the spectral theorem also in this case? Or does your question points to the fact that the OP does not specify which of the most usual fractional derivatives (Riemann-Liouville or Caputo) he's picking? $\endgroup$ Feb 22, 2017 at 8:11
  • $\begingroup$ If you are comfortable with $|\nabla|$, I do not see how you can be uncomfortable with $\sqrt{1-\Delta}$. In this context (dispersive PDEs) those symbols are always meant to be Fourier multipliers. $\endgroup$ Feb 22, 2017 at 13:17

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