In Sobolev space $H^{s}(\mathbb{R}^n):=\left\{u\in L^2(\mathbb{R}^n:\mathcal{F}^{-1}((1+|\cdot|^2)^{s/2}\mathcal{F}(u))\in L^2(\mathbb{R}^n)\right\}$
Why $\mathcal{F}^{-1}((1+|\cdot|^2)^{s/2}\mathcal{F}(u))=(Id-\Delta)^{s/2}(u)$?
I know that, if $s=2$, and because $\widehat{\Delta u}=-|\cdot|^2\widehat{u}$ (Fourier transforms properties), then \begin{align*}|(Id-\Delta)u|_{2}&=|\mathcal{F}(u-\Delta u)|_{2}\\ &=|\widehat{u}-\widehat{\Delta u}|_{2}\\ &=|\widehat{u}-(-|\cdot|^2\widehat{u})|_{2}\\ &=|\widehat{u}+|\cdot|^2\widehat{u}|_{2}\\ &=|(1+|\cdot|^2)\widehat{u}|_{2} \end{align*} and \begin{align*}|\mathcal{F}^{-1}((1+|\cdot|^2)\mathcal{F}(u))|&=|(1+|\cdot|^2)\widehat{u}|_{2} \end{align*} Therefore, $\mathcal{F}^{-1}((1+|\cdot|^2)\mathcal{F}(u))=(Id-\Delta)(u)$
But, Why $\mathcal{F}^{-1}((1+|\cdot|^2)^{s/2}\mathcal{F}(u))=(Id-\Delta)^{s/2}(u)$? The $s/2$ causes me a problem.