For $0<\alpha<2$, we define the fractional Laplacian with Fourier transform
\begin{align}
\widehat{(-\Delta)^{\frac{\alpha}{2}} u}(\xi) = |\xi|^\alpha \widehat u(\xi).
\end{align}
Consider the resolvent operator
$$
\mathrm R\left(\lambda;\ (-\Delta)^{\frac{\alpha}{2}}\right)
:=\left(\lambda I-(-\Delta)^{\frac{\alpha}{2}}\right)^{-1},\qquad 
\mathrm{Re}\,\lambda\le0.
$$
${\bf My\ question:}$. 


The following estimate
$$
\left\|\mathrm R
\left(\lambda;\ (-\Delta)^{\frac{\alpha}{2}}\right)\right\|
\_{L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)}
\leq \frac{C}{|\lambda|+1},\quad\mathrm{Re}\,\lambda\le0.
$$
or a weaker estimate
$$
\left\|\mathrm R
\left(\lambda;\ (-\Delta)^{\frac{\alpha}{2}}\right)\right\|
\_{L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)}
\leq C,\qquad\quad \mathrm{Re}\,\lambda\le0.
$$
holds true or not?


Any hint is very much appreciated.



Resolvent operator of fractional Laplacian
Denoting by $R(\lambda)$ the resolvent and $\mathcal{F}$ the Fourier transform for the sake of legibility (if needed), we have


\begin{equation}
 \mathcal{F}\left( \mathrm R(\lambda) u \right)(\xi) = \frac{\hat{u}(\xi)}{\lambda + |\xi|^{\alpha}}.
\end{equation}


Thus, 


\begin{equation}
\|\mathrm R(\lambda) u\|\_{L^2} \leq \frac{1}{|\lambda|} \|u\|\_{L^2}
\end{equation}


and this estimate is optimal, in that you may have also $\|\mathrm R(\lambda) u\|\_{L^2} \geq \frac{c}{|\lambda|} \|u\|\_{L^2}$ for some $c > 0$ if $u$ is chosen wisely.


Indeed, let $\hat{g}$ be some smooth, fast decaying function which is equal to $1$ in a neighborhood of the origin. Define $u$ by the equation


\begin{equation}
\hat{u}(\xi) = \hat{g}(\xi)|\xi|^{ \alpha - n/2}
\end{equation}


where $n$ is the dimension of the ambient space. Because of the behavior of $\hat{g}$ at infinity and $\alpha > 0$, such a $u$ is in $L^2$. But with $\lambda = 0$, you have 


\begin{equation}
\mathcal{F}\left( \mathrm R(0) u \right)(\xi) = \hat{g}(\xi)|\xi|^{- n/2},
\end{equation}


which, according to how $\hat{g}$ behaves near $0$, is not in $L^2$. Thus, any estimate in $\mathcal{L}(L^2)$ has to be performed far off from $0$ if you want it to remain uniform - otherwise, it blows up.


