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.