# Resolvent operator of fractional Laplacian

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.

• Where do you want your estimate to hold ? If you allow $\lambda$ to go to $0$, then any estimate of the kind you have written fails in the limit. Near infinity, however, the strongest one is true. Both cases are easily seen by going into Fourier space. May 19, 2015 at 14:25
• Can you show me the reason why my question fails near $0$? May 19, 2015 at 14:32
• See the answer below. May 19, 2015 at 14:45

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

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

Thus,

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

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

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

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

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

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.

As explained by Hachino, the problem is for $\lambda$ near 0. You may get a uniform estimate for low frequencies only if you put some cutoff for large $x$. Indeed, the operator $w(x)R(\lambda)w(x)$ is bounded uniformly for all $\lambda$ if $w(x)$ is a cutoff or more generally a smooth function which decays as $|x|\to\infty$ fast enough. For instance, this should be true for $w(x)=|x|^{-a/2}$, at least for $1<a<2$. See the paper "Smooth perturbations of the self-adjoint operator $|\Delta|^{a/2}$ by Watanabe, on Tokyo J. Math.14 (1991) pp.239-250.