Prove if the fractional Laplacian of a function is bounded

Question

Take $s\in (0, 1)$. I am trying to understand if $(-\Delta)^s (\log(1+x^2))$ is bounded, that is if there exists $R>0$ such that $|(-\Delta)^s (\log(1+x^2))|\le R$. Here $(-\Delta)^s$ is the fractional Laplacian, i.e. $$(-\Delta)^s u(x)=c(n, s) \ \lim_{\varepsilon\to 0}\int_{\mathbb R\setminus (x-\varepsilon, x+\varepsilon)} \frac{u(x)-u(y)}{|x-y|^{1+2s}} dy.$$

At a first glance, it seems reasonable that $(-\Delta)^s (\log(1+x^2))$ is bounded. This is because, for $s=1$, I have computed the second derivative of $\log(1+x^2)$ and it is a bounded function. But I can not get that for the fractional Laplacian.

Anyone could please help me with that?

Is the claim "easier" to prove by using the Fourier transform?

Answer 1

Yes, it is bounded. This is perhaps easiest to see via the Fourier transform as you seem to have suspected.

The Fourier transform of your $f=\log(1+x^2)$, a tempered distribution, seems to be (constants depending on your conventions) $$ \mathcal{F}[f](k) = -2\pi|k|^{-1}e^{-|k|} , $$ so that of its fractional Laplacian is $$ \mathcal{F}[(-\Delta)^sf](k) = -2\pi|k|^{2s-1}e^{-|k|} , $$ and so $$ (-\Delta)^sf(x) = -\int|k|^{2s-1}e^{-|k|}e^{ikx}dk . $$ A nice uniform bound using $|e^{ikx}|=1$ is simply $$ ||(-\Delta)^sf||_{L^\infty} \leq 2\int_0^\infty k^{2s-1}e^{-k}dk = 2\Gamma(2s) . $$ This bound is optimal at $x=0$, so it is optimal as a uniform bound.

Note that with $s>0$ the function $|k|^{2s-1}e^{-|k|}$ remains locally integrable near $k=0$. In the limit $s\to0$ this integrability breaks down, my upper bound becomes infinite, and $(-\Delta)^0f=f$ is unbounded. Things breaking down at $s=0$ should not be a surprise.