Is this statement true. A bounded half-superharmonic function in $\mathbb R$ is a constant. That is $(-\Delta)^{1/2} u\geq 0$ implies $u\equiv 0.$



Fractional super-harmonic functions
This claim true, but the details depend on your favourite definition of $(-\Delta)^{1/2} u$ and regularity assumptions on $u$.


One argument could be as follows: Suppose that $u \geqslant 0$ and $(-\Delta)^{1/2} u \geqslant 0$. Denote by $P\_r(x,z)$ the Poisson kernel for $(-\Delta)^{1/2}$ in $(-r,r)$ and by $G\_r(x,y)$ the Green function for $(-\Delta)^{1/2}$ in $(-r,r)$. Then
$$ u(x) = \int\_{(-r,r)} G\_r(x,y) (-\Delta)^{1/2} u(y) dy + \int\_{\mathbb{R} \setminus (-r,r)} P\_r(x,z) u(z) dz . $$
In particular,
$$ u(x) \geqslant \int\_{(-r,r)} G\_r(x,y) (-\Delta)^{1/2} u(y) dy . $$
Passing to the limit as $r \to \infty$ and using the monotone convergence theorem, we find that
$$ u(x) \geqslant \int\_{\mathbb{R}} \infty \times (-\Delta)^{1/2} u(y) dy .$$
It follows that $(-\Delta)^{1/2} u(y) = 0$, and, consequently, $u$ is harmonic for $(-\Delta)^{1/2}$. By Liouville's theorem, $u$ is constant.


You can find more details in Landkoff's book [1], I suppose. Liouville's theorem in full generality is due to Fall [2].


References:


[1] N. S. Landkof, *Foundations of Modern Potential Theory*, Springer, New York–Heidelberg, 1972


[2] M. M. Fall, *Entire $s$-harmonic functions are affine*, Proc. Am. Math. Soc., 144(6) (2016), 2587–2592


