I'm reading something with the following statement:
In the complex plane cut along the negative axis, the function $z \rightarrow z^{1/2}$ is analytic, hence its real part is harmonic. Moreover, because it is even in $y$, its $y$-derivative on the positive axis vanishes; this means that the half-Laplacian of $\mathcal{R}(z^{1/2}) = \sqrt{x}$ is $0$ on $\mathbb{R}_+$
Here the half-Laplacian is defined as: $$-\Delta^{1/2}u(x)=C\,\left(\text{PV}\int_{\mathbb{R}^n}\frac{u(x)-u(y)}{|x-y|^{n+1}}\right)$$ where $C$ is some constant.
I'm having trouble with the statment in bold:
- How is $\mathcal{R}(z^{1/2})$ even in $y$?
- Why does that imply the derivative is positive on the positive $x$ axis?
- Why is the half laplacian $0$ then?