An inequality about the gradient of a harmonic function Let $G$ a open and connected set. Consider a function $z=2R^{-\alpha}v-v^2$ with $R$ that will be chosen suitably small, where $v$ is a harmonic function in $G$, and satisfies 
$$|x|^\alpha\leqslant v(x)\leqslant C_0|x|^\alpha. \ \ (*)$$
Then, 
$$\Delta z+f(z)=-2|\nabla v|^2+f(z)\leqslant-C|x|^{2\alpha-2}f(0)+Kz,$$
where $K$ is the Lipschitz constant of the function $f$. ($f$ is a Lipschitz-continuous function).
I have no idea how to obtain this second inequality and I don't know if $(*)$ is really necessary.
 A: It's good that you gave a reference. In many questions of the kind "I don't follow the argument here" the underlying reason is that the OP did not notice the relevance of some information from the paper. Hence, the information included in the post is usually insufficient to give an explanation. 
Here, the missing piece is that $v$ is homogeneous of degree $\alpha$. This means $v(tx)=t^\alpha v(x)$ for all $t\ge 0$. Differentiating with respect to $t$, we find $\frac{d}{dt}v(tx) = \alpha t^{-1} v(tx)$. Specializing this to $|x|=1$ and writing $y=tx$, we obtain $\frac{\partial}{\partial n}v(y)=\alpha |y|^{-1}v(y)$ where $\frac{\partial}{\partial n}$ is the derivative in radial direction. As a consequence, $|\nabla v(y)|\ge \alpha |y|^{-1}v(y)$ for all $y$. Notice that harmonicity is not used here; only homogeneity is needed. 
Recalling that $|v(x)|\ge |x|^{\alpha}$, we get $|\nabla v(x)|\ge \alpha |x|^{\alpha-1}$. Therefore, 
$$-2|\nabla v|^2 \leqslant -2\alpha^2|x|^{2\alpha-2}$$ 
At the same time, the Lipschitz property of $f$ tells us that 
$$f(z)\le f(0)+Kz$$
Add  the inequalities:
$$-2|\nabla v|^2 + f(z) \leqslant -2\alpha^2|x|^{2\alpha-2}+f(0)+Kz$$ 
The plus sign before $f(0)$ is missing in the paper. Typos happen. If you look at the very next displayed formula in the paper, you will see the plus sign reappearing there. 
A: This is an answer to a question raised in comment to another answer. Since it is of independent interest (perhaps of more interest than the original answer), I post it as a separate answer.

Question: How do we construct homogeneous harmonic functions that are positive on a cone? 

Answer. Say, we want a positive homogeneous harmonic function $v$ on the cone $K=\{x\in\mathbb R^n:x_n>\kappa|x|\}$ where $\kappa\in (-1,1)$. Let $\alpha$ denote the degree of homogeneity of $v$: that is,  $v(rx)=r^\alpha v(x)$ for all $x\in K$. The function $v$ is determined by the number $\alpha$ and by the restriction of $v$ to the unit spherical cap $S\cap K$. The Laplacian of $v$ can be decomposed into radial and tangential terms: $$\Delta v = v_{rr}+\frac{n-1}{r}v_r+\frac{1}{r^2}\Delta_S v$$ where $\Delta_S$ is the Laplace-Beltrami operator on the unit sphere $S$. Using the homogeneity of $v$, we find $v_r = \frac{\alpha }{r}v$ and $v_{rr}=\frac{\alpha(\alpha-1)}{r^2}v$. On the unit sphere $r=1$, and  the radial term of $\Delta v$ simplifies to $\alpha(n+\alpha-2)v$. 
Therefore, the restriction of $v$ to $S\cap K$ must be an eigenfunction of $\Delta_S$: 
$$\Delta_S v + \mu v =0, \ \ \ \ \mu = \alpha(n+\alpha-2) $$
Where do we get such a thing from? 
Well, there is a well-developed theory of eigenfunctions and eigenvalues for $\Delta_S$ with the Dirichlet boundary conditions. In particular, it is known that the eigenfunction corresponding to the lowest eigenvalue $\mu_1$ has constant sign. So this is what we use for $v$. There are two drawbacks:


*

*(a) we cannot choose $\alpha$ ourselves, because $\alpha(n+\alpha-2)=\mu_1$ and $\mu_1$ is determined by the domain $S\cap K$.

*(b) the function $v$ vanishes on the boundary of spherical cap, and therefore is not bounded away from $0$. 


Concerning (a), we know that $\mu_1$ is monotone with respect to domain: larger domains (in the sense of inclusion) have lower value of $\mu_1$. (Physical interpretation: a bigger drum emits lower frequencies.) Therefore, the value of $\alpha$ decreases when the domain is enlarged. Also, when $K$ is exactly half-space $\{x_n>0\}$, we know our positive harmonic function directly: it's $v(x)=x_n$, which is homogeneous of degree $\alpha=1$. Therefore, in the cones that are smaller than half-space we have $\alpha>1$, and in the cones that are larger than half-space we have $\alpha<1$.
Concerning (b), we do the following: carry out the above construction in a slightly larger cone $K'$ and then restrict $v$ to $K$. Since the closure of $K\cap S$ is a compact subset of $K'\cap S$, the function $v$ attains a positive minimum there. We can multiply $v$ by a constant and achieve $1\le v\le C$ on $K\cap S$, hence $|x|^\alpha\le v(x)\le C|x|^\alpha$ on all of $K$. 
