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.

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 $K$. 
