The early seminal result of Bernstein in 1914 for $n=2$ is the well-known Bernstein theorem:
The only entire solutions to the minimal surface equation in $\mathbb R^3$ are the affine functions $$u(x,y)=ax+by+c,$$ where $a, b, c\in\mathbb{R}$.
Actually, Bernstein obtained his result as an application of the so called Bernstein’s geometric theorem:
If the Gauss curvature of the graph of $u\in C^\infty(\mathbb R^2)$ in $\mathbb R^3$ satisfies $K\leq 0$ everywhere and $K<0$ at some point, then $u$ cannot be bounded.
As a corollary, Bernstein proved a very general Liouville theorem:
Suppose $u$ is a solution to the elliptic equation $$\sum_{i,j=1}^2 a_{ij}u_{ij}=0\quad\text{in }\mathbb R^2$$ with $a_{ij}\in C^\infty$ such that $$|u(x)|=o(|x|) \text{ as }|x|\to+\infty.$$ Then $u$ is a constant.
Note that in the above Liouville theorem, the equation doesn't need to be uniformly elliptic, hence it is a very powerful result. What I want to know if this result has a half space version, which is like harmonic functions. More precisely, I want to obtain the following proposition:
Suppose $u\geq 0$ is a solution to the elliptic equation $$\left\{\begin{aligned}\sum_{i,j=1}^2 a_{ij}u_{ij}&=0\quad\text{in }\mathbb R^2_+,\\ u(x,0)&=0\quad \text{on }\mathbb R,\end{aligned}\right.$$ where $a_{ij}\in C^{\infty}$. Then $u$ is a linear function of form $$u(x,y)=Ay,\quad A\geq 0.$$
Note that in the question, $a_{ij}$ could be degenerate or sigular at $\infty$.
This question is motivated by seeing Mooney's notes: The Monge-Ampère equations. He used partial Legendre transform to investigate the Liouville theorem for Monge-Ampère equation in half space, and one of steps in his proof used the similar proposition for harmonic functions, and which can be proved by boundary Harnack inequality and odd extension of $u$. But it is failed for the case without uniform ellipticity.
I have searched on the internet, and I didn't find any references about this proposition. Is this propersiton true? And if there are some references that I missed? Thanks in advance.
