Suppose we have an elliptic operator $\sum_{i,j=1}^na_{ij}(x)D_{ij}$ on $\mathbb R^n$ for $n\ge 3$ which is asymptotic to the Laplacian i.e. there exists $\beta>0$ so that $$|a_{ij}(x)-\delta_{ij}|\le C|x|^{-\beta}\quad\text{as }x\to\infty.$$ We assume $a_{ij}$ are smooth functions of $x$.
Suppose $u$ is a positive solution to $$a_{ij}(x)D_{ij}u=0$$ on $\mathbb R^n\backslash\{0\}$, can we conclude that $u$ is bounded at infinity? That is can we conclude there exists $C,R>0$ so that $$|u(x)|\le C\quad \text{for all }|x|\ge R?$$
This is heuristically true because it is true for the Laplacian and $a_{ij}D_{ij}$ is a small perturbation of $\Delta$. I can prove that $u$ has at most polynomial growth since Harnack inequality is true, but I have no idea how to proceed.
Update I found an answer for my question. The original conclusion is not correct and the best conclusion we can get is: for any $\epsilon>0$, there exists $R>0,c>0$ so that $$u(x)\le c|x|^\epsilon\quad \forall |x|\ge R.$$