Recently I came into a problem and wondering if there exists such a function $f:\mathbb{R}^n\setminus\overline{B_1}\rightarrow\mathbb{R} $is convex and having zero limit at infinit. Meaning that I want to demand : $$D^2f\geq 0,\ \ \mathtt{and}\ \ \lim_{|x|\rightarrow\infty}f(x)=0$$
This is true for $n=1$ case obviously by considering $f(x)=\dfrac{1}{|x|}\ \ in\ \mathbb{R}\setminus[-1,1]$. But we cannot do a radial symmetric and hope the result holds for $n\geq 2.$
In fact, f has to be satifying $$\Delta f=\mathtt{tr}(D^2f)\geq 0\ in\ \mathbb{R}^n\setminus\overline{B_1}$$ But this is not sufficient to rule out all other cases apart from $f\equiv 0$ due to we have fundamental solutions and many other things like $$f(x)=\dfrac{1}{|x|^{n-2}},\ n\geq 3$$.
So, how can we prove the only possible convex function with zero boundary condition at infinity is the constant zero? Or is there any non-constant (better to be bounded) function satisfying these conditions for $n\geq 3$ case?
Thanks in advance!
Due to we can restrict the function to an entire line, say, $\{x_1=100,\ x_2\in\mathbb{R}\}$. Then the function f is still convex on this line. Hence f is always zero on this line. This makes f equivalent to zero (at infinity at least.)