Is a harmonic function in $\mathbb{R}^2$ which is $o(\ln |x|)$ a constant? Let $f :  \mathbb{R}^{ 2 } \rightarrow  \mathbb{R} $ be a harmonic function. Suppose
$$\lim _ { | x | \rightarrow \infty } \frac { | f ( x ) | } { \ln | x | } = 0$$
 Prove or disprove that $f$ is a constant.
I guess $f$ is a constant. Let $\epsilon>0$ then we can find $R$ large enough so that $f(x)-\epsilon\ln |x|<0$ and $f(x)+\epsilon\ln |x|>0$ for any $|x|>R$. Letting $x\rightarrow 0$ tells me that there exists $x_1, x_2$ such that $f(x_1)=\epsilon\ln |x_1|$ and $f(x_2)=-\epsilon\ln |x_2|$.
But when I got this, I can't move on. I also considered using the MVP or MP, but I just can't figure out this problem.
Any help would be appreciated.
 A: $\newcommand{\R}{\mathbb R}$
Using the inversion $q(x) = \frac{x}{|x|^2}$, we can move the singularity from the infinity to the origin. More precisely, we define 
$$
g \colon \R^2 \setminus \{0\} \to \R, 
\quad 
g(x) := f(q(x))
$$
and check that this is again a harmonic function. The condition on $f$ translates to 
$$
\lim_{x \to 0} \frac{g(x)}{\log |x|} = 0.
$$
Such singularities are removable (see here or elsewhere on MSE) in the sense that $g$ can be extended to a harmonic function $\overline{g} \colon \R^2 \to \R$. In particular, $g$ is bounded on the unit ball, which translates to $f$ being bounded outside the unit ball. Thus, $f$ is bounded everywhere and we can conclude that $f$ is constant. 

As you can see in the link, the proof of removability follows the approach you sketched (first use the assumption to get inequalities, then apply the maximum principle), but involves solving the Laplace equation. For this reason, it is convenient to switch $0$ and $\infty$ (otherwise we need to solve the equation around $\infty$). 

By avoiding complex analysis, you can easily obtain similar statements in higher dimensions with the same proof - you just need to replace $\log |x|$ with $|x|^{n-2}$. 
A: Translate the problem to the theory of complex analysis. Every 2D harmonic function can be represented as the real or imaginary part of some entire function.
WLOG suppose $f=u+iv,$ where our original function is now $u(x,y)=\Re(f).$
By the limit condition, the partial derivatives $u_x, u_y$ are bounded (if not, L'Hospitals rule would lead to a contradiction). Finish off using the Cauchy-Riemann equations.
