TITLE: A harmonic function which is bounded by $\ln(|x|)$ at infinity
QUESTION [9 upvotes]: I think I can prove that a harmonic function $u$ on $\mathbf{R}^n$ which satisfies $|u(x)|\leqslant C \ln(|x|+1)$ is constant. But what can we say about $u$ when the absolute value sign of $u$ is canceled? Can we still say that $u$ is constant?
Any comments or references will be appreciated.

REPLY [3 votes]: As a general rule, if we control the behavior of a harmonic function on one side then we have the same control on the other side by the Harnack inequality. More rigorously:
In your particular case, we have $u$ harmonic and $u \geq -\log(1+|x|)$. Assume $u(0) = a$. Suppose at a point $x$ in $B_R$ we have $u(x) = C$. Sliding $u$ up by $\log(1+2R) + 1$, we obtain a positive harmonic function in $B_{2R}$ with the value $a + 1 + \log(1+2R)$ at $0$. By the Harnack inequality, we have
$$C + \log(1+2R) + 1 \leq K(a + 1 + \log(1+2R))$$
which tells us that $u$ grows at worst logarithmically, reducing the problem to the case $|u| \leq C(1+\log(1+2R))$. The standard proof of Liouville theorem applies once we have the two-sided bound.