A harmonic function which is bounded by $\ln(|x|)$ at infinity 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.
 A: We have the following theorem (which is a slight generalisation of the classical Liouville theorem for positive harmonic functions [see, for example, chapter 3 of Axler, Bourdon and Ramey's Harmonic Function Theory]; it may help to read that proof first to get an idea of the basic approach):
Theorem Let $f:[0,\infty)\to[0,\infty)$ be a (not necessarily strictly) increasing continuous function such that $\lim_{r\to\infty} f(r)/r = 0$. Let $u:\mathbf{R}^n\to\mathbf{R}$ be harmonic, such that $u(x) \geq - f(|x|)$, then $u$ is constant. 
Proof: Observe that $u(x) + f(|x|)$ is a continuous, non-negative function. 
Consider $u(x) - u(z)$ for some fixed $x,z$. Using the mean value property for harmonic functions, we write 
$$ |B_R|(u(x) - u(z)) = \int_{B_R(x)} u(y) dy - \int_{B_R(z)} u(y) dy $$
The right hand side we rewrite
$$ = \int_{B_R(x)} u(y) + f(y) - f(y) dy - \int_{B_R(z)} u(y) + f(y) - f(y) dy $$
which is
$$ \leq \int_{B_R(x)\setminus B_R(z)} u(y) + f(y) dy + \int_{B_r(z) \setminus B_r(x)} f(y) dy $$
Writing $A\delta B$ for the symmetric set difference $(A\setminus B)\cup (B\setminus A)$, we get
$$ \leq \int_{B_R(x) \delta B_R(z)} u(y) + 2f(y) dy $$
Define $w = \max(|x|,|z|)$. Now using that $B_R(x) \delta B_R(z) \subset B_{R+ w}(0) \setminus B_{R-w}(0)$, we have
$$ \leq \int_{B_{R+ w}(0) \setminus B_{R-w}(0)} u(y) + 2f(y) dy $$
For the $u$ term, we use the mean value property again. For $f$, estimate by a supremum, you get
$$ \leq (u(0)+2f(R+w)) \left( |B_{R+ w}| - |B_{R- w}|\right) $$
Observe that $|B_{R+ w}| - |B_{R- w}| = O(R^{n-1})$, by our assumption on $f$ we have 
$$\lim_{R\to\infty} (u(0) + 2f(R+w))\frac{|B_{R+ w}| - |B_{R- w}|}{|B_R|} = 0$$
and hence
$$ u(x) - u(z) \leq 0 $$
Since the derivation is symmetric in $x$ and $z$, this implies that $u(x) = u(z)$. q.e.d.
A: In two dimensions such a function still has to be constant. Let $v(z)$ be the harmonic conjugate of $u(z)$. Then ${\displaystyle f(z) = e^{u(z) + i v(z)}}$ is an entire function, and
$$|f(z)| = e^{u(z)} \leq e^{C\ln(|z| + 1)}$$
$$\leq K(1 + |z|)^C$$
Here $K$ is some constant. It's a standard exercise to show that entire functions that grow polynomially are themselves polynomials, so $f(z)$ must be a polynomial. 
If $z_0$ is any zero of $f(z)$, then $\lim_{z \rightarrow z_0} e^{u(z)} = |f(z)| = 0$,  so $\lim_{z \rightarrow z_0} u(z) = -\infty$. Hence $u(z)$ can't be continuous at $z_0$, a contradiction. 
Thus the only possibility is that $f(z)$ is constant, so the same is true for $u(z)$.
A: 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.
