Harmonic Function bounded by a linear function Let $u$ be a harmonic function on $\mathbb C$. Suppose that for each $\epsilon > 0$, there is a constant $C_\epsilon$ such that
$$u(z) \leq C_\epsilon + \epsilon |z| .$$ 
I am trying to show that $u$ is constant. I tried the mean value property, as well as replacing $u$ by a related function, but was unsuccessful. Does anyone have any ideas?
 A: Without loss of generality, we may assume that $u(0)=0$, and it suffices to show that $u\equiv 0$.
Let us fix an arbitrary $\epsilon>0$ first. Choose $R_0>0$ such that $C_\epsilon <\epsilon R_0$. From the assumption on $u$ we know that for every $R\ge R_0$, 
$$\max_{|z|\le R}u(z)\le C_\epsilon+\epsilon\cdot\max_{|z|\le R}|z|<2\epsilon R.\tag{$*$}$$

Proof 1(using Harnack's inequality) $(*)$ implies that $2\epsilon R-u(z)$ is a positive harmonic function on $|z|<R$. Then by Harnack's inequality, when $R\ge R_0$,
$$\frac{R-|z|}{R+|z|}\cdot 2\epsilon R\le 2\epsilon R-u(z)\le\frac{R+|z|}{R-|z|}\cdot 2\epsilon R,\quad \text{ if } |z|<R.$$
It is equivalent to say when $R\ge R_0$,
$$-\frac{4\epsilon R|z|}{R-|z|}\le u(z)\le\frac{4\epsilon R|z|}{R+|z|},\quad \text{ if } |z|<R.$$
Fixing $z$, letting $R\to\infty$ and $\epsilon\to 0$, the conclusion follows.
Remark: A similar argument using Harnack's inequality still works when $u$ is a harmonic function on $\mathbb R^n$($n\ge 3$) satisfying the same condition.

Proof 2(using Schwarz lemma)
Since $u$ is harmonic on $\Bbb C$ and $u(0)=0$, there exists a unique entire function $f=u+iv$ such that $f(0)=0$. Then it suffices to show that $f\equiv 0$. 
Note that if $a>0$, the Möbius transformation $z\mapsto \frac{z}{2a-z}$ maps the half plane $\{x+iy\mid x<a\}$ conformally onto the unit disk $\Bbb D$, so by $(*)$, for $a=2\epsilon R$ with $R\ge R_0$,  the following map 
$$g:\Bbb D\to \Bbb D,\quad w\mapsto \frac{f(Rw)}{4\epsilon R-f(Rw)}$$ 
is well defined and holomorphic; moreover, $g(0)=0$. Then by Schwarz lemma, when $R\ge R_0$ and $|z|<R$, for $w=\frac{z}{R}\in \Bbb D$,  $g(z)\le |z|$, i.e.
$$\frac{|f(z)|}{|4\epsilon R-f(z)|} \le \frac{|z|}{R}\Longrightarrow |f(z)|\le |z| \cdot |4\epsilon -\frac{f(z)}{R}|, \quad \text{ if } |z|<R.$$
Fixing $z$, letting $R\to\infty$ and $\epsilon\to 0$, the conclusion follows.
A: Since $u$ is harmonic on the whole plane and is bounded linearly, we know by Liouville's theorem that $u$ is a harmonic polynomial of order less than or equal to 1. The coefficient on the linear term must be 0, since if it were anything else, then you would have a contradiction for $\epsilon$ sufficiently small. 
