How to prove Liouville's theorem for subharmonic functions I noticed this post and this paper, which gives a version of Liouville's theorem for subharmonic functions and the reference of its proof, but I think there must be an easier proof for the following version of Liouville's theorem with a stronger condition.


A subharmonic function that is bounded above on the complex plane $\mathbb C$ must be constant


I think we may need to use the fact that the maximum of a subharmonic function cannot be achieved in the interior of its domain unless the function is constant(MVP). But how do we prove that a bounded-above subharmonic function on the complex plane $\mathbb C$ can achieve its maximum at a certain point of $\mathbb C$?(Maybe we don't need to use MVP for proof)
Thanks in advance! 
 A: Yes it's  true - sorry about my stupidity earlier.
I haven't looked at that paper - you can decide whether what's below is simpler.
Say $u$ is sh in the plane and bounded above. Define $$v(z)=\frac1{2\pi}\int_0^{2\pi}u(e^{it}z)\,dt.$$Then $v$ is a radial sh function, which is to say there exists $\phi$ with $v(z)=\phi(|z|)$. Since $u$ is sh, $\phi$ is non-decreasing, so if $\phi$ is non-constant there exist $r_1<r_2$ with $\phi(r_1)<\phi(r_2)$. Choose $a$ and $b$ with $$\phi(r_j)=a+b\log(r_j)\quad(j=1,2).$$Note that $b>0$.
Now $v$ sh shows that $$\phi(r)\ge a+b\log(r)\quad(r>r_2).$$
Because if $r>r_2$ but $\phi(r)<a+b\log(r)$ then the harmonic function that equals $\phi(|z|)$ on the boundary of $\{r_1<|z|<r\}$ would be smaller than $a+b\log(r_2)$ for $|z|=r_2$, contradicting the subhamonicity of $v$.
So $b>0$ shows that $u$ is not bounded above, contradiction. So $\phi$ is constant.
So the averages of $u$ on circles centered at the origin are constant. Hence the average over disks centered at the origin are constant: $$\frac{1}{\pi r^2}\int_{|z|<r}u(z)\,dxdy=c.$$
The same applies to the bounded-above subharmonic function $u(z-p)$ for every $p$: $$\frac{1}{\pi r^2}\int_{|z-p|<r}u(z)\,dxdy=c_p.$$
But if $p$ is fixed and $r\to\infty$ we have $$\frac{m(D(p,r)\cap D(0,r))}{m(D(0,r))}\to1$$(where $D(p,r)$ is a disk and $m$ is Lebesgue measure). So letting $r\to\infty$ shows that $c_p=c_0$ for every $p$, and hence $u$ is constant (since $u$ sh shows that the average of $u$ over $D(p,r)$ tends to $u(p)$ as $r\to0$.)
A: If $v$ is subharmonic in the complex plane $\Bbb C$ then
$$ \tag 1
 v(z) \le \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} M(r_1, v)
 + \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, v)
$$
for $0 < r_1 < |z| < r_2$, where
$$
  M(r, v) := \max \{ v(z) : |z| = r \} \quad .
$$
That is the "Hadamard three-circle theorem" for subharmonic
functions, and follows from the fact that the right-hand side of
$(1)$ is a harmonic function which dominates $v$ on the boundary
of the annulus $\{ z : r_1 < |z| < r_2 \}$ .
(Remark: It follows from $(1)$ that $M(r, v)$ is a convex function of $\log r$.)
Now assume that $v(z) \le K$ for all $z \in \Bbb C$.
Then $M(r_2, v) \le K$, and $r_2 \to \infty$ in the inequality $(1)$
gives
$$ \tag 2
 v(z) \le M(r_1, v)
$$
for $0 < r_1 < |z|$. It follows that
$$
 v(z) \le \limsup_{r_1 \to 0} M(r_1, v) = v(0)
$$
because $v$ is upper semi-continuous. Thus $v$ has a maximum
at $z=0$ and therefore is constant.
Remark: As noted in the comments, the condition “$v$ is bounded above”
can be relaxed to
$$\liminf_{r \to \infty} \frac{M(r, v)}{\log r} = 0 $$
 which  is
still sufficient to conclude $(2)$ from $(1)$.
