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!
Best Answer
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 $(1)$.