I got stuck with the following problem: For any subharmonic function $U(z)$, set $m_r \colon= \underset{|z| = r}{\mathrm{sup}} ~U(z)$. Suppose that $\underset{R \to \infty}{\mathrm{lim}} \frac{m_R}{log R} = 0$. Then it follows that $U(z) \leq m_r ~(r < |z|)$ and $U(z)$ is constant.
Question. Why does this hold?
I cannot understand the hint that I should find the harmonic functions $u$ or $v$ on ${\Bbb D}_{r, R}$ such that $u(z) = 0 ~(|z| = R)$ and $u(z) = 1~ (|z| = r)$ or $v$ satisfying the condition vice versa, where ${\Bbb D}_{r,R} \colon= \{r < |z| < R\}$ is the annulus domain.