Let $u$ be a negative subharmonic function on $D(0,1)$. Then $\limsup_{r\to 1^-} \frac{u(rz_0)}{1-r}< 0$ for each $z_0\in \partial D(0,1)$.

Question

This exercise has a hints to follow which is apply maximum principle $f(z)=u(z)+c\log|z|$ on the domain $\frac{1}{2}<|z|<1$. Now if I choose $c>0$ then since $u<0$ in the whole domain so on both the boundary of annulas $f(z)<0$ and since $f(z)$ is subharmonic so $f(z)<0$ in the interior of the annulas too. But I can't proceed to the result asked for after that.

Answer 1

The idea is to consider $f(z)=u(z)+c\log|z|$ with some negative $c$ which is chosen such that $f(z) \le 0$ still holds for $|z|=1/2$.

A subharmonic function is upper semi-continuous, and therefore attains its supremum on every compact set. Therefore we can define $$ M = \max_{|z|=1/2} u(z) $$ and $M$ is negative. Now consider the function $$ f(z)=u(z)+\frac{M}{\log(2)}\log|z| $$ in the annulus $1/2 \le |z| \le 1$. $f$ is subharmonic, and $\le 0$ on both boundary components. It follows that $f(z) \le 0$ inside the annulus, so that $$ u(z) \le -\frac{M}{\log(2)}\log|z| $$ and therefore $$ \frac{u(rz_0)}{1-r} \le \frac{M}{\log(2)}\frac{-\log(r)}{1-r} \, . $$ For $r \to 1-$ the right-hand side converges to $M/\log 2 < 0$, so that the desired conclusion follows.