If $\Omega\subset\mathbb{C}$ is the punctured disk $0< |z| < 1$ and if $f$ is given by $f(\zeta)=0$ for $|\zeta|=1$, $f(0)=1$, show that all functions $v\in B(f)$ are $\le 0$ in $\Omega$.
$B(f)$ is the class of functions $v$ such that
(a) $v$ is subharmonic
(b) $\overline{\lim\limits_{z\rightarrow \zeta}}\, v(z)\le f(\zeta)$ for all $\zeta$ in the boundary of $\Omega$
I know how to show that all the functions $v$ are $\le 1$ in $\Omega$, but I am unsure how to show they are less than or equal to 0. Any suggestions will be greatly appreciated.
If $v\in B(f)$ and $\epsilon>0$ define $$v_\epsilon(z)=v(z)+\epsilon\log|z|.$$Then $v\in B(f)$ implies $v_\epsilon\in B(0)$, so $v_\epsilon\le0$. Let $\epsilon\to0$.