This question was asked in my complex analysis quiz and I was unable to solve it. So, I am asking it here
Question: Let $\{ f_n \}\subset H(D) $ , where $D$ is an open unit disc. Assume that $f_n(D) \subset D\setminus \{0\}$ for every $n$ and that $\lim_{n\to \infty}$ $f_n(a) =0$ where $a$ is the center of disc $D.$ Then prove that $\lim_{n\to \infty} f_n (z) =0 $ uniformly on compact subsets of $D.$
Attempt: $f_n$ is bounded on this disc as domain. But that doesn't ring any bell as to why limit should go to $0$ uniformly.
Can anyone please shed some light on which result should be used?
Thank you!!
Assume, WLOG, that $a=0$. First proof: It is easy to see that $u_n(z)=-\ln(|f_n|)$ is a positive harmonic function, and thus we can apply Harnak's inequalities in the disc of radius $r<R<1$:
$$\frac{R-r}{R+r}u_n(0)\le u_n(z)\le \frac{R+r}{R-r}u_n(0)\\ \frac{R-r}{R+r}\ln(|f_n(0)|)\ge\ln(|f_n|)\ge\frac{R+r}{R-r}\ln(|f_n(0)|)\\ |f_n(0)|^{C_r}\ge |f_n(z)|\ge |f_n(0)|^{c_r}$$
convergence easily follows.
Second proof: Assume that $f_n(0)\in\mathbb{R}^+$; $f_n(\mathbb{D})\neq \mathbb{D}-\{0\}$ and thus we can write $g_n(z):=\ln(f_n)$; $g_n(0)<0$.
We can now apply Herglotz inequalities (see the answer here for an hint about the derivation of the inequalities), obtaining
$$\frac{1-r}{1+r}\le \left|\frac{g_n(z)}{g_n(0)}\right|\le \frac{1+r}{1-r}$$
Since $g_n(0)\to -\infty$, for every compact subset of $\mathbb{D}$, $g_n(z)\to -\infty$ uniformly on the compact subsets of $\mathbb{D}$, and $f_n\to 0$