If $f: D(0,1)\to \mathbb C$ is holomorphic, then there is a sequence $\{z_n\}$ in $D(0,1)$ such that $|z_n| \to 1$ and $\{f(z_n)\}$ is bounded

Question

Let $f: D(0,1)\to \mathbb C$ be a holomorphic function. How to show that there exists a sequence $\{z_n\}$ in $D(0,1)$ such that $|z_n| \to 1$ and $\exists M>0$ such that $|f(z_n)|<M,\forall n \ge 1$ ?

My try: If not, then $\lim_{|z|\to 1} |f(z)|=\infty$. So in particular, $f$ has finitely many zeroes in $D(0,1)$. Also, $1/f$ is meromorphic in $D(0,1)$ with $\lim _{|z|\to 1}\dfrac {1}{|f(z)|}=0$. I am not sure where to go from here. Please help.

Answer 1

The way to do it is to first take out the (finitely many) zeros with a finite Blaschke product $B$ - which has absolute value $1$ on the circle- so get $g = \frac{f}{B}$ analytic, no zeros, $g \to \infty$ on the circle, so then $\frac{1}{g}=\frac{B}{f}$ is an analytic function in the unit disc that goes to zero at the boundary and that means it is zero by maximum modulus, so you get a contradiction and are done.

Answer 2

Edit: You are on the right track/ Start by removing the zeroes.

Let $w_1,.., w_m$ be the zeroes of $f$ with multiplicity $k_1,..,k_m$. Let $P(z):= (z-w_1)^{k_1}....(z-w_m)^{k_m}$. Then $g(z):= \frac{f(z)}{(P(z)}$ is holomorphic on $D(0,1)$ and non-vanishing on $D(0,1)$.

Then $\frac{1}{g(z)}$ admits a continuous extension to $\{ z : |z| \leq 1 \}$, namely the function which is $0$ on the boundary.

By compactness, $|\frac{1}{g(z)}|$ has a local maximum on this domain. Show that this local maximum appears at some point in $D(0,1)$ and use the maximum modulus principle.