When I learnt that there exists a holomorphic function on the unit disc $D$ that cannot be continuously extended to a domain that is strictly larger $D$, I was taught the example $$z\mapsto\sum_{n=1}^\infty z^{n!}.$$ The reason why this function cannot be continuously extended is for at any root of unity $\xi$, there is a sequence in $D$ that converges to $\xi$ whose function values go to infinity and the roots of unity are dense on $\partial D$. This lead me to thinking that there should not be a holomorphic function $f$ on the unit disc such that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow 1$, for otherwise such an $f$ would clearly be a holomorphic function on $D$ that cannot be continuously extended, and why bother with the above series?
So I tried to prove that there is not a holomorphic function $f$ on the unit disc such that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow 1$, but could not get anywhere. Can anyone prove this assertion? Or is it false? Many thanks!
I don't think so.
if $f$ exists consider $g(z)=\frac{1}{f(z)}$ defined in open unit disk. It should be meromorphic and extendable to the closed unit disk with the value $0$ at the boundary.
Since it's zero on a line (has an accumulation point) it must be zero everywhere. So for $f$ to exist it would have to be $\infty$ everywhere.
This is but a sketch but i think it can be made rigorous.