# To show a holomorphic map on the open unit disk is Constant

Q. Let $$f$$ be a holomorphic function on the unit disk $$\mathbf{D}$$ such that $$|f(z)|\rightarrow 1$$ as $$|z|\rightarrow 1$$, and suppose $$f(z)\neq 0 \hspace{1ex}\forall z$$. Show that $$f$$ is constant.

Using maximum principle we get $$|f(z)|\leq 1$$. Now if $$1$$ is attained we get $$f$$ is constant. So,we can assume $$f:\mathbf{D} \rightarrow \mathbf{D}$$. Then how to proceed? I tried to compose f with an automorphism of $$\mathbf{D}$$ and applied Schwartz lemma, didn't help much.

• Do you mean $|f(z)|=1$ for every z with $|z|\rightarrow 1$? If so then since $f(z)\neq 0$, we must also have $\frac{1}{|f(z)|}\leq 1$ i.e. $|f(z)| \geq 1$ and equality follows. – Diger Apr 10 at 13:26
• I think it's clear. As $z\rightarrow$ some point in $S^1, f(z)\rightarrow$ some point in $S^1$ – Larsson Apr 10 at 13:29
• @Diger would you explain a bit more what you want to say – Larsson Apr 10 at 13:32
• I think it's clear. ;) – Diger Apr 10 at 13:33
• Well you already started with $f$ must obey maximum principle, but $f(z)\neq 0$ then also $1/f$ must obey the maximum principle. – Diger Apr 10 at 13:34

Hint: $$|f|$$ attains a minimum somewhere.
• $\frac{1}{f}$ satisfies same conditions as $f$ – Conrad Apr 10 at 14:47