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.

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    $\begingroup$ 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. $\endgroup$ – Diger Apr 10 at 13:26
  • $\begingroup$ I think it's clear. As $z\rightarrow$ some point in $S^1, f(z)\rightarrow $ some point in $S^1$ $\endgroup$ – Larsson Apr 10 at 13:29
  • $\begingroup$ @Diger would you explain a bit more what you want to say $\endgroup$ – Larsson Apr 10 at 13:32
  • $\begingroup$ I think it's clear. ;) $\endgroup$ – Diger Apr 10 at 13:33
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    $\begingroup$ Well you already started with $f$ must obey maximum principle, but $f(z)\neq 0$ then also $1/f$ must obey the maximum principle. $\endgroup$ – Diger Apr 10 at 13:34

Hint: $|f|$ attains a minimum somewhere.

  • $\begingroup$ Can you explain a bit.. I can't really see it. $\endgroup$ – Larsson Apr 10 at 13:24
  • $\begingroup$ $\frac{1}{f}$ satisfies same conditions as $f$ $\endgroup$ – Conrad Apr 10 at 14:47

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