$f$ is non-vanishing holomorphic function on open unit disc such that $|f(z)|\to1$ as $|z|\to1$. Is $f$ constant?

Question

The whole question looks like-

$f$ is a holomorphic function on the interior of the open unit disc (say $D$) around $0$ and $f(z)\ne0\ \forall z\in D$. Suppose, $|f(z)|\to1$ as $|z|\to1$. Is f a constant function? Now, if we remove the non-vanishing condition then is $f$ constant?

The 2nd part is easy if we take $f(z)=z$, then $|f(z)|\to1$ as $|z|\to1$ but $f(0)=0$.
For the first part, I tried to show $f'(z)=0\ \forall z\in D$
From application of Cauchy intrgral formula, we know that $f'(z)={1\over 2\pi i}\int_{C} \frac{f(\zeta)}{(\zeta-z)^2}d\zeta$ where $z\in D$ and $C$ is a circle around $z$ inside $D$.
So, $|f'(z)|\le\sup\{\frac{f(\zeta)}{(\zeta-z)^2}|\zeta\in C\}\operatorname{Radius}(C)$
Can use these concepts and the condition $|f(z)|\to1$ as $|z|\to1$ to show $f$ is constant? I even don't know the answer, so may be the $f$ is non-constant. But I think it will be constant.
Can anybody solve the problem? Thanks for assistance in advance.

Answer 1

You have to show that the given conditions imply that $|f(z)| \le 1$ in the unit disk $\Bbb D$. Then the same reasoning can be applied to $1/f$ (since $f$ has no zeros). It follows that $|f(z)| = 1$ in $\Bbb D$, and the maximum modulus principle (or open mapping principle) implies that $f$ is constant.

So it remains to show that:

Let $f$ be holomorphic in the unit disk $\Bbb D$ with the property that $|f(z_n)| \to 1$ for each sequence $(z_n)$ in $\Bbb D$ with $|z_n| \to 1$. Then $|f(z)| \le 1$ for all $z \in \Bbb D$.

If $f$ were continuous on the closure $\overline{\Bbb D}$ then this would be a simple application of the maximum modulus principle. For the general case we can consider the maximum modulus on an increasing sequence of disks, and proceed as follows:

For $0 \le r < 1$ let $$ M(r) = \max \{ |f(z)| : |z| = r \} $$ be the maximal modulus of $f(z)$ on the circle $|z| = r$. It follows from the maximum modulus principle that $M(r)$ is increasing in $r$. Now choose a sequence of increasing radii converging to one: $$ 0 < r_1 < r_2 < r_3 < \ldots \, , \quad r_n \to 1 \, . $$ Then $$ M(r_1) \le M(r_2) \le M(r_3) \le \ldots \, . $$ For each circle $|z| = r_n$, $|f(z)|$ attains the maximum value $M(r_n)$ at a point $z_n$, i.e. there is a sequence $(z_n)$ in $\Bbb D$ such hat $$ |z_n| = r_n \text{ and } |f(z_n)| = M(r_n) $$ for all $n$. In particular $|z_n| \to 1$, which implies $$ M(r_n) = |f(z_n)| \to 1 $$ for $n \to \infty$. It follows that $M(r_n) \le 1$ for all $n$, and consequently $M(r) \le 1$ for all $0 \le r < 1$.

Answer 2

INCORRECT (1 does not imply 2, since in Cauchy's inequality $R$ gets smaller and smaller as we go close to the boundary. We do get uniform continuity on any sub disc of radius $r < 1$ for any $r$). I unable to add this as a comment so writing here:

1. Show that $f$ is bounded.

2. Use 1. to show that $f$ is uniformly continuous.

3. So, $f$ can be extended to the closed unit disc and we have $|f(z)| = 1$ whenevar $|z| = 1$.

4. Now extend $f$ to all of $\mathbb{C}$ by a reflection type construction.

5. Now use Louville's theorem.