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

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.

• If $\min_{z\in D} |f(z)|$ is either $1$ (in which case $f$ is constant) or $<1$ and it is attained inside $D$. If $f$ doesn't vanish on $D$, then $1/f$ is holomorphic and $1/|f|$ attains it maximum inside $D$. Therefore, by the maximum modulus property, it is constant. – user647486 Apr 26 at 16:03
• How you say $1/|f|$ attains maximum inside $D$ and where have you used $|f(z)|\to1$ as $|z|\to1$? – Biswarup Saha Apr 26 at 16:21
• $1/|f|$ is maximum where $|f|$ is minimum. The limit condition is used in proving that either $|f|=1$ on $D$ or its minimum is $<1$. – user647486 Apr 26 at 16:58

## 1 Answer

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 $$0 \le r < 1$$.