Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ is a non-vanishing continous function on $\overline{D(0,1)} $ and holomorphic on ${D(0,1)} $ such that $$|f(z) | = 1$$ whenever $$|z | = 1$$

Then I have to prove that f is constant.

We can extend $f$ to all $\mathbb{C}$ by setting $$f(z) = \frac{1}{\overline{f(\frac{1}{\bar{z}})}}$$ and the resulting function is holomorphic on ${D(0,1)} \ $, $\mathbb{C} - \overline{D(0,1)}$ and continous on $\partial D(0,1)$.

But how can we say that the resulting function is holomorphic in $z \in \partial D(0,1)$ ?

share|cite|improve this question
Use Morera's theorem for, say, a rectangle around some point $z$ on the boundary. – Your Ad Here Jan 23 '14 at 18:22
Can you be more precise ? – WLOG Jan 23 '14 at 18:25
Yes, I will write it down below. – Your Ad Here Jan 23 '14 at 18:26
up vote 1 down vote accepted

Let $z$ be a point on the boundary. Consider a small rectangle around $z$. We would like to show that the contour integral of $f$ around that rectangle is $0$, to conclude by Morera's theorem that $f$ is holomorphic at $z$.

Cut the rectangle into two parts on either side of the boundary, where you make both parts (i.e. their boundary curves) to have some distance $\epsilon>0$ from the boundary of $D$. By Cauchy's theorem the contour integral around both of these individually is $0$, since the function is holomorphic on the complement of $\partial D$.

Now notice that you can write the contour integral around the rectangle as a limit of the sum of the two contour integrals as $\epsilon\rightarrow 0$. But each of these summands is $0$, therefore the contour integral is $0$ and the function is holomorphic at $z$.

This is a famous trick, which you can read up in greater detail e.g. in the book of S. Lang (it may even include a picture).

share|cite|improve this answer

To prove that $f$ is identically constant, you'd better employ the maximum principle, according to which the maximum of $|f|$ on $\overline{D}$ equals 1. But $f$ is non-vanishing on $\overline{D}$, hence $\frac{1}{f}$ is holomorphic on $D$ while $\bigl|\frac{1}{f}\bigr|=1$ on $\partial{D}$. By the same maximum principle now follows that the minimum of $|f|$ on $\overline{D}$ also equals 1. Hence $|f|=1$ on $\overline{D}$, whence readily follows the desired result.

share|cite|improve this answer
Yes, but the OP asked for completion of his ansatz. – Your Ad Here Jan 23 '14 at 18:54
TooOldForMath: The OP is aimed at proving that $f$ is identically constant. The way chosen in the OP is wrong – mkl314 Jan 23 '14 at 19:37
Yea, he says he must do so, but his question was to complete his ansatz :-D. J/k, I know what you mean. – Your Ad Here Jan 23 '14 at 19:50
TooOldForMath: Such $f$ cannot be generally extended to be holomorphic at $|z|=1$. For instance, take $$f(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^2}$$ which is holomohic in the unit circle and continuous on its closure. But such $f$ has at least one singular point at $|z|=1$ since it is the boundary of its convergence circle. – mkl314 Jan 23 '14 at 20:41
Your $f$ is not one of his kind. His $f$ has no zeros. – Your Ad Here Jan 23 '14 at 20:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.