# Extending a holomorphic function defined on a disc

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 $z \in \partial D(0,1)$ ?

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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

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).

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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.

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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