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I have been working through Stein and Shakarchi's Complex Analysis and I'm stuck on Exercise 15 from Chapter 2. The question states:

Suppose $f$ is a non-vanishing continuous function on $\overline{D}$ that is holomorphic in $D$. Prove that if $$|f(z)| = 1$$ whenever $|z|=1$, then $f$ is constant. [Hint: Extend $f$ to all of $\mathbb{C}$ by $f(z) = 1/\overline{f(1/\overline{z})}$ whenever $|z|>1$, and argue as in the Schwarz Reflection Principle.]

I have made some progress with the hint. Namely I have shown that $1/\overline{f(1/\overline{z})}$ is holomorphic whenever $|z|>1$. Furthermore, since $|f(z)|=1$ whenever $|z|=1$, we know that the following function is well defined: $$ g(z) = \begin{cases} f(z) & |z|<1 \\ f(z) = 1/\overline{f(1/\overline{z})} & |z|=1 \\ 1/\overline{f(1/\overline{z})} & |z|>1 \end{cases} $$ However, since we only have continuity at $|z|=1$, I'm not sure how we can use something like the Symmetry Principle to conclude that $g$ is a holomorphic function on $\mathbb{C}$. Once we have this, I'm pretty sure that we can deduce by Liouville's Theorem that $g$ is constant since it would be bounded and entire.

Any hints or suggestions would be greatly appreciated.

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  • $\begingroup$ Does your reflection principle work only for upper half plane? $\endgroup$ Sep 26, 2021 at 16:47
  • $\begingroup$ Have you looked at the proof of the symmetry principle (theorem 5.5)? Perhaps you can adapt those ideas to this situation (i.e. Morera's theorem) $\endgroup$
    – Legendre
    Sep 26, 2021 at 16:48
  • $\begingroup$ It turns out that you can use a modified proof of the Symmetry Principle given in the text. All you have to do is consider the intersection of a triangular contour with the unit disk. Essentially, you break the arched segment of intersection into smaller and smaller pieces until, in the limit, they cancel entirely. $\endgroup$
    – slowspider
    Sep 30, 2021 at 2:08
  • $\begingroup$ I have shown that $g(x)$ entire, but why is it bounded? $\endgroup$
    – quuuuuin
    Nov 8, 2021 at 2:41

2 Answers 2

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Since this seems to be a popular question, here is my full-credit solution:

Consider the following function $$g(z) = \begin{cases} f(z) & \text{ if }|z|\leq 1 \\ \frac{1}{\overline{f(1/\overline{z})}} & \text{ if }|z|\geq 1 \end{cases} $$ Note that $g$ is well defined since $\frac{1}{\overline{z}} = z$ for $z$ such that $|z| = 1$, $|f(z)| = 1$ whenever $|z| = 1$, and $$\frac{1}{\overline{f(1/\overline{z})}} = \frac{1}{\overline{f(z)}} = f(z)$$ We now verify that $\frac{1}{\overline{f(\frac{1}{\overline{z}}})}$ is holomorphic for $|z|>1$. Since $f(z)$ is holomorphic and non-vanishing on $D$, we know that $\frac{1}{f(\frac{1}{z})}$ is holomorphic for $z \in \mathbb{C}$ such that $|z|>1$ as it is the composition of holomorphic functions. Thus, by Theorem 4.4 in Chapter 2 of Stein, for every $z_0 \in \mathbb{C}$ such that $|z_0|>1$, there is a disc centered at $\overline{z_0}$ such that $\frac{1}{f(\frac{1}{z})} = \sum_{n=0}^\infty a_n(z-\overline{z_0})^n$. Then, $$\frac{1}{\overline{f(\frac{1}{\overline{z}}})} =\sum_{n=0}^\infty\overline{a_n}(z-z_0)^n$$ Thus, $\frac{1}{\overline{f(\frac{1}{\overline{z}}})}$ is holomorphic on $\big\{z \in \mathbb{C}:|z|>1\big\}$. We claim that $\int_T g(z)dz =0$ for every triangle in the complex plane. By Cauchy's Theorem, we know that if a triangle is completely inside $D$ or $\big\{z:|z|>1\big\}$, $\int_Tg(z)dz = 0$. Consider now a triangle as that depicted below: enter image description here

Zooming in on the region of intersection, enter image description here

We then see by Cauchy's theorem that the only contributor to $\int_T g(z)dz$ is the following contour

enter image description here

Note here that the treatment of the case where the triangle touches the unit disk at one point is in the proof Stein provides of the Symmetry Principle in Chapter 2. If we continue this process for both triangles in the above diagram, and continue doing so for those that follow, the resultant limiting line integrals will be zero. Thus, we conclude that $$\int_T g(z)dz = 0$$ for all triangles in the complex plane. Thus, by Morera's Theorem, $g$ is holomorphic in $\mathbb{C}$. Furthermore, as $g$ is clearly bounded, we know by Liouville's Theorem that $g$ is constant. We conclude then that $f$ is constant.

For those who potentially need it, there are lots of free copies of Stein's text online. Here is a link that I use: Complex Analysis

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  • $\begingroup$ Thank you for your answer, as I hardly see anyone tackling Morera's theorem on the circle. However, I have some doubts on it. First how do you justify that we can subdivide the triangles into smaller ones with that eventually do not touch the circle. Second, the proof of Schwarz reflection principle argues by lifting the triangle up then use a continuity argument, but since we have curves as the boundary, the lifting may not go to zero. Third, if we use a different triangle, say one that covers the circle entirely, does the argument still works? Appreciate if you can reply. $\endgroup$ Jul 6, 2023 at 18:08
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Why not use maximal modules thm in D?We want to prove that |f(z)|=1 in D,then f(z) is a constant.If |f(z0)|<1 ,then we can see 1/f,and use the maximal modules thm to contradict.

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  • $\begingroup$ When do you mean by “we can see $1/f$”? $\endgroup$ Mar 17, 2022 at 8:00
  • $\begingroup$ the mod of 1/(f(z0) >1 ,if |f| is not a constant,this is a contradiction to the maximal thm $\endgroup$
    – why
    Mar 17, 2022 at 8:14
  • $\begingroup$ And how do you get that contradiction? $\endgroup$ Mar 17, 2022 at 8:15
  • $\begingroup$ 1/f is mod 1 on the boundary of D and 1/(f(z0) >1 $\endgroup$
    – why
    Mar 17, 2022 at 8:17
  • $\begingroup$ We are only assuming that $f$ is holomorphic on $D$. Therefore, we cannot apply the maximum modulus principle to $\overline D$. Besides, the inequality $\frac1{f(z_0)}>1$ makes no sense. $\endgroup$ Mar 17, 2022 at 8:21

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