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$f(z)$ is analytic in the open unit disk and continuous on its edge.

Can you prove that if $f(z)=1$ on the upper half of the unit circle (for $z=e^{i\theta}, 0\le\theta\le\pi$) then $f(z)$ in constant in the unit disk?

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I guess contrary... considering the Dirichlet problem. –  sos440 Mar 15 '13 at 9:55
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the open unit circle? Do you mean the unit disk? The unit circle is not open. –  Heitor Fontana Mar 15 '13 at 10:50
    
Right :) I edited the question. –  Ran Kashtan Mar 15 '13 at 10:59
    
I think you might have changed one too many of the "circles" into "disks" - the question as it now stands is rather trivial (or at least the bit in brackets contradicts the words just before). –  Old John Mar 15 '13 at 11:01
    
Let $D$ be the open unit disk. Set $f(z) = 1$ if $z = e^{i \theta}$ with $0 \le \theta \le \pi$, $f(z) = e^{2 i \theta} = z^2$ if $z = e^{i \theta}$ with $\pi < \theta \le 2\pi$, and extend $f$ with Cauchy's formula $f(z) = (2 \pi i)^{-1} \int_{\partial D} f(s)/(z-s) ds$ into the interior of $D$. This seems to be a counterexample. –  Hans Engler Mar 22 '13 at 2:32

2 Answers 2

up vote 2 down vote accepted
+25

Consider the function $g(z) = \overline{f(\bar z)}$. The function $g(z)$ is analytic in the unit disk and equals $1$ on the lower half of the unit circle. Now let $h(z) = (f(z) - 1) (g(z) -1)$. We have that $h(z) = 0$ both on the lower and upper half of the unit circle. By the maximum modulus principle, $h(z)$ is identically equal to $0$. Therefore, for every $z$ either $f(z) = 1$ or $g(z) =1$ (or both). Since functions $f(z)$ and $g(z)$ are continuous, one of them must be equal to $1$ in some neighborhood of $0$, and thus be equal to $1$ identically.

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Nice argument! $g(z) = f(-z)$ would work too. Btw, I understood the last sentence like this: If one of the functions isn't $1$ at some point, then by continuity - it isn't $1$ at an open neighborhood of the point, which means the other is $1$ in this neighborhood. –  Yoni Rozenshein Mar 25 '13 at 2:26
    
@YoniRozenshein: You are right $g(z) = f(-z)$ also works. Your explanation for the last sentence is shorter than what I had in mind! –  Yury Mar 25 '13 at 2:33

By the reflection principle the function $$g(z):=\cases{f(z)\quad&$\bigl(|z|\leq 1\bigr)$\cr &\cr \overline{f\bigl(1/\bar z\bigr)} &$\bigl(|z|>1\bigr)$\cr}$$ is analytic in an open neighborhood $U$ of the point $i$ and constant on an arc through this point. It follows that $f$ is constant on $U\cap D$; therefore $f$ has to be constant on all of $D$.

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Can you use the claim "is analytic in a neighborhood of the point i and constant on an arc through this point. Therefore it has to be constant all over.." if g(z) isn't analytic at |z|=1? –  Ran Kashtan Mar 15 '13 at 15:24
    
@Ran Kashtan: My wording was indeed imprecise. I have edited it. –  Christian Blatter Mar 15 '13 at 15:49
    
This proves a stronger statement: It suffices that $f$ extends continuously to only a piece of arc where it is equal to $1$. It does not have to extend a priori to other parts of the unit circle at all. –  WimC Mar 24 '13 at 20:04

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