If $f$ is a complex valued function which takes the unit disc $U$ to itself and $f(\frac43)=\frac43$ while $f'(\frac23)=\frac43$, how can we find $f$ if it exists?
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1I think something is sketchy here - $f$ is supposed to take the unit circle to itself, but $f(\frac{2}{3})=\frac{4}{3}$? – Miel Sharf Dec 15 '14 at 15:49
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@MielSharf: I think it is $f'(2/3)=4/3$. – Clayton Dec 15 '14 at 15:52
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@ Miel Sharf This is quite possible to have $|f(z)|= 1$ whenever $|z|=1$, and $|f(z)|>1$ when $|z|<1$. For example, $f(z)=\frac{1}{z}$. – Tim Raczkowski Dec 15 '14 at 15:53
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Isn't the unit disc all of $|z|<1$ (or $|z|\le1$)? – Miel Sharf Dec 15 '14 at 15:53
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1@Tim: The unit disc contains $2/3$ and not $4/3$. Thus it doesn't take the unit disc to itself. However, the point is superfluous as it is the derivative and not the function itself. – Clayton Dec 15 '14 at 15:54
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I am almost certain that all $\frac{4}{3}$s should in fact be $\frac{3}{4}$s. It's a common exercise then. – Daniel Fischer Dec 15 '14 at 15:55
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Sorry I misread. – Tim Raczkowski Dec 15 '14 at 15:57
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@DanielFischer If that's the case, this might be a duplicate of: http://math.stackexchange.com/questions/1063584/existence-of-holomorphic-function-application-of-schwarz-lemma Which you already answered. – hjhjhj57 Dec 16 '14 at 06:48