Suppose $f$ is analytic in the unit disc $D(0,1)$ and maps the unit circle into itself. Show then that $f$ maps the entire disc onto itself.
So the outline wants us to use the Max Modulus Theorem to show that $f$ maps $D(0,1)$ into itself. Then, use the fact that we proved that if $f:S \to T$, $f$ non-constant and analytic on $S$, and if $f(z)$ is a boundary point, $z$ is a boundary of $S$ to show that the mapping is onto. I'm not sure if mapping the unit circle into itself means that $|f|=1$ on the unit circle. Also is the unit disc compact? Thanks!
Show that $f$ maps the entire unit disc onto itself.
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complex-analysis
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0The closed unit disk is compact. Also: what is T? – 2012-03-06
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0That's what I thought too but D(0,1) is the open unit disc so I don't understand how it could be compact, unless the book meant the closed unit disc. Also T=f(S)/ – 2012-03-06
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0In the complex plane, just as in $R^2$ , K is compact iff K is closed and bounded. Also: can you use winding number? Maybe you can show that the winding number about any point on the disk is non-zero. – 2012-03-06