# Show that $f$ maps the entire unit disc onto itself.

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!

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The closed unit disk is compact. Also: what is T? – AQP Mar 6 '12 at 5:26
That'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)/ – caligurl11 Mar 6 '12 at 5:42
In 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. – AQP Mar 6 '12 at 6:50

If $w \in D$ and $w \not \in f(D)$ then $$z \mapsto \frac{1}{f(z)-w}$$ is holomorphic on $\overline{D}$. By the maximum modulus principle, for any $z \in D$ $$\left|\frac{1}{f(z)-w}\right| \leq \max_{\omega \in \partial D} \left| \frac{1}{f(\omega)-w} \right| \leq \frac{1}{\left||f(\omega)|-|w|\right|} = \frac{1}{1-|w|}$$ so $|f(z) - w| \geq 1-|w|$. This means that $D \setminus f(D)$ is open. By the open mapping theorem either $f$ is constant or $f(D)$ is open. The latter implies that $f(D)=D$ since $D$ is connected. Moreover, the image of a compact set under a continuous function is compact. Therefore if $f$ is not constant then $f(\overline{D}) = \overline{D}$.

Or, alternatively, this bound shows that $w$ can be moved in a straight line towards $0$ while the radius of the "image free" disc around it increases. For $w=0$ the bound becomes $|f(z)| \geq 1$ so that $f(D) \cap D = \emptyset$. So either $f(D)$ contains all of $D$ or avoids it entirely. In the latter case $f$ maps into the unit circle. This would mean that $\overline{f} = f^{-1}$ but $\overline{f}$ can be holomorphic (complex differentiable) only if $f'$ vanishes identically. The conclusion is that either $f(D)=D$ or $f$ is constant.

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Can you explain a little more. I don't understand why that means that D\f(D) is open. Thanks! – caligurl11 Mar 7 '12 at 23:58
I think you need to review basic point set topology. He means that $\forall w\in D-f(D)$ has a neighborhood that does not include any point from $f(z)$. Thus $D-f(D)$ must be open. The close mapping theorem suggested that $f(D)$ must be open unless $f'=0$ everywhere, and since $D=(D-f(D))\cup f(D)$, while $D$ is connected, $f(D)=D$ or $f(D)=\emptyset$. The later is not possible. Thus $f(D)=D$. Since $\bar{D}$ is compact, we should have $\bar{f(D)}=f(\bar{D} )=\bar{D}$. – Kerry Mar 8 '12 at 2:48

A proof:

Suppose $f$ does not map $D^{2}$ to itself, then it must map a certain point $p\in D^{2}$ to the outside of $D^{2}$. Use the maximal modulus principle we may argue that $f$'s maximal value must be on the boundary of $S^{1}$. But this is impossible by our hypothesis. Thus $f$ must map $D^{2}$ to itself.

You may question why $f$ must map a point in $D^{2}$ to the outside. In fact, $f$ cannot keeps the boundary while leaving any point in $D^{2}$ uncovered because $f$ is largely locally conformal(otherwise $f$ would be a constant function and the assertion would be trivial). Suppose $p$ is some point uncovered by $f$, and $d(p, f(D^{2})\not=0$, then $f$ would leave a 'hole' and this cannot hold if $f$ is conformal. The othercase $d(p,f(D^{2}))=0$ can be handled similarly. In the isolated points $f'=0$ the behavior is well controlled since $f$ is analytic. Thus $f$ must map at least one point from $D^{2}$ to $\mathbb{C}-D^{2}$ if $f$ is not an automorphism.

For your questions: I'm not sure if mapping the unit circle into itself means that $|f|=1$ on the unit circle.

Yes, this implies $|f(S^{1})|=1$.

Also is the unit disc compact?

This should be trivial. I guess your confusion stems from the fact that $f$ is only supposed to be analytical at the open disc (thus could have no analytical continuation on the boundary of the disc). But this does not violate the statement of the theorem, if you check it carefully.

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Does this prove that the mapping is onto though? Thanks! – caligurl11 Mar 7 '12 at 19:23
Yes. See the second part of the reasoning. It might be not rigor enough, though. – Kerry Mar 8 '12 at 1:23
I'm actually not familiar with what conformal means. Is there another way to do it? – caligurl11 Mar 8 '12 at 2:20
I recommend you to learn it, wikipedia is a good source. – Kerry Mar 8 '12 at 2:39

Let $D$ denote $D(0,1)$, and $cl(D)$ the closed disk $\{|z| \leq 1\}$. The maximum modulus theorem immediately gives that $f$ maps $D$ into itself. So the issue is showing that it's onto. The following is a topological proof.

By the open mapping theorem $f(D)$ is open, so $f(D) \cap D$ is open. Since $cl(D)$ is compact, so is $f(cl(D))$. Hence $f(cl(D))$ is closed. Hence $f(cl(D)) \cap D$ is a relatively closed subset of $D$. Since $f$ takes $\{|z| = 1\}$ to $\{|z| = 1\}$, $f(cl(D)) \cap D = f(D) \cap D$. Hence $f(D) \cap D$ is an open, closed, and nonempty subset of $D$. Thus by connectedness of $D$, you have that $f(D) \cap D = D$, which means $f$ is onto as well.

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