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Suppose that $f$ is a one-to-one analytic mapping of the unit disc onto a domain $\Omega$. Show that if $g$ is any other analytic map of the unit disc into $\Omega$ such that $g(0) = f(0),$ then $g(D_r(0))\subset f(D_r(0))$ for all $0<r<1$. (Here $D_r(0)$ denotes the open disc of radius $r$ about $0$.)

This seems reminiscent of problems which use the Schwarz Lemma, but the only meaningful composition for that seems to be $f^{-1}\circ g,$ but I not don't think this is well-defined.

My other thought was Since $f$ is a 1-1 analytic mapping of $D_1(0)$ onto $\Omega$, we have $$f\colon D_r(0)\to f(D_r(0))$$ is also 1-1 and onto. If we suppose that $g(D_r(0))\supset f(D_r(0))$, then maybe this will lead to the conclusion $g=f$? I'm just not sure how to proceed. Suggestions?

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  • $\begingroup$ Why do you doubt that $f^{-1}\circ g$ is well-defined? It is, and then the Schwarz lemma quickly finishes the argument. $\endgroup$ – Daniel Fischer Feb 17 '17 at 21:10
  • $\begingroup$ I had confused myself into thinking that we want $f^{-1}\circ g$ to be bijective. I see now that it is well-defined. I know the Schwarz lemma gives $|f^{-1}(g(z))|\le |z|$ and $|g'(0)|\le |f'(0)|.$ To show the inclusion I want, I think I just need $|g(z)|\le |f(z)|, $ but I don't see how to get there with the $|\,\cdot\,|$. $\endgroup$ – user346096 Feb 17 '17 at 21:28
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Since $f$ is a biholomorphism, we have $f^{-1}\circ g \colon D_1(0) \to D_1(0)$, and from $g(0) = f(0)$ it follows that $(f^{-1}\circ g)(0) = 0$. The Schwarz lemma then says

$$\lvert (f^{-1}\circ g)(z)\rvert \leqslant \lvert z\rvert$$

for all $z\in D_1(0)$. From that we have

$$(f^{-1}\circ g)(D_r(0)) \subset D_r(0)$$

for all $r \in (0,1]$, and hence

$$g(D_r(0)) = f\bigl((f^{-1}\circ g)(D_r(0))\bigr) \subset f(D_r(0)).$$

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    $\begingroup$ Thank you. As soon as I read this it seemed glaringly obvious. $\endgroup$ – user346096 Feb 17 '17 at 21:40

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