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?

  • $\begingroup$ 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

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

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

  • 1
    $\begingroup$ Thank you. As soon as I read this it seemed glaringly obvious. $\endgroup$ – user346096 Feb 17 '17 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.