# If $f,g: U \rightarrow \Omega$ are holomorphic, $f(0)=g(0)$ and $f$ is 1-1&onto, then $f$ has larger image of a disk than that of $g$.

I'm working on the RCA of rudin but having a difficulty in the following problem:

Suppose $f$ and $g$ are holomorphic mappings of $U$(the unit circle centered at 0) into $\Omega$, $f$is one to one and $f(U)= \Omega$, and $f(0)=g(0)$. Prove that

$g(D(0;r)) \subset f(D(0,r))$ for each $0 < r < 1$.

I tried to use the fact that both images are open and espeially, $f(D(0,r))$ is a simply connected region, but have no idea to begin. Can anyone give me a hint?

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Words like "problem" and "question" in the title are redundant, since everything that's asked here is a problem or a question. The tags are visible wherever the title is displayed, so having a tag as the only substantial part of the title makes the title entirely redundant. The title should summarize the question more specifically. –  joriki Dec 12 '12 at 12:13
Thanks joriki. It was my mistake that I did not obey the essential rules! I edited the title and tags. Please let me know if there is another mistake:) –  gaouls Dec 12 '12 at 12:33

Hint: Apply Schwarz lemma to $f^{-1}\circ g$.
Thank you very much! I solved the prolem with you hint! It is not so difficult to show the result with this help, but it was not easy for me to consider the function $f^{-1} \circ g$. Appreciate your help. –  gaouls Dec 12 '12 at 12:57
@richard Believe it or not, I figured your hint on my own but I cannot proceed to show $|z_1| \le |z_2| \implies |f(z_1)| \le |f(z_2)|$. Would you please add the complete solution? –  PeterM Feb 28 '13 at 0:08
@richard Schwarz lemma gave me $|f^{-1}\circ g(z)| < |z|$. I thought of applying $f$ to both sides to reach |g(z)| < |f(z)| and prove the exercise. I don't know what else to do. Thanks. –  PeterM Feb 28 '13 at 9:10