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Suppose $\phi_{1}, \phi_{2} : D -> D$ are 2 holomorphic functions from unit disk into itself such that $\phi_{1}(0) = \phi_{2}(0) = 0$ and $\phi_{1}(D)$ contains $\phi_{2}(D)$. I want to show that in this case $ \vert \phi_{1} \prime (0) \vert \leq | \phi_{2} \prime (0) |$. I know that I should use Schwarz lemma, but I don't know how to proceed?

Thank you in Advance.

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    $\begingroup$ I am a bit confused by your desired conclusion $\phi_{1} \prime (0) \geq \phi_{2} \prime (0)$. Are these not complex numbers where ordering does not work? Maybe you need some modulus signs in there? $\endgroup$ – Old John Dec 7 '13 at 23:22
  • $\begingroup$ yes, there should be modulus signs. :) $\endgroup$ – Meryl Dec 7 '13 at 23:31
  • $\begingroup$ Thanks. Just made a tiny edit to the title to make it match the question - hope that is OK. $\endgroup$ – Old John Dec 7 '13 at 23:36
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Hint:

Let $g(z)=\frac {φ_1(z)}{z}$ if $z\neq 0$ and $g(0)=φ_1'(0)$ .

Let $h(z)=\frac {φ_2(z)}{z}$ if $z\neq 0$ and $h(0)=φ_2'(0)$.

Then $g,h$ are continuous.

Also $\Bbb D$ is connected. And so do $φ_1(\Bbb D),φ_2(\Bbb D)$.

Think. If $\vert \phi_{1} \prime (0) \vert >| \phi_{2} \prime (0) |$ then $|g(z)|>|h(z)|$ near $0$.Can this be done since $φ_1(\Bbb D)\supset φ_2(\Bbb D)$?

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  • $\begingroup$ Thank you so much. And this will be the contradiction of maximum modulus principle and maybe we should add that $g(0):=\phi_{1} \prime (0)$ and $h(0):=\phi_{2} \prime (0)$. $\endgroup$ – Meryl Dec 8 '13 at 10:20
  • $\begingroup$ @Hovher yes correct ,$g(0)=φ_1'(0)$ same for $φ_2'(0)=h(0)$. i wrote it wrong. Typo mistake. $\endgroup$ – Haha Dec 8 '13 at 10:39

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