Suppose that for some $\epsilon>0$ the function $f$ is holomorphic on $B(0,1+\epsilon)$ such that $f(a) = 0$ and $|f(z)|\leq1$ if $|z| \leq 1$. Prove for $|z| \leq 1$:

$$|f(z)|\leq \left|\frac{z-a}{1-\overline{a}z}\right|.$$

I tried using the lemma of Schwarz which states that on the unit sphere, if $f(0) = 0$ and $|f(z)|\leq1$ then $|f(z)|\leq |z|$.

I think I am missing here a smart translation or something, any tips?

  • $\begingroup$ Can you find the inverse of $z\mapsto \frac{z-a}{1-\bar a z}$? $\endgroup$ – AlexR May 21 '15 at 7:01
  • $\begingroup$ how would that help me? $\endgroup$ – Kees Til May 21 '15 at 7:16
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    $\begingroup$ If you then look at $f\circ g^{-1}$ you can check if this function satisfies the properties of Schwarz lemma to get $$|f\circ g^{-1}(z)| \le |z|$$ and thus $$|f(z)|\le |g(z)|$$ $\endgroup$ – AlexR May 21 '15 at 7:18
  • $\begingroup$ is there an easy way to determine inverses of mobius-transformations? $\endgroup$ – Kees Til May 21 '15 at 7:36
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    $\begingroup$ Yes is a formula, look at the wiki $\endgroup$ – AlexR May 21 '15 at 7:37

Consider the function :$h(z)=\frac{f(z)}{\varphi(z)}$ which: $\varphi(z)=\frac{z-a}{1-\overline az}$. Since $f(a)=0$, $h(z)$ is analytic on $B(0,1+ϵ)$ and because of when $|z|=1$, we have: $|\varphi(z)|=1$, we can write: $|h(z)|=|f(z)|$ for every complex number which: $|z|=1$. From here: $|h(z)|\le 1$ for every complex number which: $|z|\le 1$. Thus: $|f(z)|\le |\frac{z-a}{1-\overline az}|$.


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