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Let $D$ be a convex domain in $\mathbb{C}$, with $D \neq \mathbb{C}$, and let $F : \mathbb{D} \to D$ be a conformal mapping, being $\mathbb{D}$ the unit disc in $\mathbb{C}$. Let $f$ an holomorphic funcion in $\mathbb{D}$, with $f(0) = F(0)$, such that $f(\mathbb{D}) \subset D$. Obviously, exist two sequences, ${\{a_n\}}_{n = 0}^{\infty}$ and ${\{A_n\}}_{n = 0}^{\infty}$, such that $$ f(z) = \sum_{n = 0}^{\infty} a_n z^n \qquad \mbox{ and } \qquad F(z) = \sum_{n = 0}^{\infty} A_n z^n \tag{1} $$ for all $z \in \mathbb{D}$. Prove that $|a_0| \leq |A_0|$ and $|a_n| \leq |A_1|$ for all $n = 1 , 2 , \ldots$.


My thought: by hypothesis, it isn't difficult to prove that $f \prec F$ ($f$ is subordinated to $F$: exists an holomorphic function $w$ in $\mathbb{D}$, with $w(\mathbb{D}) \subset \mathbb{D}$, such that $w(0) = 0$ and $f = F \circ w$). When it happens, we can obtain that $f(0) = F(0)$; in fact, $f(0) = F(w(0)) = F(0)$ and also we can obtain the next affirmations:

  1. $f(0) = F(0)$.
  2. $f(\mathbb{D}) \subset F(\mathbb{D})$.
  3. $f\left(\overline{D(0 , R)}\right) \subset F\left(\overline{D(0 , R)}\right)$ for all $R \in (0 , 1)$.
  4. $\sup_{|z| \leq R} |f(z)| \leq \sup_{|z| \leq R} |F(z)$ for all $R \in (0 , 1)$.
  5. $|f'(0)| \leq |F'(0)|$.
  6. $\sup_{|z| \leq R} (1 - {|z|}^2) |f'(z)| \leq \sup_{|z| \leq R} (1 - {|z|}^2) |F(z)$ for all $R \in (0 , 1)$.

Using $(1)$ and 1., we obtain that $|a_0| = |A_0|$ because $$ a_0 = f(0) = F(0) = A_0\mbox{.} $$ Now, to prove $|a_n| \leq |A_1|$ for all $n = 1 , 2 , \ldots$, I think I must use 2., 3., 4., 5. or 6., and the next property: if $z_1 , \ldots , z_n \in D$ and ${\lambda}_1 , \ldots , {\lambda}_n \in [0 , 1]$, with $\sum_{j = 1}^n {\lambda}_j = 1$, then $\sum_{j = 1}^n {\lambda}_j z_j \in D$, but I can't obtain any result. Any help? Thank you very much.

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  • $\begingroup$ Is $F$ 1-1 and onto? $\endgroup$ – zhw. May 22 '17 at 19:17
  • $\begingroup$ Yes because $F$ is a conformal mapping $\endgroup$ – joseabp91 May 22 '17 at 19:24

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