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Suppose $f$ is a holomorphic function on the open unit disk $\mathbb{D}$ with $f(0)=0$ and $| f(z) + zf^{'}(z)| <1$ for all $z \in \mathbb{D}$. I have to show that $|f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$.

I have tried to apply Schwarz Lemma but failed to obtain the inequality.

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Let $\phi(z)=zf(z)$, then $\phi'(z)=f(z)+zf'(z)$.

  1. Apply Schwarz lemma to $\phi'$ to get $|\phi'(z)|\le |z|$.
  2. Fix $z\in\Bbb D$ and let $\psi(r)=\phi(rz)$, $r\in[0,1]$. Integrate $$ \phi(z)=\psi(1)=\int_0^1\psi'(r)\,dr=\int_0^1\phi'(rz)z\,dr $$ and use the estimate $$ |zf(z)|=|\phi(z)|\le\int_0^1|\phi'(rz)||z|\,dr\le\int_0^1|rz||z|\,dr=\frac{|z|^2}{2}. $$
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You can not show that $ |f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$ !

Take $c \in \mathbb D, c \ne 0$ and look at $f(z)=c$.

Then we have $| f(z) + zf^{'}(z)| <1$ for all $z \in \mathbb D$.

But we do not have $ |f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$.

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  • $\begingroup$ Is it true if we assume $f$ to be non-constant? $\endgroup$ – AVATAR Jun 22 '18 at 9:14
  • $\begingroup$ No. Take $f(z)=\frac{1}{4}(z+1)$. $\endgroup$ – Fred Jun 22 '18 at 9:32
  • $\begingroup$ Indeed, thanks! $\endgroup$ – AVATAR Jun 22 '18 at 9:42
  • $\begingroup$ @AVATAR Note that $|f(z)|\le\frac{|z|}{2}$ implies $f(0)=0$. Any holomorphic $f$ with $f(0)\ne 0$ and bounded $f(z)$, $f'(z)$ in $\Bbb D$ can be scaled to make a counter-example. $\endgroup$ – A.Γ. Jun 22 '18 at 10:36
  • $\begingroup$ @Fred Sorry. I forgot to put the condition $f(0)=0$. I think the problem is correct now. $\endgroup$ – AVATAR Jun 22 '18 at 14:58

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