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Let $f(z)$ be holomorphic in the unit disk, with $f(0)=0$, and $|f(z)+zf'(z)|<1$. Show that $|f(z)|\leq |z|/2$.

So this should be an easy problem, but I got stuck. I defined $g(z)=f(z)+zf'(z)$ which is holomorphic in the unit disk and satisfies $g(0)=0$ so by an application of Schwarz lemma we have $|f(z)+zf'(z)|\leq |z|$ for all $z$, and I can also conclude $|f'(0)|\leq \frac{1}{2}$, but I have been unable to get the desired inequality from here.

Any hints on how to proceed? Thanks in advance!

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Isn't $g(z)$ the derivative of $z\mapsto z f(z)$ ?

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  • $\begingroup$ Ahhhh, I did not see that, that makes it easy. Thanks :) $\endgroup$ – TheManWhoNeverSleeps Jan 16 '15 at 19:43
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Put: $g(z)=(zf(z))'$. We have $g(0)=0$. By Schwars lemma: $$|g(z)\leqslant |z| \Rightarrow (zf(z))'\leqslant |z|\quad (*)$$ We can write: $$zf(z)=zf(z)-0f(0)=\int_0^z(tf(t))'dt\displaystyle\mathop{\leqslant}^{(*)} \int_0^{|z|}tdt=\dfrac{|z|^2}{2}$$ So $$|zf(z)|\leqslant \dfrac{|z|^2}{2}\Rightarrow |f(z)|\leqslant \dfrac{|z|}{2}$$

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