# A complex analysis problem to prove an inequality

$\textbf{Problem.}$ Suppose $f$ is a holomorphic function on $\{z\in\mathbb{C}:|z|<1\}$, the open unit disk, with the property that Re$f(z)>0$ for every point $z$ in the disk. Prove that $|f'(0)|\leq 2\text{Re}f(0)$.

This is a problem that appeared on a qualifying exam in some graduate school. What I tried is, define $\displaystyle\varphi(z)=\frac{z-1}{z+1}$, then this sends the open right half plane onto the open unit disk, so let $\hat{f}=\varphi\circ f(z)$, then its image is contained in the open unit disk. And I wanted to use the Schwarz lemma, but then the origin had to be fixed, so define $\displaystyle\psi(z)=\frac{\hat{f}(0)-z}{1-\overline{\hat{f}(0)}z}$ and let $\tilde{f}=\psi\circ\hat{f}$, then the image of this is still contained in the open unit disk and it fixes the origin so the Schwarz lemma $|(\tilde{f})'(0)|\leq 1$ could be applied. But after all the calculation, I got $$|f'(0)|\leq\frac{|f(0)+\overline{f(0)}+2|f(0)|^{2}|^{2}}{2|f(0)+1|^{2}|f(0)|}$$ but, for example, if the magnitude of the imaginary part of $f(0)$ is quite bigger than the magnitude of the real part of $f(0)$, then this doesn't result the desired inequality but actually means that the desired inequality is wrong instead. I also thought about composing some different function to the RHS of $\hat{f}$ and use the Schwarz lemma but it didn't seem to work.

Maybe I should try something else, but what could be tried instead?

## 1 Answer

Let $w=f(0)$, and consider the map $\phi(z)=\frac{z-w}{z+\overline{w}}$. Since $-\overline{w}$ has the same imaginary part as $w$ and lies in the left half-plane, $\phi$ maps $\{z:\Re z>0\}$ into the unit disk, with $\phi(w)=0$.

Therefore if $g(z)=\phi(f(z))$, then $g$ maps the unit disk into itself, with $g(0)=0$. Therefore $|g^{\prime}(0)|\leq 1$ by the Schwarz lemma.

However, $g^{\prime}(0)=\phi^{\prime}(w)f^{\prime}(0)$ and $$\phi^{\prime}(z)=\frac{z+\overline{w}-(z-w)}{(z+\overline{w})^2}=\frac{2\Re w}{(z+\overline{w})^2}$$ hence $\phi^{\prime}(w)=\frac{2\Re w}{(2\Re w)^2}=\frac{1}{2\Re w}$. Therefore $|g^{\prime}(0)|\leq 1$ implies that $|f^{\prime}(0)|\leq |2\Re w|=2\Re(f(0))$.