First of all, notice that the complex function $$\phi_a(z)=\frac{z-a}{1-\bar a z}$$ is a bijection from the unit closed disk in $\mathbb C$ in itself.
a) Prove now that, if $f$ is holomorphic in a domain $D$ including the closed unit disk, and $|f(z)|<1$ for all $|z|<1$, and exists $z_0$, $|z_0|<1$, such that $f(z_0)=f(-z_0)=0$, then $|f(0)|\leq |z_0|^2$.
b) Determine then all functions $f$ as in (a) that satisfy $f(0)=|z_0|^2$.
First of all, $g=f\circ \phi_{z_0}$ and $h=f\circ \phi_{-z_0}$ are in the hypothesis of Schwarz Lemma. We know then that, for example, $|g(z)|\leq |z|$ and then $|f(0)|=|g(z_0)|\leq |z_0|$.
But we want $|z_0|^2$, so, since I do not know anything about the derivatives of $g$ and $h$, i.e. about the order of $0$ as zero, I tried to modify Schwarz Lemma in the following way. If $z_0=0$ we are done. If not, choose a ball around $z_0$ which does not include $z_0^2$ and which remains disjoint from a circle of radius $r$, $|z_0|<r<1$. Let $D_r$ be the domain whose boundary is represented by the $r-circle$ and the circle around $z_0$. The function $g=f\circ \sqrt\phi_{-(z_0)^2}$ (we can choose a branch) is holomorphic in any $D_r$, and satisfies $g<1$. Therefore, since $g(0)=f(\pm z_0)=0$ (in any choice of one of the two branches, which differ only by the sign), we have that $g(z)/z$ is holomorphic on $D_r$. Now I would like to apply the maximum module principle and say that $|g(z)/z|\leq g(y)/y<1/r$, for some $y$ on the $r-circle$, and therefore, taking $r$ to $1$, say that $|g(z)/z|\leq 1$ (this is how one proves Schwarz Lemma). Then choosing $z=z_0^2$ we would end. But the boundary of $D_r$ is not just the $r$-circle.
Is there some way to conclude this proof? Or is there a better and simpler proof?
Thank you in advance.
