Let $\Delta$ be the unit disc centered at $0$ in the complex plane and consider an analytic function $g:\Delta \to \Delta $ such that $g(0)=0$. I have proved the following claim (using Schwarz Lemma).
For all $z\in \Delta,$ we have $|g(z)+g(-z)|\leq 2|z|^2.$
I would like to show that
If $|g(a)+g(-a)|= 2|a|^2$, for some $0<|a|<1$, then there is a constant $c$ with $|c|=1$ such that $g(z)=cz^2$, for all $z\in \Delta$.
My Attempt: Let $h(z)=\frac{g(z)+g(-z)}{2}$. I've managed to show that $h$ has the form $h(z)=k(z^2)$, for some function $k$. Since $h(z)\leq 1$, for all $z\in\Delta$ and $h(0)=0$, it follows by Schwarz Lemma that there exists a constant $c$ with $|c|=1$ such that $|k(z^2)|=cz^2$, so it follows that $\frac{g(z)+g(-z)}{2}=cz^2\Longrightarrow g(z)+g(-z)=2cz^2.$
My Question: How do I proceed from here to show that $g(z)=cz^2$?
Any help/hint will be very useful. Thanks in advance.