An Application of Schwarz Lemma Let $f:\mathbb{D} \rightarrow \mathbb{D}$ be a holomorphic mapping satisfying $f(0) = 0$. Here $\mathbb{D} $ is the unit disk centred at $0$.
(i)Show that $|f(z)+f(-z)| \leq 2|z|^{2}$.
(ii) Show that if the equality holds for some $z_0 \in \mathbb{D} \setminus \{0\}$, then there exist $\theta \in \mathbb{R}$ such that $f(z) = e^{i \theta} z^{2}$.
I have figured out a solution with $f(z)$ expanded in power series form. But I believe that Schwarz Lemma is also applicable here. I am just not sure of how to apply it. 
I tried defining $g(z) = \frac{f(z)+f(-z)}{2z}$, but this doesn't seem to work since $g(0)$ is not well defined, and hence not equal to $0$.
Appreciate any help. Thanks in advanced. 
 A: Hint for (i). Write $h(z) = \frac{1}{2}(f(z) + f(-z))$ and show that $h(z)$ is of the form $k(z^2)$.
EDIT: For (ii), here is the intuition I have, and then I'll write it more nicely.
You can write $f(z) = e^{i\theta} z^2 + m(z)$, where $m(z)$ is odd. If the function $m(z)$ is not identically zero, there is a sequence $z_n$ with $|z_n| \to 1$ and $|m(z_n)| \geq \alpha$, say. By considering both $f(-z_n)$ and $f(z_n)$, you can obtain a contradiction with the bound $|f(z)| < 1$. (Once you get close enough to the boundary of the circle, you can no longer add two vectors of length $\alpha$ in opposite directions and stay within the disk.)
All right, now that you've read that, here's a version of the same thing that's a bit prettier. We have
$$\left|\frac{f(z) - f(-z)}{2}\right|^2 = \frac{1}{2}|f(z)|^2 + \frac{1}{2}|f(-z)|^2 - |h(z)|^2 \leq 1 - |h(z)|^2 = 1 - |z|^4 \xrightarrow[|z| \to 1]{} 0,$$
which shows that $f(z) - f(-z) = 0$ by the maximum modulus principle. (The first equality is the parallelogram law.)
A: Let $g(z)=\frac{f(z)+f(-z)}{2}$, then $g'(0)=0.$ Apply Schwarz lemma to $g'(z)$ and apply Schwarz lemma to $g'(z)$, we get $|g'(z)|\le|z|$ and $|g(z)|\le|z|$. So let $|\phi(z)|=|\frac{g(z)}{z}|\le1$ if $z\ne0$ and $\phi(0)=0$. Easy to check $\phi$ is holomorphic. Then Apply Schwarz lemma to $\phi(z)$, we get (i).  
