If $f(0)=0$ then prove that $|f(z)+f(-z)| \leq 2|z|^2$

Let , $\Delta = B(0,1)$.

If $f:\Delta \rightarrow \Delta$ is holomorphic, and $f(0)=0$, prove that $|f(z)+f(-z)| \leq 2|z|^2$.

My attempt so far:

$\displaystyle f(z) = \sum_{n=0}^{+\infty} a_n z^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} + \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}$

$\displaystyle f(-z) = \sum_{n=0}^{+\infty} a_n (-z)^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} - \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}$

Then, $\displaystyle f(z)+f(-z) = 2 \sum_{n=0}^{+\infty} a_{2n} (z^2)^n =: 2 g(z^2)$

If $g(\Delta) \subset \Delta$, as $g(0) = 0$, I can invoke Schwarz Lemma and I win. However, I don't know how to prove this. Could anyone help me? Thanks!

Assume $g(z)\not \in \Delta $ for some $z\in \Delta$. Then we can take some square root $w\in \Delta$ of $z$ to get $$ \frac{1}{2}\left(f(w)+f(-w)\right) = g(z). $$ Taking absolute values on the left hand side we get $$ \left|\frac{1}{2}\left(f(w)+f(-w)\right)\right|\leq \frac{1}{2}\left(\left| f(w)\right|+\left|f(-w)\right|\right) \leq \frac{1}{2}(1+1)=1. $$ This contradicts our assumption that $g(z)\not \in \Delta$. So $g(\Delta)\subseteq \Delta$.

Hint :

Apply Schwarz lemma on $$g(z)=\begin{cases}\frac{f(z)+f(-z)}{2z}&\text{ if } z\not=0\\0 &\text{ if } z=0\end{cases}$$

Applying Schwarz lemma on $f$ we get , $|f(z)|\le |z|$ for all $z\in \Delta$. Then ,

$$|g(z)|\le \frac{|f(z)|+|f(-z)|}{2|z|}\le \frac{2|z|}{2|z|}=1.$$That is , $g:\Delta \to \Delta$ analytic with $|g(z)|\le 1$ and $g(0)=0$. So you are able to apply Schwarz lemma on $g$.