Schwarz' lemma: Prove that the inequality is strict unless the function f is of a certain form. the question (not homework) I am trying to answer is, in part:

Let $f$ be an analytic function that maps the open unit disk $D$ into
  itself and vanishes at the origin.  Prove that the inequality $$|f(z)| + |f(−z)| ≤ 2 |z^2| $$ is strict, except at the origin, unless f has the
  form $f(z) = λz^2$ for some $λ$ a constant of absolute value one.

I applied Schwarz' lemma to obtain the inequality.  Below is my answer:

It is clear that the inequality holds at the origin.  The hypotheses given for $f$ are the same as those required for Schwarz' lemma to apply to $f$; the lemma clearly applies to both $f(z)$ and $f(-z)$.  Thus I have
  $$|f(z) + f(-z)| \leq |f(z)| + |f(-z)| \leq |z| + |z| = 2|z|$$
  Divide both sides by $|2z|$ (I have assumed $z\neq 0$):
  $$\frac{|f(z) + f(-z)|}{|2z|} \leq 1$$
  This fact shows that the function $(f(z) + f(-z))/ 2z$ has a removable singularity at $z = 0$ (since it is bounded in a neighbourhood of that point).  Calling the analytic continuation $g(z)$, $g$ is a holomorphic map from $D$ to $D$ and vanishes at the origin; to see why, expand $f(z) + f(-z)$ into the sum of two power series, note that the first two terms vanish, and conclude that $g$ has a zero of order at least one at the origin.  So Schwarz' lemma applies to $g(z)$, and in particular $|g(z)| \leq |z|$.  But this fact directly implies the desired inequality.

The problem is that I have come most of the way to proving the strict inequality, but cannot prove that the given form is the only possible form for $f(z)$.  I proved that if $|f(c) + f(-c)| = 2|c^2|$ for some $c$ not the origin, then the constructed function $g(z)$ is a rotation by Schwarz' lemma, which means that $f(z) + f(-z) = \lambda z^2$.  This means that all even-index coefficients of the power series of $f$ must be zero, but it does not rule out the possibility that there are odd-index coefficients.  None of the standard tricks like Cauchy inequalities work since the domain is the unit disc.  I also tried looking at the other implications of Schwarz (i.e., that $|f'(0)| < 1$) and that, too, led nowhere.  What am I missing here?
 A: Apply the Schwarz Lemma to
$$
g(z)=\frac{f(z)+f(-z)}{2z^2}
$$
to deduce that $|g|\le 1$. If equality holds somewhere on $|z|<1$, then $g$ is constant, by the Schwarz Lemma or the maximum principle, so
$$
f(z)+f(-z)=2\lambda z^2
$$
for some $|\lambda|=1$. This says that $f(z)=\lambda z^2 + h(z)$, with $h$ odd. We want to show that $h\equiv 0$, and this follows by considering points near $|z|=1$: Fix $\alpha$ and consider
$$
f(\pm re^{i\alpha/2}) = \lambda r^2 e^{i\alpha} \pm h(re^{i\alpha/2}) .
$$
The first term approaches a limit of absolute value $1$ as $r\to 1-$, so we must have that $\lim_{r\to 1-} h(re^{i\beta})=0$ for all $\beta$, or $f$ would not take values inside the unit disk. A bounded holomorphic function with (radial) boundary value zero is well known to be identically zero.
A more elementary argument is also possible in this last step: the convergence is uniform in $\beta$, so the maximum principle also shows that $h\equiv 0$.
A: You alredy have that $f(z) +f(-z) = 2\lambda z^2$. Let $h:D \longrightarrow \mathbb{C}$ be a function defined by $h(z) = \lambda z^2 -f(-z)$. Hence, $h$ is analytic and $f(z) = \lambda z^2 +h(z)$. We are going to show that $h \equiv 0$.
First, $h$ is odd because
\begin{align*}
        h(-z) &= \lambda (-z)^2 -f(z) \\
            &= \lambda (z)^2 -\big( \lambda z^2 +h(z) \big) \\
            &= -h(z).
\end{align*}
On the other hand, $f$ satisfafies the Schwarz Lemma hypotheses , then $\forall z \in D, \ |f(z)| < |z|$ and
\begin{equation}        \label{07:eq:01}
    \left| \lambda z^2 +h(z) \right| = |f(z)| \leq |z|.
\end{equation}
Futhermore $|f(-z)| < |z|$, thus
\begin{equation}        \label{07:eq:02}
    \left| \lambda z^2 -h(z) \right|
    = \left| \lambda (-z)^2 +h(-z) \right|
    = |f(-z)|
    \leq |z|.
\end{equation}
Because of the Parallelogram Law, we have
\begin{align*}
    2|\lambda z^2|^2 +2|h(z)|^2
        &= \left| \lambda z^2 +h(z) \right|^2 + \left| \lambda z^2 -h(z) \right|^2 \\
        & \leq 2|z|^2 .
\end{align*}
It follows that $\forall z \in D, \ |h(z)| \leq \sqrt{|z|^2 -|z|^4}.$ The square root is well-defined because $z \in D$ implies $0<|z|<1$, then $0 < |z|^4 < |z|^2$.
Finally we are going to do something similar to the Schwarz Lemma proove. Let $r \in \mathbb{R}$ such that $0<r<1$ and $B = B_{r}(0)$. If $z \in \partial B$, then $|z|=r<1$, so $h$ is well-defined in all $\overline{B}$. By the Maximum Modulus Principle we assure that
$$
    \max_{z \in \overline{B}} |h(z)|
    = \max_{z \in \partial B} |h(z)|
    \leq \sqrt{r^2 -r^4}.
$$
As $r \rightarrow 1$ we get $|h(z)|\leq 0$ for all $z \in D$. Thus $h \equiv 0$ and $f(z) = \lambda z^2$.
