$|f(f(z))-z^2|$ must be large somewhere in the disc $\mathbb{D}$? I wish to prove the statement shown in the following block. I thought this may have appeared in the Math Stack before; sorry if I failed to find it (see "Research" below).
The proposition seems to call for a proof via H.A. Schwarz Lemma, but I am interested in any proof of it.

Let $\sup$ always mean the supremum in the unit open disc
$\mathbb{D}$. Let $f$ map $\mathbb{D}$ into $\mathbb{D}$ analytically, and fix the origin. Prove that $$\sup \left|f(f(z))-z^2\right| \,\,\geq\,\,\frac{1}{4}.$$

My Attempt.
If it happens that $f$ is a rotation, $z \mapsto {e^{i\psi}}z,$ then the result follows from the choice $z=1/2 \in $ the disc:
$$\left|f(f(\frac{1}{2}))-(\frac{1}{2})^2\right| \,\,\geq\,\,\left|f(f(\frac{1}{2}))\right|-\left|\frac{1}{4}\right|\,\,=\,\,\left|f(\frac{1}{2})\right|-\frac{1}{4}\,\,=\,\,\left|\frac{1}{2}\right|-\frac{1}{4}\,\,=\,\,\frac{1}{4},$$
since rotation means $|z|\,=\,|f(z)|$ for any $z$ in $\mathbb{D}$.
Therefore we assume $f$ is not a rotation. By Schwarz Lemma, we know $f'(0) \in \mathbb{D}$, and we know $|f(z)|<|z|<1$ throughout the disc.
(Starting here I pursue an idea; I am not sure if it is helpful...) Define the function $$\phi(z)\,\,=\,\,\frac{f(f(z))-z^2}{2},$$
and note that it also satisfies the hypotheses of the Schwarz Lemma. It is easy to check that $\phi$ is not a rotation when $f$ is not a rotation. So now our goal is to show

$$\sup |\phi(z)| \,\,\geq\,\,\frac{1}{8}.$$

Remarks.
That's what I have done. The derivative of $\phi$ is $\frac{1}{2}(f'(f(z))f'(z)-2z)$, and using this we can know that $|\phi'(0)|<\frac{1}{2}.$ Of course we know $|\phi(z)|<|z|<1$ throughout the disc.
Another idea is to pass to series expansions of $f$ and $\phi$.
Research.

*

*Approach Zero search results.


*Schwarz Lemma search results: https://math.stackexchange.com/search?page=11&tab=Relevance&q=schwarz%20lemma
 A: The following solution is very similar to what Conrad wrote. Instead of Parseval's identity we use an estimate which can be obtained from the Schwarz lemma and its corollary, the Schwarz-Pick theorem:

Let $f(z) = a z + b z^2 + \dots$ be holomorphic in the unit disk $\Bbb D$ with $f(0) = 0$ and $f(\Bbb D) \subset \Bbb D$. Then $|a|^2 + |b| \le 1$.

Proof: If $f$ is a rotation then $|a| = 1$ and $b = 0$ and we are done. Otherwise the function $g(z) = f(z)/z$ maps $\Bbb D$ into itself and the Schwarz-Pick theorem can be applied to $g$:
$$
\frac{|g'(z)|}{1-|g(z)|^2} \le \frac{1}{1-|z|^2} \, .
$$
Setting $z=0$ gives $|g'(0)| \le 1-|g(0)|^2$, which is exactly the desired estimate.

Now we can can proceed as follows: Let
$$
 f(z) = az + bz^2 + \dots
$$
be a holomorphic function which maps the unit disk into itself and fixes the origin and assume that
$$
 \sup \{ |f(f(z))-z^2| : z \in \Bbb D \} < \frac 1 4 \, .
$$
Then
$$
 F(z) = 4 \bigl(f(f(z)) - z^2\bigr) = 4 a^2 z + 4 \bigl(a(a+1)b - 1\bigr) z^2 + \dots
$$
also maps the unit disk into itself and fixes the origin. The above lemma gives
$$
 16 |a|^4 + 4 |a(a+1)b - 1| \le 1 \, .
$$
It follows that $|a| \le 1/2$ and
$$
 \frac 14 \ge |a(a+1)b - 1| \ge 1 - |a||a+1||b| \ge 1- \frac 1 2 \cdot \frac 3 2 \cdot 1 = \frac 1 4.
$$
So equality holds everywhere in the above inequality chain. In particular, $|a| = 1/2$ and $|b| = 1$, which is a contradiction to $|a|^2 + |b| \le 1$.
A: If $f(z)=az+bz^2+..$, by Parseval (integrating $|f|^2$ on $|z|=r<1$ and letting $r \to 1$) we get that $|a|^2+|b|^2+..\le 1$ so $|b| \le 1$
But now $f(f(z))-z^2=a^2z+(ab+a^2b-1)z^2+...$ and if the result would be false, we would get again by Parseval that:
$|a|^4+|ab+a^2b-1|^2 \le 1/16$ so $|a| \le 1/2, |ab+a^2b| \le 3/4$ hence $|ab+a^2b-1|^2 \ge 1/16$ so we must have equality in the inequalities above or $|a|=1/2, |b|=1$ and that contradicts $|a|^2+|b|^2+..\le 1$ so we are done!
