$|f(f(z))-z^2|$ must be large somewhere in the disc $\mathbb{D}$?

Solution 1:

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!

Solution 2:

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 $|a|^2 + |b| \le 1$.