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

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


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 $\phi$.


  1. Approach Zero search results.

  2. Schwarz Lemma search results: https://math.stackexchange.com/search?page=11&tab=Relevance&q=schwarz%20lemma

  • 2
    $\begingroup$ Thank you very much for your efforts towards confirming the lack of a duplicate, and +1. I'd love to know where you found this problem(not for the sake of context, but in the hope that there will be great problems there!) $\endgroup$ Jul 16, 2021 at 17:32
  • 1
    $\begingroup$ Sure thing Teresa. This is from a past qualifying exam at a U.S. college. I can try to find the link... $\endgroup$
    – 311411
    Jul 16, 2021 at 17:54
  • $\begingroup$ No need! This is just a magnificent question all round. I hope it will receive a great answer. $\endgroup$ Jul 16, 2021 at 17:56
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    $\begingroup$ Doubt that this will be of any help at all but the $\frac14$ reminds me of the following theorem: Suppose $f : \Bbb D \to \Bbb C$ is analytic, injective, and satisfies $f(0) = 0$ and $f'(0) = 1$. Then, the image of $f$ contains the open disc $D(0, \frac14)$. (A reference is Theorem 14.14 of Rudin's RCA.) $\endgroup$ Jul 16, 2021 at 18:14
  • $\begingroup$ Unless I made an error, the lower bound $1/4$ can be replaced by $1/2$. I posted a follow-up question asking for more precise estimates: math.stackexchange.com/q/4217283/42969. $\endgroup$
    – Martin R
    Aug 5, 2021 at 9:17

2 Answers 2


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!

  • $\begingroup$ I will need to study some new (for me) material in order to understand your answer; thanks Conrad. I assume you are using this theorem: en.wikipedia.org/wiki/Parseval%27s_theorem $\endgroup$
    – 311411
    Jul 16, 2021 at 18:47
  • $\begingroup$ How do you get $|ab + a^2b| \le 3/4$? I can see you deducing $|ab + a^2b - 1| \le 1/4$ but that doesn't imply the former inequality. $\endgroup$ Jul 16, 2021 at 19:35
  • $\begingroup$ Very nice! – If I understand your proof correctly, $1/4$ is not the best possible bound. $\endgroup$
    – Martin R
    Jul 16, 2021 at 19:47
  • $\begingroup$ @Martin looks like that $\endgroup$
    – Conrad
    Jul 16, 2021 at 19:52
  • $\begingroup$ @Arya we use that $|a| \le 1/2, |b| \le 1$ and the triangle inequality to get that at most $1/2+1/4=3/4$; then $|1-ab-a^2b| \ge 1-3/4=1/4$ $\endgroup$
    – Conrad
    Jul 16, 2021 at 20:01

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

  • $\begingroup$ This is fantastic. Thank you for diving into this problem with me. It gave me a good reason to study Schwarz-Pick and its proof. There were only two moments when I had to pause and fill in some details. For the sake of fellow asses trying to cross these bridges, I will add comments below. I will follow your follow-up question with interest: math.stackexchange.com/q/4217283 $\endgroup$
    – 311411
    Aug 5, 2021 at 14:54
  • $\begingroup$ Details on concluding that $|a|=1/2$ and $|b|=1$: by the inequality(equality) chain, we know $0<|a||a+1||b|=3/4.$ Suppose $|a|<1/2$ or $|b|<1$. Then $0<|ab|<1/2.$ Hence $3/2 < |a+1|$. But that cannot be, for $|a+1|\leqslant |a|+1\leqslant 3/2.$ $\endgroup$
    – 311411
    Aug 5, 2021 at 15:03
  • $\begingroup$ Details on why the function $g(z) = f(z)/z$ maps $\Bbb D$ into itself: this conclusion appears in the proof of the ordinary Schwarz Lemma. It follows from the Maximum Principle (continuous on a compact version) that $g(z)$ maps $\Bbb D$ into $\overline{\Bbb D}$. But if $\zeta\in\Bbb D$ with $g(\zeta)\in \partial \Bbb D$ then $g$ must be constantly $g(0)=f'(0)\in \partial \Bbb D,$ and $f$ is a rotation, a possibility already ruled out in the proof of the estimate. (Here "$g$ constant" follows from the other form of Maximum Principle, the one dealing with a domain.) $\endgroup$
    – 311411
    Aug 5, 2021 at 15:21
  • $\begingroup$ @311411: The Schwarz lemma already states that – unless $f$ is a rotation – $|f(z)| < |z|$ for $z \ne 0$, and $|f'(z)| < 1$. That implies $|g(z)| < 1$ for all $z$. $\endgroup$
    – Martin R
    Aug 5, 2021 at 16:11
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    $\begingroup$ Aha, quite right. No need to prove Schwarz Lemma again; we just apply (the "furthermore" part of) it to $f$. $\endgroup$
    – 311411
    Aug 5, 2021 at 16:17

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