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Let $f:\mathbb{D} \rightarrow \mathbb{D}$ be a holomorphic mapping satisfying $f(0) = 0$. Here $\mathbb{D} $ is the unit disk centred at $0$.

(i)Show that $|f(z)+f(-z)| \leq 2|z|^{2}$.

(ii) Show that if the equality holds for some $z_0 \in \mathbb{D} \setminus \{0\}$, then there exist $\theta \in \mathbb{R}$ such that $f(z) = e^{i \theta} z^{2}$.

I have figured out a solution with $f(z)$ expanded in power series form. But I believe that Schwarz Lemma is also applicable here. I am just not sure of how to apply it.

I tried defining $g(z) = \frac{f(z)+f(-z)}{2z}$, but this doesn't seem to work since $g(0)$ is not well defined, and hence not equal to $0$.

Appreciate any help. Thanks in advanced.

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Hint for (i). Write $h(z) = \frac{1}{2}(f(z) + f(-z))$ and show that $h(z)$ is of the form $k(z^2)$.

EDIT: For (ii), here is the intuition I have, and then I'll write it more nicely.

You can write $f(z) = e^{i\theta} z^2 + m(z)$, where $m(z)$ is odd. If the function $m(z)$ is not identically zero, there is a sequence $z_n$ with $|z_n| \to 1$ and $|m(z_n)| \geq \alpha$, say. By considering both $f(-z_n)$ and $f(z_n)$, you can obtain a contradiction with the bound $|f(z)| < 1$. (Once you get close enough to the boundary of the circle, you can no longer add two vectors of length $\alpha$ in opposite directions and stay within the disk.)

All right, now that you've read that, here's a version of the same thing that's a bit prettier. We have $$\left|\frac{f(z) - f(-z)}{2}\right|^2 = \frac{1}{2}|f(z)|^2 + \frac{1}{2}|f(-z)|^2 - |h(z)|^2 \leq 1 - |h(z)|^2 = 1 - |z|^4 \xrightarrow[|z| \to 1]{} 0,$$ which shows that $f(z) - f(-z) = 0$ by the maximum modulus principle. (The first equality is the parallelogram law.)

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  • $\begingroup$ This works nicely for me. However, for (ii) I can only deduce that $f(z) + f(-z) = 2e^{i \theta}z^{2}$, which implies that $a_{2k} = 0$ for $k > 1$ and $a_{2} = e^{i \theta}$. What about those $a_{2k+1}$ terms? $\endgroup$ – theflyingwolves Nov 9 '14 at 10:34
  • $\begingroup$ I've edited my answer. $\endgroup$ – user187373 Nov 9 '14 at 11:46
  • $\begingroup$ Thank you soooooo much. That's an elegant solution. $\endgroup$ – theflyingwolves Nov 9 '14 at 13:33
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Let $g(z)=\frac{f(z)+f(-z)}{2}$, then $g'(0)=0.$ Apply Schwarz lemma to $g'(z)$ and apply Schwarz lemma to $g'(z)$, we get $|g'(z)|\le|z|$ and $|g(z)|\le|z|$. So let $|\phi(z)|=|\frac{g(z)}{z}|\le1$ if $z\ne0$ and $\phi(0)=0$. Easy to check $\phi$ is holomorphic. Then Apply Schwarz lemma to $\phi(z)$, we get (i).

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  • $\begingroup$ Could you explain why the Schwarz lemma is applicable to $g'(z)$? $\endgroup$ – user187373 Nov 9 '14 at 8:02

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