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Let $D={z\in\mathbb{C}:|z|<1}$ and $f:D\to D$ be analytic with f(0)=0

(i) Show that $|f(z)+f(-z)|\leq2|z|^2$

(ii) Suppose that $|f(z_0)+f(-z_0)=2|z_0|^2$ for some $z_0\in\mathbb{C}\setminus{0}$. Show that $f(z)=\alpha z^2$ for all $z\in \text D$ where $\alpha$ is a constant of modulus 1

Thoughts thus far:

(i) We know that by the Schwarz lemma $|f(z)|\leq|z|$ and $|f(-z)|\leq|-z|=|z|$, but what we want is $|f(z)+f(-z)|\leq|f(z)|+|f(-z)|\leq2|z|^2$, so we want $|f(-z)|=|f(z)|\leq|z|^2$ which I am not sure how to do since $|z|\geq|z|^2$ within the unit circle.

(ii) Proposed solution: $|f(z)+f(-z)|\leq|f(z)|+|f(-z)|\leq2|z|^2$,so we know that $|f(z)+f(-z)|=|f(z)|+|f(-z)|=2|z|^2$ and thus by inspection that $|f(z)=|f(-z)|=|z|^2$, which means that $f(z)=\alpha z^2$ for some $\alpha\in\mathbb{C}$ such that $|\alpha |=1$

Edit: Proposed solution using power series that a friend recently suggested. This has the advantage of making the inspection in part (ii) more rigorous.

Part (i)

Consider the power series of $f(z)=\sum\limits _{n=0}^{\infty}a_{n}z^{n}$ and $f(-z)=\sum\limits _{n=0}^{\infty}(-1)^{n+1}a_{n}z^{n}$. Then, we know that

$|\sum\limits _{n=0}^{\infty}a_{n}z^{n}+\sum\limits _{n=0}^{\infty}(-1)^{n+1}a_{n}z^{n}|=|\sum\limits _{n=0}^{\infty}a_{n}z^{n}+\sum\limits _{n=0}^{\infty}(-1)^{n+1}a_{n}z^{n}|=|[\sum\limits _{k=0}^{\infty}a_{2k}z^{2k}+\sum\limits _{k=0}^{\infty}a_{2k+1}z^{2k+1}]+[\sum\limits _{k=0}^{\infty}a_{2k}z^{2k}-\sum\limits _{k=0}^{\infty}a_{2k+1}z^{2k+1}]|=|2\sum\limits _{k=0}^{\infty}a_{2k}z^{2k}|\leq2|z|^{2}$


Part (ii)

If $|f(z_{0})+f(-z_{0})|=2|z_{0}|^{2}$ for some $z_{0}\in D\backslash\{0\}$, then we know that $|f(z_{0})+f(-z_{0})|\leq|f(z_{0})|+|f(-z_{0})|\leq2|z_{0}|^{2}\implies|f(z_{0})+f(-z_{0})|=|f(z_{0})|+|f(-z_{0})|=2|z_{0}|^{2}$, which means that $2|\sum\limits _{k=0}^{\infty}a_{2k}z_{0}^{2k}|=|\sum\limits _{k=0}^{\infty}a_{2k}z_{0}^{2k}+\sum\limits _{k=0}^{\infty}a_{2k+1}z_{0}^{2k+1}|+|\sum\limits _{k=0}^{\infty}a_{2k}z_{0}^{2k}-\sum\limits _{k=0}^{\infty}a_{2k+1}z_{0}^{2k+1}|=2|z_{0}|^{2}$, which by inspection implies the $a_{2k+1}$ terms reduce to zero. Thus, $f(z_{0})=f(-z_{0})=\sum\limits _{k=0}^{\infty}a_{2k}z_{0}^{2k}$, which means that that from part a we know that $|f(z_{0})+f(-z_{0})|=|f(z_{0})|+|f(-z_{0})|=2|f(z_{0})|=|2\sum\limits _{k=0}^{\infty}a_{2k}z_{0}^{2k}|=2|z_{0}|^{2}$, which implies that the power series terminates after the first second term where the first constant term is equal to zero. This means that we have $2|\alpha z_{0}^{2}|=2|z_{0}|^{2}\implies f(z_{0})=|\alpha||z_{0}|^{2}=|z_{0}|^{2}$ and hence $|\alpha|=1$. Since f is analytic and we know that $z_{0}\neq0$, we also know that the above is true for all $z\in D$ and thus $f(z)=\alpha z^{2}$ where $\alpha$ is a constant of modulus 1.

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up vote 3 down vote accepted

$|f(z)|\le |z|^2$ is incorrect. For example, it fails for $f(z)=z$.

You can consider the function $g(z)=\frac{f(z)+f(-z)}{2z}$ instead and show that Schwarz Lemma is applicable to $g$.

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Thank you for your response. That seems to work quite nicely. – ABC Bach Oct 28 '12 at 5:50

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