Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

$\therefore|f(z)+f(-z)|\leq2|z|^{2}$

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.

share|improve this question

1 Answer 1

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 $g$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.