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If $f \in \operatorname{Hol}(D),f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$

$D = \{ z \in \mathbb{C} : |z| < 1 \} $

My thoughts so far: Let's say $f(0) = a$. Define $g = \frac{z-a}{1-\bar{a}z}$ and $h(z) = (g(f(z))$ Now all the conditions for the Shwarz lemma are met, and I can conclude that $|h(\frac{1}{2})| \leq \frac{1}{2}$ and $|h(-\frac{1}{2})| \leq \frac{1}{2}$. The idea would then be to multiply the two inequalities together and try to somehow separate $a$, but the algebra gets really messy and I feel like I'm doing something wrong. Any help would be appreciated!

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1 Answer 1

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$(f(z) + f(-z))/2$ is an even function in $D$. It follows (see below) that there exists a holomorphic function $g$ in $D$ such that $$ g(z^2) = \frac{f(z)+f(-z)}{2} \, . $$ $g$ satisfies $|g(z)| < 1$ in $D$ and $g( \frac 14) = 0$. Then $$ h(z) = g \bigl(\frac{z + \frac 14}{1+ \frac 14 z} \bigr) $$ satisfies $|h(z)| < 1$ in $D$ and $h(0) = 0$.

It follows from Schwarz lemma that $|h(z)| \le |z|$ in $D$ and in particular $$ \frac 14 \ge |h(-\frac 14)| = |g(0)| = |f(0)| \, . $$

The example $$ f(z) = \frac{z^2 - \frac 14}{1 - \frac 14 z^2} $$ with $f(\frac 12) = f (-\frac 12) = 0$ and $f(0) = -\frac 14$ shows that the bound $|f(0)| \le \frac 14$ is best possible.


Existence of $g$: If $F$ is an even holomorphic function in the unit disk $D$ then its power series has only terms with even exponents: $$ F(z) = \sum_{n = 0}^\infty a_{2n} z^{2n} \, . \tag 1 $$ Now define $g$ as $$ g(z) = \sum_{n = 0}^\infty a_{2n} z^n \, . \tag 2 $$ It is easy to see that if $R$ is the radius of convergence of $(1)$ then $R^2$ is the radius of convergence of $(2)$. Therefore $g$ is holomorphic in $D$ and satisfies $g(z^2) = F(z) \, $.

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  • $\begingroup$ Thanks for this answer. Why does g exist? $\endgroup$
    – John
    Mar 5, 2015 at 22:10
  • $\begingroup$ @John: The power series of an even function has only terms with even exponents: $a_0 + a_2 z^2 + a_4 z^4 + ...$. Then you can define $g(z) = a_0 + a_2 z^1 + a_4 z^2 + ...$. (There are probably other methods as well.) $\endgroup$
    – Martin R
    Mar 5, 2015 at 22:14
  • $\begingroup$ Do you mean $g(1/2)=0$ $\endgroup$
    – Elaqqad
    Mar 5, 2015 at 22:57
  • $\begingroup$ @Elaqqad: No. Setting $z=\frac 12$ in the first equation gives $g( \frac 14) = (f( \frac 12) + f(- \frac12))/2 = 0$. Writing the rhs as a function of $z^2$ is the important point here, otherwise you would only get $|f(0)| \le \frac 12$. $\endgroup$
    – Martin R
    Mar 5, 2015 at 22:59
  • $\begingroup$ @Micah: Thanks for fixing my typo! $\endgroup$
    – Martin R
    Mar 5, 2015 at 23:02

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