# Let f be analytic on ∆

The problem is: let $f$ be an analytic function on $\Delta$ and satisfy $|f|<1$. Prove that if $f(1/2)=f(−1/2)=0$, then $|f'(0)|\le 1/4$.

I tried to expand $f$ at $0$ and then plug in $1/2$ and $-1/2$ to evaluate the bound. It is quite straight forward if I use $f^{(n)}(0)/n! < 1$. But I got the bound to be $1/3$. Is there some key I am missing here?

• What is $\Delta$?
– Keba
Apr 12, 2015 at 16:12
• possible duplicate of If $f \in \operatorname{Hol}(D)$, $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$ – Ooops, no, that is different. Sorry! Apr 12, 2015 at 16:13
• Hint: consider the functions $T_a \colon z \mapsto \frac{z-a}{1-\overline{a}z}$ for $\lvert a\rvert < 1$. What do you know about these functions? How can $T_{\pm 1/2}$ help? Apr 12, 2015 at 16:14
• Is it really $|f'(0)| \leq \frac{1}{4}$, or is it $|f(0)| \leq \frac{1}{4}$ ? Apr 12, 2015 at 16:18
• @Keba it is the open unit disk. Apr 12, 2015 at 17:34

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

It is actually sufficient to require that $f(\frac 12)=f(−\frac 12)$ instead of $f(\frac 12)=f(−\frac 12)=0$.

$(f(z) - f(-z))/2$ is an odd function in $\mathbb D$ (the unit disk). It follows that there exists a holomorphic function $g$ in $\mathbb D$ such that $$z \, g(z^2) = \frac{f(z)-f(-z)}{2} \, . \tag 1$$ Taking the derivates gives $$g(z^2) + 2 z^2 g'(z^2) = \frac{f'(z)+f'(-z)}{2}$$ and for $z= 0$ it follows that $$g(0) = f'(0) \, .$$

$f(\frac 12)=f(−\frac 12)$ implies $g( \frac 14) = 0$, and $|g(z)| < 1$ in $\mathbb D$ follows from the Schwarz lemma applied to the right-hand side of $(1)$.

Then $$h(z) = g \bigl(\frac{z + \frac 14}{1+ \frac 14 z} \bigr)$$ satisfies $|h(z)| < 1$ in $\mathbb D$ and $h(0) = 0$.

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

The example $$f(z) = z \, \frac{z^2 - \frac 14}{1 - \frac 14 z^2}$$ 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.

• @S.Panja-1729: By taking the derivatives in (1) and then setting $z = 0$. – I have extended the answer. Apr 12, 2015 at 20:17
• @S.Panja-1729: The RHS of (1) – let's call it $r(z)$ – satisfies $r(0) = 0$ and $|r(z)| < 1$. It follows from the Schwarz Lemma that $|r(z)| \le |z|$. $|r(z)| = |z|$ for some $z$ would imply $r(z) = az$ for some $a$ with $|a|=1$, this contradicts $r(1/2) = 0$. – Therefore $|r(z)| < |z|$ for all $z$ and $|g(z^2)| = |r(z)|/|z| < 1$. Apr 12, 2015 at 21:13