Let f be analytic on ∆

Question

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?

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! — Martin R, Apr 12 '15 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? — Daniel Fischer, Apr 12 '15 at 16:14
Is it really $|f'(0)| \leq \frac{1}{4}$, or is it $|f(0)| \leq \frac{1}{4}$ ? — Martin R, Apr 12 '15 at 16:18
@DanielFischer I dont see how everything connects to $f'(o)$ — hoper, Apr 12 '15 at 18:09
@hoper: Then I was wrong in flagging as a duplicate, and I apologize for not reading your question correctly. – However, I think that one can proceed similarly, see below. — Martin R, Apr 12 '15 at 18:46

score 5 · Accepted Answer · edited Apr 13 '17 at 12:21

(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. — Martin R, Apr 12 '15 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 $|g(z^2)| = |r(z)|/|z| < 1$. — Martin R, Apr 12 '15 at 21:13

Let f be analytic on ∆

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