0
$\begingroup$

Let $f$ be analytic on $\Delta$ (unit disc) and satisfy $|f| < 1$. Prove that if $f( \frac{1}{2} ) = f(-\frac{1}{2} ) = 0$, then $|f^{\prime}(0)| \leq \frac{1}{4}$.

  • 0
    So, I tried to find the power series expansion of $f=\sum_k a_kz^k$ around 0 at $1/2$ and $-1/2$. I got $\sum_k a_k 2^{-k}=\sum_k (-1)^k a_k 2^{-k}$. I am fairly sure I can't conclude that $f$ is an even function from this. Is there any way I can show $0$ maps to $0$, because then I think I can show the result. – 2017-02-24
  • 0
    [This problem](https://math.stackexchange.com/questions/81649/first-derivative-bounded-by-supremum-of-difference-of-values-in-disc) is similar in some ways. – 2017-02-24
  • 0
    Hint: 3 applications of the Schwarz Lemma – 2017-02-24
  • 0
    @zokomoko To which disks do you intend to apply Schwarz's lemma? – 2017-02-24

1 Answers 1

1

So, I think I have a solution: We can factor $f(z)=\frac{z-1/2}{1-1/2 z} \cdot \frac{z+1/2}{1+1/2 z} \cdot f_1(z)$ where $f_1(z)$ is holomorphic; (like Blaschke factorization except that $f_1(z)$ may have other zeroes). Using a limiting argument, we can conclude that $|f_1(z)|\leq 1$ on $|z|=1$. Using Cauchy integral estimates we have $|f_1^{\prime}(0)|\leq 1$. Differentiating $f(z)=\frac{z-1/2}{1-1/2 z} \cdot \frac{z+1/2}{1+1/2 z} \cdot f_1(z)$ and plugging our estimates for $z=0$, gives us the desired result.

  • 0
    (+1) This looks like the right way to do it. You could use Schwarz's lemma on $f_1$ to show $\lvert f_1'(0) \rvert <1$, rather than estimating integrals, which looks a bit cleaner. Well done, this was a tough one. – 2017-02-24
  • 1
    Hmm.. but do I know 0$\mapsto 0$? If I do, then I can use Schwarz, else I might have to work extra i think? – 2017-02-24
  • 0
    Ooh, good point... – 2017-02-24