Prove that $|f^{'''}|\le 48$ for an analytic function $f$

Question

Let $\Omega$ be an open connected subset of $\Bbb C$ containing $U=\{z\in \Bbb C : |z|\le 1/2\}$ and let, $$\mathscr F:=\{f:\Omega \to \Bbb C: f \text{ is analytic and } \sup_{z,w\in U}|f(z)-f(w)|=1\}$$ Then prove that, $|f^{'''}(0)|\le 48$ for all $f\in \mathscr F$.

We have by Cauchy's integral formula, \begin{align} |f^{'''}(0)|&=\left|\frac{3!}{2\pi i}\int_U \frac{f(z)}{z^4}\,dz\right|\\ &\le \frac{6}{2\pi}\sup_{z\in U}|f(z)|\cdot 2^4\cdot 2\pi\cdot\frac 12\\ &=48 \sup_{z\in U}|f(z)| \end{align}

So, I have to prove $\sup_{z\in U}|f(z)|\le 1$. How to prove this?

Any help, please?

Answer 1

Hint: $f(x)$ and $g(x) := f(x)-f(0)$ have the same (first, second, third,...) derivative, and $\sup_{x\in U}|g(x)|\le1$.

Answer 2

Actually one can show the better estimate $|f'''(0)|\le 24$:

Using the “trick” of substituting $z$ by $-z$ in Cauchy's integral formula (seen here),

$$ f^{'''}(0)=\frac{3!}{2\pi i}\int_C \frac{f(z)}{z^4}\,dz = \frac{3!}{2\pi i}\int_C \frac{-f(-z)}{z^4}\,dz $$ where $C$ is the circle with radius $1/2$. So $$ 2 f'''(0)=\frac{3!}{2\pi i}\int_C \frac{f(z)-f(-z)}{z^4}\,dz $$ and therefore $$ 2 |f'''(0)| \le \frac{6}{2\pi}\sup_{z\in U}|f(z)-f(-z)|\cdot 2^4\cdot 2\pi \frac 12 \le 48 \, . $$

The bound is sharp, equality holds e.g. for the function $f(z) = 4z^3$.