# Prove $2|f'(0)| \leq \sup_{z_1,z_2\in D} |f(z_1)-f(z_2)|$

Let $f:D\to \mathbb{C}$ be a holomorphic function where $D$ is the open unit disk.

Then prove

$$2|f'(0)| \leq \sup_{z_1,z_2\in D} |f(z_1)-f(z_2)|$$

I can show that $$2f'(0) = \frac{1}{2\pi i} \int_{\gamma_R} \frac{f(w)-f(-w)}{w^2} dw$$

where $\gamma_R$ is a circle with radius $R<1$.

Then by using the standard tools I can arrive that $$2|f'(0)| \leq \frac{1}{R} \sup_{\gamma_R} |f(w)-f(-w)|$$

$$\sup_{\gamma_R} |f(w)-f(-w)| \leq \sup_{z_1,z_2\in D} |f(z_1)-f(z_2)|$$ Then I'm stuck, as I can't just let $R$ be 1, and it is not obvious why $\frac{1}{R} \sup_{\gamma_R} |f(w)-f(-w)|$ can't be a constant larger than $\sup_{\gamma_R} |f(w)-f(-w)|$ for any $R$.

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So $$2 |f'(0)| \leq \frac{1}{R}\sup_{\gamma_R} |f(w)-f(-w)| \leq \frac{1}{R}\sup_{z_1, z_2 \in D} |f(z_1)-f(z_2)|.$$ You can let $R \to 1^{-}$, since the right-hand side and the supremum are indipendent of $R$.
Wow. Thanks for the quick answer. I can't believe I go stuck on this part... I tried half an hour to try to find function that bound $(1-1/R)\sup_{\gamma_R} |f(w)-f(-w)|$ and wondering about if there are continuous extension to extend it to the closed disk and invoke the maximum modulus principle. – Chao Xu May 11 '12 at 13:19
@Siminore I just ran into this problem reading Stein's Complex Analysis. It also passingly mentions that this is a strict equality in case f is linear: $$f(z)=a_0+a_1z$$ In that case, I can see why the second inequality from one supremum to the next is actually a strict equality, but I can't seem to be able to show it for $2|f'(0)|$ because the inequality is inherent in passing the absolute under the integral. Should I go back to the Minkowski inequality and see the cases where the absolute value passed into the integral doesn't cause an inequality? Thanks. – Craig Apr 9 at 10:55