Showing $2 |f'(0) | \leq d$, where $f$ is some holomorphic function and $d$ is the diameter of $f$'s image?

Question

Suppose $f: \mathbb{D} \rightarrow \mathbb{C}$ is holomorphic. Show that the diameter

$$d = \sup_{z, w \in \mathbb{D}} | f(z) - f(w) |$$

of the image of $f$ satisfies:

$$2 | f'(0) | \leq d$$

Here's my attempt at the start of a proof:

Since $f$ is holomorphic in $\mathbb{D}$, we can take an open set $\Omega$ slightly larger than $\mathbb{D}$ and apply Cauchy Integral Theorem, defining $f: \Omega \rightarrow \mathbb{C}$.

Then,

$$f^{(n)}(z) = \frac{n!}{2\pi i}\int_C \frac{f(\xi)}{(\xi - z)^{n+1}}d\xi$$ for all $z$ in the interior of $C$ for $C \subset \Omega$, where $C$ is a circle whose interior is contained in $\Omega$.

We let $n = 1$ and $z = 0$, and the above equation becomes:

$$2f'(0) = \frac{1}{\pi i}\int_C \frac{f(\xi)}{\xi^2}d\xi$$

How do I get from here to:

$$2 | f'(0) | \leq sup_{z, w \in \mathbb{D}}| f(z) - f(w) | ?$$

I'm thinking that I could let the contour $C$ be some disc with radius $r$, setting the integral $\frac{1}{\pi i}\int_C \frac{f(\xi)}{\xi^2}d\xi$ to be $\frac{1}{\pi i}\int_{| z | = r} \frac{f(\xi)}{\xi^2}d\xi$, but not quite sure how to proceed...

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I'm aware that this link contains a very similar problem, but it states that if $2 | f'(0) | \leq d$, then $f$ is linear, which is different from what I'm trying to prove.

Answer 1

The question is from many years ago, but I think it can be useful to post an answer. You only have to define a function $g(z)=f(z)-f(-z)$. Observe that you have $g'(0)=2f'(0)$. From the Cauchy Integral Theorem, and the fact that if f is holomorphic in $\mathbb{D}$, so it is $g$ and in an open neighbourghood of the unit disc, we have that for all $z\in\mathbb{D}^{\circ}$: $$g'(z) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}} \frac{g(\xi)}{(\xi - z)^2}d\xi \longrightarrow 2f'(0) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}} \frac{f(\xi)-f(-\xi)}{\xi^2}d\xi$$ We now need to observe two things. The boundary of the unit circle has lenght $2\pi$, and the points in there have modulo 1. Moreover, we know from the properties of the integrals in $\mathbb{C}$ that, for a path $\Gamma$ (which is class $C^1$): $$\int_{\Gamma}f(z) dz \leq \text{lenght($\Gamma$)} \sup_{z\in{\Gamma}}|f(z)|$$ Thus: $$2|{f'(0)}| = \frac{1}{2\pi}|\int_{\partial \mathbb{D}} \frac{f(\xi)-f(-\xi)}{\xi^2}d\xi| \leq \frac{1}{2\pi} \text{lenght($\partial \mathbb{D}$)} \sup_{\xi\in\partial\mathbb{D}}|\frac{f(\xi)-f(-\xi)}{\xi^2}|\leq \sup_{\xi\in\partial\mathbb{D}}|\frac{f(\xi)-f(-\xi)}{\xi^2}| \leq \sup_{z,w\in\partial \mathbb{D}}|f(z)-f(w)|$$ which is the inequality we were looking for.