$f: D\to \mathbb C$ is holomorphic such that $\sup_{u,v\in D}|f(v)-f(u)|=d$, then $|f'(0)|=d/2\implies f$ is linear.
Here, let's suppose that $D$ is a closed unit disk.
By Cauchy's integral formula, since $f(z)$ and $f(-z)$ are both holomorphic, we have for every circle $C_r$ centered at $0$ and with radius $r<1,$ $$2|f'(0)|=\frac 1{2\pi}|\int_{C_r}\frac{f(z)-f(-z)}{z^2}|\le d/r$$
Since $r$ can be brought arbitrarily close to $1$, we have $2|f'(0)|\le d$.
We have the equality in case $f(z)=a+bz$. But I'm not sure how to show the converse i.e., if $2|f'(0)|= d$, then $f$ is linear.