True or false: in an open set \Omega of C where the distances between points have supremum d ...

Question

Edit: FALSE. Numerous counterexamples in the comments. Thanks everyone!

... there exists a point $w\in\Omega$ such that $$|z - w| < d / 2$$ for all $z\in\Omega$.

(In other words, we can always encompass an open complex set of diameter d with a disc radius d/2.)

I have a terrible feeling that this question has an obvious answer, but I just can't find it. I tried proving the contrapositive but can't get the inequalities to face the right direction.

In case anyone was interested, the question in -- well, question -- involves proving that $$2|f'(0)| \leq d$$ for any holomorphic function $f:\mathbb{D}\rightarrow\mathbb{C}$, $d$ being the diameter of $f$ on $\mathbb{D}$ AND that equality holds if and only if $f$ is a linear function. The proof seems straightforward enough if I can prove that we can "centre" the image of $f$ in a disc around the origin of radius $d / 2$; then we can apply Schwarz lemma.

Any ideas? Thanks in advance!

Answer 1

I don't think so? If I'm reading your question right, then we just take the union of open sets $|z| < 1$ and $|z-100| < 1$ to be $\Omega$. Then $\sup_{z_1,z_2\in \Omega}\{|z_1 - z_2|\} = 102.$ But any $w$ in $\Omega$ belongs to one and only one disk, so there always exists some $z\in\Omega$ such that $|z-w| > 98 > 51 = \frac{102}{2}$.

Answer 2

Counter-example, with $d=2$: Let $\Omega=\{z: |z|<1\}$ \ $\{0\}.$ Then for all $x,y \in \Omega$ we have $|x-y|<2$ but for $x=re^{it}\in \Omega$ with $0<r<1$ and $t\in \mathbb R$ we have $(1-r/2)e^{i(t+\pi)}\in \Omega,$ and $$|x-(1-r/2)e^{i(t+\pi)}|=1+\frac {r}{2}>1.$$