$B=\{b(s): 0\leq s \leq 1 \}$ (Brownian motion in $\mathbb{R}^2$).
$ \displaystyle r\left(\theta \right)=\underset{0\le s\le 1}{\mathrm{sup}}({b}_{s}.{e}_{\theta})-\underset{0\le s\le 1}{\mathrm{inf}}({b}_{s}. {e}_{\theta})$ where $ {e}_{\theta}=(\mathrm{cos}\theta ,\mathrm{sin}\theta ) $
For any $a,h>0$, $ \mathbb{E}\lvert X_1-X_2\rvert \ge 2h(\mathbb{P}(X\le a )).\mathbb{P}(X\ge a+h) $, where $X_1$ and $X_2$ are independent copies of $X:=r(0)$.
Proof. We have
$ \mathbb{E}\lvert X_1-X_2\rvert\ge \mathbb{E}\left[\lvert X_1-X_2\rvert 1\left\{X_1\le a,X_2\ge a+h\right\}\right]+ \mathbb{E}\left[\lvert X_1-X_2\rvert 1\left\{X_2\le a,X_1\ge a+h\right\}\right]$ $ \mathbb{E}\lvert X_1-X_2\rvert\ge h(\mathbb{P}(X_1\le a )).\mathbb{P}(X_2\ge a+h) +h(\mathbb{P}(X_2\le a )).\mathbb{P}(X_1\ge a+h) ?$
I would like to know how you get the last inequality (Thanks for your support).
We use the reversed triangular inequality: if $ x_1 \leqslant a$ and $ x_2 \geqslant a+h$, then $$ \left\lvert x_1-x_2\right\vert\geqslant \left\lvert \left\lvert x_1 \right\vert -\left\lvert x_2\right\vert \right\vert \geqslant \left\lvert x_2\right\vert-\left\lvert x_1\right\vert\geqslant a+h-a=h $$ Therefore, the following inequality holds almost surely; $$ \left\lvert X_1-X_2\right\vert\mathbf 1\left\{ X_1 \leqslant a, X_2 \gt a+h\right\}\geqslant h\cdot \mathbf 1\left\{ X_1 \leqslant a, X_2 \gt a+h\right\}. $$ Then integrate.