 Strict comparison of one-sided and absolute Brownian maxima

For $t>0$, let $M_t={\sup }_{s\le t}{B}_s$ and $A_t={\sup }_{s\le t}|B_s|$. The Brownian reflection principle gives $M_t\stackrel{d}{=}|B_t|$, hence $\Vert M_t{\Vert }_2=\sqrt{t}$. Also $A_t\ge M_t$, with strict inequality on the positive-probability event $\{B_t<-\lambda ,M_t<\lambda \}$: reflection gives its probability as $\mathbb{P}(B_t>\lambda )-\mathbb{P}(B_t>3\lambda )>0$. The Doob L2 maximal inequality bounds $\Vert A_t{\Vert }_2\le 2\sqrt{t}$, so $\sqrt{t}<\Vert A_t{\Vert }_2\le 2\sqrt{t}$. At $t=0$ all quantities vanish and the lower inequality is equality.

// Target: probability-theory.bigb