= Strict comparison of one-sided and absolute Brownian maxima
For $t>0$, let $M_t=\sup_{s\leq t}B_s$ and $A_t=\sup_{s\leq t}|B_s|$. The <Brownian reflection principle> gives $M_t\overset d=|B_t|$, hence $\|M_t\|_2=\sqrt t$. Also $A_t\geq M_t$, 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 $\|A_t\|_2\leq2\sqrt t$, so $\sqrt t<\|A_t\|_2\leq2\sqrt t$. At $t=0$ all quantities vanish and the lower inequality is equality.
// Target: probability-theory.bigb
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