Doob maximal bounds for a centered random walk

In the natural filtration, $S_{k-1}$ is known and ${\xi }_k$ is independent of the past with mean zero. Thus F1 gives $\mathbb{E}[S_k\mid {\mathcal{F}}_{k-1}]=S_{k-1}+\mathbb{E}{\xi }_k=S_{k-1},$ so $S$ is a square-integrable martingale.

Expanding the square gives $\mathbb{E}{S}_n^2={\sum }_{k=1}^n\mathbb{E}{\xi }_k^2+2{\sum }_{j<k}\mathbb{E}({\xi }_j{\xi }_k)\mathrm{.}$ For $j<k$, independence and centering give $\mathbb{E}({\xi }_j{\xi }_k)=\mathbb{E}{\xi }_j\mathbb{E}{\xi }_k=0$; hence $\mathbb{E}{S}_n^2={\sum }_k{\sigma }_k^2$.

Apply F3's $L^1$ inequality to the nonnegative submartingale $S_k^2$ at level ${\lambda }^2$. The event is exactly $\{{\max }_{k\le n}\mathrm{\mid }{S}_k\mathrm{\mid }\ge \lambda \}$, so the first bound follows from step 2.1. Apply the $p=2$ inequality to $\mathrm{\mid }{S}_k\mathrm{\mid }$ to get $\mathrm{\parallel }{\max }_{k\le n}\mathrm{\mid }{S}_k\mathrm{\mid }{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{S}_n{\mathrm{\parallel }}_2;$ squaring and using step 2.1 gives the second. AC is used exactly through F4.