Doob Lp maximal inequality

Put $X_N^{\ast }={\max }_{k\le N}{X}_k$ and fix $r>0$. Layer cake and F1 give $\begin{array}{rl}{\displaystyle \mathbb{E}(X_N^{\ast }\wedge r{)}^p}& {\displaystyle ={\int }_0^{r}p{\lambda }^{p-1}\mathbb{P}(X_N^{\ast }\ge \lambda )d\lambda }\\ {\displaystyle }& {\displaystyle \le {\int }_0^{r}p{\lambda }^{p-2}\mathbb{E}[X_N{1}_{\{X_N^{\ast }\ge \lambda \}}]d\lambda }\\ {\displaystyle }& {\displaystyle =q\mathbb{E}\text{\hspace{0.17em}⁣}[X_N(X_N^{\ast }\wedge r{)}^{p-1}]\mathrm{.}}\end{array}$ The last identity follows by integrating $p{\lambda }^{p-2}$ up to $X_N^{\ast }\wedge r$.

Hölder bounds the last expression by $q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p\mathrm{\parallel }{X}_N^{\ast }\wedge r{\mathrm{\parallel }}_p^{p-1}\mathrm{.}$ If the truncated norm is nonzero, divide by its $(p-1)$st power; if it is zero, the desired inequality is immediate. Thus $\mathrm{\parallel }{X}_N^{\ast }\wedge r{\mathrm{\parallel }}_p\le q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p$.

Let $r\uparrow \mathrm{\infty }$. F4 yields $\mathrm{\parallel }{X}_N^{\ast }{\mathrm{\parallel }}_p\le q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p$, including the case where a priori the maximal moment might be infinite.

If $M$ is a martingale, $\mathrm{\mid }{M}_k\mathrm{\mid }$ is a nonnegative submartingale by F5 and its terminal value is $\mathrm{\mid }{M}_N\mathrm{\mid }$. Applying step 3.1 proves the second display. AC is exactly the inherited dependence in F6.