Doob's martingale inequality. Why does $(2)$ hold true?

Question

Doob's Martingale Inequality Let $M=\left(M_n\right)_{n\ge0}$ be a martingale or a positive submartingale. Set $M^*_n=\sup_{j\le n}|M_j|$. Then $$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|\right\}}{\alpha}\tag{1}$$

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Does $(1)$ imply that for all $p\ge1$: $$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|^{\color{red}{p}}\right\}}{\alpha^{\color{red}{p}}}\tag{2}$$?
If so, does that simply follow from the fact that: $$\mathbb{P}\left(M_n^*\ge \alpha\right)=\mathbb{P}\left((M_n^*)^p\ge (\alpha)^p\right)\le\frac{\mathbb{E}\left\{|M_n|^\color{red}{p}\right\}}{\alpha^\color{red}{p}}\tag{3}$$?

Answer 1

$\phi(x)=|x|^p$ is a positive convex function therefore $( |M_n|^p,n \ge 0)$ is also a submartingale (as long as it is in $L^p$). Hence, your implication is true.