= Solution
For a nonnegative <submartingale> $(M_k)_{0\leq k\leq n}$ and $M_n^*=\max_{0\leq k\leq n}M_k$, the weak form of the <Doob maximal inequality> is
$$
\boxed{u\,\mathbb P(M_n^*\geq u)\leq\mathbb E[M_n\mathbf1_{\{M_n^*\geq u\}}]\leq\mathbb E M_n\qquad(u>0).}
$$
To prove it, let $A_k$ be the event that $k$ is the first index with $M_k\geq u$. These events are disjoint, $A_k\in\mathcal F_k$, and their union is $A=\{M_n^*\geq u\}$. The <submartingale> property gives $\mathbb E[M_n\mid\mathcal F_k]\geq M_k$. Multiplying by $\mathbf1_{A_k}$, taking expectations and summing yields $\mathbb E[M_n\mathbf1_A]\geq\sum_k\mathbb E[M_k\mathbf1_{A_k}]\geq u\mathbb P(A)$. Nonnegativity gives the second inequality.
The associated <Doob Lp maximal inequality>, for $p>1$ and $M_n\in L^p$, is
$$
\boxed{\|M_n^*\|_p\leq\frac p{p-1}\|M_n\|_p.}
$$
Indeed $M_k\leq\mathbb E[M_n\mid\mathcal F_k]$, so conditional <Jensen inequality> first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $pu^{p-2}\,du$ and using the <Tonelli theorem> gives
$$
\mathbb E(M_n^*)^p\leq\frac p{p-1}\mathbb E[M_n(M_n^*)^{p-1}]
\leq\frac p{p-1}\|M_n\|_p\|M_n^*\|_p^{p-1}.
$$
The last step is <Holder inequality>. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary <martingale>, its absolute value is a nonnegative <submartingale> by conditional <Jensen inequality>, so both corresponding absolute-maximum bounds follow.
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