Doob Lp maximal inequality excludes p equals one

Fix $n$. On $[0,1]$ put $A_k=(0,2^{-k}]$ for $1\le k\le n$, $A_0=[0,1]$, and ${\mathcal{F}}_k=\sigma (A_1,\dots ,A_k)$. For $X=2^n{1}_{A_n}$, direct averaging on $A_k$ gives $M_k:=\mathbb{E}[X\mid {\mathcal{F}}_k]=2^k{1}_{A_k}(0\le k\le n)$ up to the null endpoint. F1 verifies the martingale, and $\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }=2^n{2}^{-n}=1$.

On the shell $A_j\setminus A_{j+1}$ for $0\le j<n$, the maximum is $2^j$ and the shell has measure $2^{-(j+1)}$. On $A_n$ the maximum is $2^n$ and the measure is $2^{-n}$. Therefore $\mathbb{E}{\max }_{k\le n}\mathrm{\mid }{M}_k\mathrm{\mid }={\sum }_{j=0}^{n-1}{2}^j{2}^{-(j+1)}+2^n{2}^{-n}=\frac{n}{2}+1.$

If a horizon-independent $C$ existed, step 1.1, step 2.1 would give $n\mathrm{/}2+1\le C$ for every $n$, impossible. Thus the $p>1$ restriction cannot be extended to this strong $L^1$ form. AC has exactly the inherited role in F3.