One-sided Davis inequality for (F4) filtrations

The classical Davis inequality $\mathbb{E}Mf\simeq \mathbb{E}Sf$, where $(Sf{)}^2={\sum }_k{\mid f_k-f_{k-1}\mid }^2$ is the square function and $Mf={\sup }_n\mid f_n\mid$ is the maximal function, is true with a universal constant for any martingale $f$ on any filtration. A natural analog in the setting of (F4) doubly indexed filtrations, i.e. ${\left({\mathcal{F}}_{i,j}\right)}_{i,j}$ such that the operators $\mathbb{E}(\cdot \mid {\mathcal{F}}_{i,\mathrm{\infty }})$ and $\mathbb{E}(\cdot \mid {\mathcal{F}}_{\mathrm{\infty },j})$ commute and their product is $\mathbb{E}(\cdot \mid {\mathcal{F}}_{i,j})$, is the conjecture $$\mathbb{E}\underset{n,m}{\sup }\mid f_{n,m}\mid \simeq \mathbb{E}{\left(\sum _{i,j}{\mid \mathrm{\Delta }{f}_{i,j}\mid }^2\right)}^{\frac{1}{2}},$$ where $\mathrm{\Delta }{f}_{i,j}=f_{i,j}-f_{i-1,j}-f_{i,j-1}+f_{i-1,j-1}$. It was known to be true only with some highly restrictive additional assumptions, e.g. regularity of the filtration ($g_{n,m}\gtrsim g_{n+1,m},g_{n,m+1}$ for any positive martingale $g$) or $f$ being a strong martingale ($\mathbb{E}(\mathrm{\Delta }{f}_{i,j}\mid {\mathcal{F}}_{i-1,j}\vee {\mathcal{F}}_{i,j-1})=0$). We prove the inequality $\lesssim$ assuming just the (F4) condition.