In the paper [http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf](http://www.mat.univie.ac.at/%7Eschachermayer/pubs/preprnts/prpr0154.pdf)


you can find a trajectorial version of Doob's inequality. It is given by:



> 
> $$\bar{s}^2\_T+4\sum\_{k=0}^{T-1}\bar{s\_k}(s\_{k+1}-s\_k)\le 4s^2\_T$$
> 
> 
> $$ \bar{s}\_k=\max (s\_1,...,s\_k) $$
> 
> 
> 


The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.



Trajectorial version of Doob's $L^2$ inequality
I think induction over $T$ will be hard. But you can prove the inequality by considering the times $n$ at which the maximum $\bar s\_n$ increases. For example, let
$
1 = k\_1, \ldots, k\_r \le T
$
be the different times at which $s\_k = \bar s\_k$ attains a maximum (with respect to all previous times). Then it holds
$$
\sum\_{l=k\_j}^{k\_{j+1}-1} \bar s\_l (s\_{l+1} - s\_l) = \sum\_{l=k\_j}^{k\_{j+1}-1} \bar s\_{k\_j} (s\_{l+1} - s\_l) = \bar s\_{k\_j} (\bar s\_{k\_{j+1}} - \bar s\_{k\_j} ).
$$
Therefore we see that
$$
\bar s\_T^2 + 4\sum\_{n=1}^{T-1} \bar s\_n (s\_{n+1} - s\_n) = \bar s\_{k\_r}^2 + 4\sum\_{j=1}^{r-1} \bar s\_{k\_j} (\bar s\_{k\_{j+1}} - \bar s\_{k\_j} ) + 4\bar s\_{k\_r} (s\_T - \bar s\_{k\_r})
$$
Since the $\bar s\_{k\_j}$ are increasing, we estimate the right hand side by
$$
\bar s\_{k\_r}^2 + 2\sum\_{j=1}^{r-1} (\bar s\_{k\_{j+1}} + \bar s\_{k\_j}) (\bar s\_{k\_{j+1}} - \bar s\_{k\_j} ) + 4\bar s\_{k\_r} (s\_T - \bar s\_{k\_r})
$$
which by the third binomial formula yields
$$
\bar s\_{k\_r}^2 + 2\sum\_{j=1}^{r-1} (\bar s\_{k\_{j+1}}^2 - \bar s\_{k\_j}^2) + 4\bar s\_{k\_r} (s\_T - \bar s\_{k\_r}) = \bar s\_{k\_r}^2 + 2(\bar s\_{k\_{r}}^2 - \bar s\_{1}^2) + 4\bar s\_{k\_r} (s\_T - \bar s\_{k\_r}).
$$
Obviously, this is bounded by
$$
-\bar s\_{k\_r}^2 + 4\bar s\_{k\_r} s\_T = -(\bar s\_{k\_r} - 2s\_T)^2 + 4s\_T^2 \le 4s\_T^2,
$$
which finishes the proof. For comparison, also refer to the proof of Lemma 2.2 in the paper you quoted.


