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

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$\bar{s}_k=\max (s_k) s_1,...,s_k)$$

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

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# Trajectorial version of Doob's $L^2$ inequality

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

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$\bar{s}_k=\max (s_k)$$

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