If $Y\_n=\sum\_{i=1}^n X\_i$ is a martingale, where $X\_i$ is a martingale difference sequence, $\mathbb{E}[X\_n\mid \mathcal{F}\_{n-1}]=0$ for all $n$, we know that
$$ \mathbb{E}\big[Y\_n^2-Y\_{n-1}^2\big]=\mathbb{E}X\_n^2.$$
A similar property, but now as an inequality, holds if we replace the square with the absolute value,
$$ \mathbb{E}\big[|Y\_n|-|Y\_{n-1}|\big]\le\mathbb{E}|X\_n|.$$
Does something analogous hold for other powers? Namely, something along the lines of
$$ \mathbb{E}\big[|Y\_n|^r-|Y\_{n-1}|^r\big]\le C\mathbb{E}|X\_n|^r,$$
for $1<r<2$ and some $C>0$?


I guess if the distribution of $X\_n$ given $Y\_{n-1}$ were symmetric about zero, then this would be a direct consequence of von Bahr-Essen bounds (*Inequalities for the rth Absolute Moment of a Sum of Random Variables*, $1 \le r\le 2$, The Annals of Mathematical Statistics, 36), with $C=1$. Does it also hold under weaker assumptions? The von Bahr-Essen paper states the first equation (equality in case of $r=2$) as a special case.



Inequality for increments of $r$th absolute moments of martingales, $1<r<2$
Yes, this inequality, with the best possible $C$ ($\le 2$), was proved in [this paper](https://projecteuclid.org/journals/annals-of-functional-analysis/volume-6/issue-4/Best-possible-bounds-of-the-von-Bahr--Esseen-type/10.15352/afa/06-4-1.full); see e.g. inequality (1.11) there.


Indeed, that inequality implies that
$$E\Big|\sum\_{i=1}^m U\_i\Big|^r\le E|U\_1|^r+C\_r\sum\_{i=2}^m E|U\_i|^r,$$
where the $U\_i$'s are martingale differences and $C\_r\in[1,2]$. Taking here $m=2$, $U\_1:=Y\_{n-1}$, and $U\_2:=X\_n$, we get
$$E|Y\_n|^r\le E|Y\_{n-1}|^r+C\_rE|X\_n|^r,$$
as desired.


