Suppose $(M_t)_{t \geq0}$ is a martingale w.r.t. a filtration $(\mathcal{F}_t)_{t \geq0}$. Suppose that $$ E[(M_t-M_s)^2 \mid \mathcal{F}_s] $$ is uniformly bounded by some constant. I want to prove that $$ E[(M_t-M_s)^4 \mid \mathcal{F}_s] \quad \quad (*) $$ is uniformly bounded as well.
It would certainly suffice to show that $$ E[(M_t-M_s)^4 \mid \mathcal{F}_s] \leq E[(M_t-M_s)^2 \mid \mathcal{F}_s]^2 $$ but all inequality-results I can conjure up (Jensens, Cauchy Schwartz, Hölder, etc.) all bound in the "wrong direction". I also suspect this bound is too good to be true. But is $(*)$ bounded regardless? We probably need to use the martingale property somewhere but I can't find any useful results regarding the fourth moment of a martingale.
Any help is appreciated!
Suppose that $M_t=0$ for $t \in [0,1)$ and $M_1$ has density $(3/8) \cdot \min\{1,|x|^{-4}\}$. For $t>1$ let $M_t=M_1$. Then $$ E[(M_t-M_s)^2 \mid \mathcal{F}_s] $$ is uniformly bounded by a constant, but for $0\le s<1 \le t$, $$E[(M_t-M_s)^4 \mid \mathcal{F}_s] =\infty \,.$$