Does this setup imply that $E[(M_t-M_s)^4 \mid \mathcal{F}_s]$ is bounded? 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 $0\le s<1 \le t$,
$$E[(M_t-M_s)^4 \mid \mathcal{F}_s] =\infty \,.$$
