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Show that $\mathbb{E}[\sup_{0 \leq s \leq t} M^4_s] = \frac43 \mathbb{E}[M_t \sup_{0 \leq s \leq t} M^3_s] $, where $M = (M_t)_{t \geq 0}$ is a non-negative, continuous martingale starting from $0$.

I have already shown that :

$a \mathbb{P}(\sup_{s \in [0, t]} M_s \geq a) = \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ] \,\,\, \forall (a, t) \in \mathbb{R}^2_{+} - \{0,0\}$

It's supposed to help me prove the equality in the title.

So I integrated both sides :

\begin{align} & \frac12\int_{0}^{\infty} 2a \mathbb{P}(\sup_{s \in [0, t]} M_s \geq a) da =\int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ] da \\ \iff & \frac12 \mathbb{E}[(\sup_{s \in [0, t]} M_s)^2] = \int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ] da \\ \iff & \frac12 \mathbb{E}[\sup_{s \in [0, t]} M_s^2] = \int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ] da \\ \end{align}

Since the property of being a martingale is weakened down to being just a sub-martingale when squaring the martingale then we can't replace $M$ by $M^2$ and thus I don't how else to make the to the fourth power term appear, I also have no idea how to deal with the RHS.

Any help will be greatly appreciated, thanks !