# Why is this a martingale?

In our homework assignment, we are supposed to prove:

If $M$ is a countinuous local martingale and if for each $T > 0, E[\sup_{t \leq T } |M_t|] < + \infty$ and $H^T$ is a bounded predictable process, then $H \cdot M$ is a true martingale.

In the hint it says that we should use the previous exercise, which was:

If $M$ is a countinuous local martingale and $H$ a predictable process such that
for each $T > 0$ we have

$E[\sqrt{\int_0^T H_s^2 d \langle M \rangle_s }] < + \infty$,

then $H \cdot M$ is a true martingale.

Well, I started like this:

$| H^T | \leq C \ \Longrightarrow \ E[\sqrt{\int_0^T H_s^2 d \langle M \rangle_s }] \ \leq C E[\sqrt{ \langle M \rangle_T }]$

Moreover I know that $E[\sup_{t \leq T } |M_t|] < + \infty, \forall T > 0,$ implies that $M$ is a true martingale (but I don't know if we need this at all).

Anyway, I don't know how to proceed from here, or if the above was the right way to start at all.

Thanks for all your efforts! Regards, Si

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Try to look at the Burkholder-Davis-Gundy inequality. –  Kolmo Apr 30 '12 at 14:55
@Kolmo: Well, I used the BDG-inequality to prove the previous exercise. But what shall I do with it here? Could you give me a hint? –  Mad Si Apr 30 '12 at 15:17
Well the inequality says that $c E\left[ [M]_T^{\frac{1}{2}}\right ] \leq E[\sup_{t\leq T}|M_t|]$ –  Kolmo Apr 30 '12 at 16:00
@Kolmo: Ah, I see ;-) Thanks –  Mad Si Apr 30 '12 at 16:18