I stumbled across a claim I couldn't verify. Let $M_t$ be a continuous local martingale, $M_0=0$ a.s. and $\lambda>0$. Then $$ \mathbb{E}\left( \exp \left( \lambda M_t \right) \right) \leq \sqrt{\mathbb{E}\left(\exp\left(2\lambda^2 \left<M_t\right>\right)\right)} .$$
Can anybody point me to a proof of this or a similar result? I looked up some Martingale inequalities like Doobs, Bernstein, etc. but they didn't help me at all. Any advice on how to proof it would be also appreciated.
Write, for $b>0$, $$ \Bbb E[\exp(\lambda M_t)]=\Bbb E[\exp(\lambda M_t-b\langle M\rangle_t)\cdot\exp(b\langle M\rangle_t)]. $$ Use Cauchy-Schwarz $$ \Bbb E[\exp(\lambda M_t)]\le\sqrt{\Bbb E[\exp(2\lambda M_t-2b\langle M\rangle_t)]\cdot\Bbb E[\exp(2b\langle M\rangle_t)]}. $$ Now make a judicious choice of $b$, namely, $b=\lambda^2$. This is a judicious choice because $\exp(2\lambda M_t-2\lambda^2\langle M\rangle_t)$ is an exponential martingale, hence a positive supermartingale (with initial value equal to $1$), and so has mean value less than or equal to $1$.