Hölder's inequality with $p=1,q=\infty$ and SDEs

Question

I have an expression of an expectation of stochastic processes at time $T$, $$\mathbb{E}[X_TM_T]$$ $M_T$ is a martingale such that $\mathbb{E}[M_T] = \mathbb{E}[M_0] = M_0 = 1 $. I am tempted to apply Hölder's inequality to write $$\mathbb{E}[X_TM_T]\le \mathbb{E}[|X_T|]E[|M_T|^\infty]^{\frac{1}{\infty}}$$

My question is: is it possible to evaluate $E[|M_T|^\infty]^{\frac{1}{\infty}}$? As far as I know this is the infinity norm, which is usually defined as the max value of the elements, but I rarely see it with this expected value notation. What I hope is that we can use the martingale property of $M_t$ but after thinking more it seems maybe the martingale property doesn't matter for norms besides the 1-norm.

Answer 1

A remark, your Holder's inequality should be $$\mathbb{E}[X_TM_T]\le \mathbb{E}[|X_T|]E[|M_T|^\infty]^{\frac{1}{\infty}}$$ ($\mathbb{E}[|X_T|]$, not $\mathbb{E}[X_T]$)

About your question, it's all about the convention. What is the definition of $E[|M_T|^\infty]^{\frac{1}{\infty}}$? According to the Hölder's inequality convention, this is the supremum of $M_T$. Hence, by definition $$E[|M_T|^\infty]^{\frac{1}{\infty}} := \sup |M_T|$$

Besides, without Holder's inequality, we have $$E(X_TM_T) \le E(|X_T| \sup |M_T|) = E(|X_T|)\sup |M_T| $$

Answer 2

If you use Holder's with $p=1,q=\infty$ you are assuming that $X_T \in \mathcal{L}^1(\mathbb{P})$ and $M_T \in \mathcal{L}^\infty(\mathbb{P})$. Then you get that $\mathbb{E}[X_TM_T]< \infty$ and $$|\mathbb{E}[X_TM_T]|\leq\mathbb{E}[|X_TM_T|]=\int|xm|d\mathbb{P}\leq\underbrace{\bigg(\int|x|d\mathbb{P}\bigg)}_{\|X_T\|_1=\mathbb{E}[|X_T|]}\|M_T\|_{\infty}=\mathbb{E}[|X_T|]C, \ \ \ C > 0$$ Mind the infinity seminorm is not really $(\int|m|^\infty\mathbb{P})^{\frac{1}{\infty}}$, if you say $M_T \in \mathcal{L}^\infty(\mathbb{P})$ it just means that $M_T$ is a.e. bounded by some $c$ and $\|M_T\|_{\infty}$ is the greatest lower bound of the set of possible $c$'s.