we consider, on a probability space $(\Omega,\mathcal{A},P)$, two random variable $X$ and $Y$ and let $\mathcal{H} \subset \mathcal{A}$ be a $\sigma$-algebra. Let $p,q>1$ such that $\frac{1}{p}+\frac{1}{q}=1.$ Prove :$$E[|XY||\mathcal{H}] \leq (E[|X|^p|\mathcal{H}])^{1/p}(E[|Y|^q|\mathcal{H}])^{1/q}$$
I tried to use the Holder inequality for integral, I mean if we have :
$$\forall B \in \mathcal{H},\int_B|XY|dP \leq \int_B(E[|X|^p|\mathcal{H}])^{1/p}(E[|Y|^q|\mathcal{H}])^{1/q}dP$$
then the problem is solved.
So I am stuck in proving the integral inequality.
Best Answer
OH! I came up a similar but much shorter proof:
Let $\epsilon\geq 0$, define $U_{\epsilon}:=\Big[\mathbb{E}(|X|^{p}|\mathcal{F})+\epsilon\Big]^{1/p}$ in $L_{p}(\Omega,\mathcal{F},\mathbb{P})$ and $V_{\epsilon}:=\Big[\mathbb{E}(|Y|^{q}|\mathcal{F})+\epsilon\Big]^{1/q}$ in $L_{q}(\Omega,\mathcal{F},\mathbb{P})$. Then, for each $\epsilon>0$, both $U_{\epsilon}$ and $V_{\epsilon}$ are uniformly bounded by $0$ from below.
Recall from the proof of regular Holder that we have, for all $x,y\geq 0$, that $$\dfrac{x^{p}}{p}+\dfrac{y^{q}}{q}-xy\geq 0,$$ for which we consider the first two derivatives in $x$ of the function on the LHS above.
This implies that for all $\omega$ and $\epsilon>0$, we have $$\Big|\dfrac{X(\omega)Y(\omega)}{U_{\epsilon}(\omega)V_{\epsilon}(\omega)}\Big|\leq\dfrac{1}{p}\Big|\dfrac{X(\omega)}{U_{\epsilon}(\omega)}\Big|^{p}+\dfrac{1}{q}\Big|\dfrac{Y(\omega)}{V_{\epsilon}(\omega)}\Big|^{q}.$$
Since both $1/U_{\epsilon}$ and $1/V_{\epsilon}$ are uniformly bounded, the expectation of both sides conditional upon $\mathcal{F}$ is well defined, and then it follows from the monotonicity that for almost every $\omega$, we have $$\dfrac{\mathbb{E}(|XY||\mathcal{F})}{U_{\epsilon}V_{\epsilon}}\leq\dfrac{1}{p}\dfrac{U_{0}^{p}}{U_{\epsilon}^{q}}+\dfrac{1}{q}\dfrac{V_{0}^{q}}{V_{\epsilon}^{q}}\leq\dfrac{1}{p}+\dfrac{1}{q}=1.$$
To finish, multiply both side by $U_{\epsilon}V_{\epsilon}$, and take $\epsilon\searrow 0$, then we have the desired inequality $$\mathbb{E}(|XY||\mathcal{F})\leq U_{0}V_{0}.$$