# Expectation of $XY$ bounded for all bounded $Y$ implies $X$ is $L^p$

I'm trying to prove:

Let $$X$$ be a real random variable, $$p, q \in (1,\infty)$$, $$\frac 1 p + \frac 1 q = 1$$. If there is $$C < \infty$$ such that $$|\mathbb E[XY]| \leq C ||Y||_q$$ for any bounded random variable $$Y$$, then $$X$$ is in $$\mathcal L^p$$.

My idea is to use the fact that $$\left(L^q(\mathbb P)\right)' \cong L^p(\mathbb P)$$, and to show $$F : L^q(\mathbb P) \to \mathbb R$$ defined by $$F(Y) = \mathbb E[XY]$$ is continuous. For then, the isomorphism in particular is the isometry $$\kappa(f) = \left( Y \mapsto \mathbb E[fY]\right)$$, so we must have $$f = X$$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $$F$$ is continuous over $$L^q(\mathbb P)$$; only over bounded functions in $$L^q(\mathbb P)$$. Any suggestions?

Hint: This is about how you can apply Riesz representation theorem saying that $$[L^q]^*=L^p$$. Let $$\Omega$$ denote the underlying space. First note that by choosing an appropriate $$\theta(\omega)$$ for each $$\omega$$ (in a measurable way), we can make $$|X(\omega)||Y(\omega)|= X(\omega)Y(\omega)e^{i\theta(\omega)}.$$ By letting $$Y'(\omega)=Y(\omega)e^{i\theta(\omega)}$$, we can improve the inequality to $$E[|X||Y|]\le C\|Y\|_{L^q}$$ for all bounded $$Y$$. Then we can use monotone convergence theorem to conclude $$E[|X||Y|]\le C\|Y\|_{L^q}$$ for all $$Y\in L^q$$. Now deduce the conclusion that $$X\in L^p$$.
• Still working on showing that $X \in L^p$ at this point. Since $F(Y) = \mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f \in L^p$ so that $\mathbb E[fY] = \mathbb E[XY]$ for all $Y \in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X \in L^p$. – D Ford Jan 11 at 21:30
• In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Y\in L^q$, then we can test it for $Y_1=1_{\{X-f>0\}}$ and $Y_2=1_{\{X-f<0\}}$. See what we can say about $X-f$. – Song Jan 11 at 21:37
Hint: for each fixed $$n$$, apply the assumption to the random variable $$Y=Y_n= \operatorname{sgn}\left(X\right)\left\lvert X\right\rvert^{p-1}\mathbf 1\{\left\lvert X\right\rvert\leqslant n\},$$ where $$\operatorname{sgn}\left(X\right)=1$$ if $$X$$ is positive, $$-1$$ if $$X$$ is negative and $$0$$ for $$X=0$$. This will give a bounded on $$\mathbb E\left[\left\lvert X\right\rvert^{p}\mathbf 1\{\left\lvert X\right\rvert\leqslant n\}\right]$$ which does not depend on $$n$$.
Indeed, let $$X_n:=\left\lvert X\right\rvert^{p}\mathbf 1\{\left\lvert X\right\rvert\leqslant n\}$$ and $$x_n:=\mathbb E\left[X_n\right]$$, Then we got that $$x_n\leqslant cx_n^{1/q}$$.