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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?

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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 $X\in L^p$.

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  • $\begingroup$ 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 $X \in L^p$. $\endgroup$ – D Ford Jan 11 at 21:30
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    $\begingroup$ 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 $X-f$. $\endgroup$ – Song Jan 11 at 21:37
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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 $x_n\leqslant cx_n^{1/q}$.

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