If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?

Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren’t given that such a $C$ exists?


So that $L^p$ and $L^q$ are duals of each other, let $1\lt p,q\lt\infty$, as well as $\frac1p+\frac1q=1$.

Furthermore, so that we can apply the Riesz Representation Theorem, we should assume that we are working in a measure space where the measure is countably additive on a regular, locally compact Hausdorf space.

Without these conditions, the counterexample given by Danny Pak-Keung Chan and David C. Ullrich in Danny’s answer shows that the answer is no.

Suppose that $fg\in L^1$, but there is no $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Without loss of generality, we can assume all functions are positive. Suppose we have a sequence of $L^q$ functions $\{g_k:\|g_k\|_{L^q}=1\}$ where $\int|fg_k|\;\mathrm{d}x>3^k$. Set $g=\sum\limits_{k=1}^\infty2^{-k}g_k$. $\|g\|_{L^q}\le1$ yet $fg\not\in L^1$. Thus, there must be a $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. Then, as you say, apply the Riesz Representation Theorem.

Source : Link , Question Author : user15464 , Answer Author : robjohn

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