# 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?

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Is the measure finite? –  Dylan Moreland Sep 2 '11 at 19:22
I recommend checking out this related question for inspiration. –  Jonas Meyer Sep 2 '11 at 19:29
Hint: the closed graph theorem. –  Nate Eldredge Sep 2 '11 at 19:42
I recommend checking out this related question for inspiration... :-) –  Did Sep 2 '11 at 21:23
@Davide: in case you haven't seen robjohn's answer here: for the duality between $L^p$ and $L^q$ with $1 \lt p \lt \infty$ you don't need any hypotheses on the measure space. For the duality between $L^1$ and $L^\infty$ you need something, $\sigma$-finiteness is enough. In fact you can show that the natural map $L^\infty \to (L^1)^\ast$ is an isomorphism if and only if the measure space is "localizable" (this condition allows you to patch together an arbitrary family of measurable functions to a measurable function defined on the entire space). –  t.b. Nov 24 '11 at 16:30
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.
@robjohn: do you mind elaborating on your proof a bit. the part where you showed that $\|fg\|_{L^1} \leq C \|g\|_{L^q}$? Thanks –  Kuku May 7 '12 at 22:08
@kuku: This is an indirect argument. I started by assuming there was no such $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$ for all $g\in L^q$. Using that assumption, I derived a contradiction, which shows that the assumption was false. Thus, there must be a $C$ so that $\|fg\|_{L^1}\le C\|g\|_{L^q}$. –  robjohn May 7 '12 at 22:37
@kuku: Since $\|g_k\|_{L^q}=1$, \begin{align} \|g\|_{L^q} &\le\sum_{k=1}^\infty 2^{-k}\|g_k\|_{L^q}\\ &=1 \end{align} Since $\int fg_k\;\mathrm{d}x>3^k$ (we were assuming all functions were non-negative) \begin{align} \int fg\;\mathrm{d}x &=\sum_{k=1}^\infty2^{-k}\int fg_k\;\mathrm{d}x\\ &>\sum_{k=1}^\infty2^{-k}3^k \end{align} which diverges. –  robjohn May 7 '12 at 23:17