# 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?

#### Solutions Collecting From Web of "If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$?"

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.

I would like to add another answer to this question. Consider the linear functional $T:L^q\to\mathbb{C}$ defined by $$Tg=\int gf.$$

It is sufficient to prove that $T$ is bounded. To this end, assume that $g_n\in L^q$ is such that $\|g_n\|_q\to 0$.

We can extract a subsequence of $g_n$ (not relabeled) such that $$g_n(x)\to 0,\ |g_n(x)|\le h(x),\ \mbox{a.e.},\tag{1}$$

where $h\in L^q$ (this partial converse of Lebesgue theorem, can be found, for example, in Rudin’s book “Real and Complex Analysis”, Theorem 3.12, or in Brezis book “Functional Analysis”, Theorem 4.9). It follows from $(1)$ that $$g_n(x)f(x)\to 0,\ |g_n(x) f(x)|\leq h(x)|f(x)|,\ \mbox{a.e}.\tag{2}$$ Note that by hypothesis, $h|f|\in L^1$, therefore, we can apply Lebesgue theorem to conclude that $$Tg_n\to 0.\tag{3}$$

As every subsequence of $g_n$, has a subsequence which satisfies $(1)$, we conclude that $(3)$ is true for the whole sequence.