# $fg\in L^1$ for all $g\in L^q$ $\Longrightarrow f\in L^p$

Let $(X,\mathscr{M},\mu)$ be a $\sigma$-finite measure space and $f$ be a $\mathscr{M}$-measurable function. Let $p,q$ be Hölder conjugates of each other where $1\leq p\leq\infty$. Then is it true that

$fg\in L^1$ for all $g\in L^q$ $\Longrightarrow f\in L^p$ ?

I've managed to prove this for $1\leq p<\infty$. I argued as follows:

First, it is easy to check that $f$ must be finite almost everywhere. Now define $\nu(A)=\int_A |f|^p \,d\mu$ for each $A\in\mathscr{M}$. Then it is also easy to check that $(X,\mathscr{M},\nu)$ is a $\sigma$-finite measure space, and the assertion is that this is actually a finite measure space. Assume $\nu(X)=\infty$ for contradiction. Then there exist a measurable function $h$ such that $$h\in L^p(X,\mathscr{M},\nu)\text{ for all }p>1,\text{ but }h\notin L^1(X,\mathscr{M},\nu).$$ Then the function $g=h|f|^{p-1}$ is in $L^q(X,\mathscr{M},\mu)$ but we have $fg\notin L^1(X,\mathscr{M},\nu)$, a contradiction.

The case $p=\infty$ must be handled separately, but I don't see how I should proceed...

This is an exercise from Jones' Lebesgue Integration on Euclidean Space, and the book does not contain the theory of Banach spaces, dual spaces, Riesz representation theorem, etc. The proof I'm looking for is therefore the one avoiding such advanced theories. Can someone show me how to handle the case $p=\infty$ in a (relatively) elementary manner? Any advice is welcome. Please enlighten me.

• I am interested in the case where X is a closed interval [a,b] and p is equal to 2. I do not understand the part of the proof where the questioner says: "...Then there exist a measurable function ℎ such that ...". How can a function defined over a closed interval be in $L^2$ but not in $L^1$? Commented Mar 12 at 19:00

Suppose $$f$$ is unbounded. We would like to show that there exists nonnegative $$g\in L^1$$ such that $$f\cdot g \notin L^1$$. We may assume $$f$$ is nonnegative and we may replace $$f$$ with a smaller function such that the following is true: There exists a sequence of disjoint sets of positive measure $$\{A_k\}_{k\in\mathbb{N}}$$ such that $$k\leq f(x) whenever $$x\in A_k$$. Now for $$x\in A_k$$ define $$g (x):=\frac{1}{k^2\mu(A_k)}$$ and define $$g(x) := 0$$ for $$x\notin \cup_{k=1}^\infty A_k$$. Then $$g$$ has the desired properties.
• @Dilemian The $L^1$ norm of $fg$ is not affected if we replace f with $|f|$. That's why we can assume it's positive. To answer your second question: since $f$ is unbounded, there exists a sequence $A_k$ of disjoint sets of positive measure st $f(A_k)\subset [a_k, a_k+1)$ where $a_k$ is some strictly increasing sequence of positive integers. You can "fill in"'the integers the sequence omits by chopping off parts of $A_k$ and redefining $f$ to be equal to the omitted integers on those parts you chopped off. ... Commented Jan 10, 2017 at 21:50
• Just make sure the new values are smaller than $a_k$. This gives a smaller function, so the $L^1$ norm of $fg$ is smaller. I.e. If $fg$ is non integrable, so is the old function Commented Jan 10, 2017 at 21:50