If $fg\in L_1$ for every $f\in L_p$, show that $g\in L_q$

Question

This is from Rudin's book:

Suppose $1\leq p \leq \infty$ and $q$ such that $1/p+1/q=1$. Suppose $\mu$ is a positive $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L_1$ for every $f\in L_p$. Prove that $g\in L_q$

I'm trying to follow the steps in this post Riesz representation theorem, part of the proof (2)., I'll just paste here the steps:

$(X,A, \mu)$ is $\sigma$-finite. For $|g|$, exists an increasing sequence $(\phi_n)$ of simple functions, such that converges to $|g|$. Since $\sigma$-finite, exists an increasing sequence of sets $(E_n)$ with finite measure such that $\bigcup_{n=1}^\infty E_n=X$. Define $g_n=\phi_n\cdot\chi_{E_n}$, we have that $(g_n)$ is an increasing sequence of simple function such that converges to $|g|$, even more \begin{align*} \int |\text{sig}(g)(g_n)^{q-1}|^p\;d\mu & = \int |(g_n)^{q-1}|^p\;d\mu \\ &= \int \chi_{E_n}|(\phi_n)^{q-1}|^p\;d\mu \\ &= \int_{E_n}|(\phi_n)^{q-1}|^p\;d\mu \\ & \leq \mu(E_n)\max(|(g_n)^{q-1}|^p)<\infty, \end{align*} hence $(g_n)^{q-1}\text{sig}(g)\in L_p$ for all $n$, even more \begin{align*} \int (g_n)^q\;d\mu &= \int (g_n)^{q-1}g_n\;d\mu\\ &\leq \int (g_n)^{q-1}|g|\;d\mu\\ &\leq \int (g_n)^{q-1}\text{sig}(g)g\;d\mu\\ \end{align*}

Now I want to proceed with the argument as follows: since $\mbox{sig}(g)(g_n)^{q-1}\in L_p$, by hypothesis it follows that the above integral $\int(g_n)^q d\mu <+\infty $. Now the problem is: I can't ensure that those integrals are "equally bounded", because I can't guarantee that $f\mapsto \int fg d\mu$ is a bounded operator as in the hypothesis of the question linked here. So, it may be the case that $\int (g_n)^q d\mu \rightarrow \infty$ and I can't proceed using the monotone convergence theorem to sate that $g\in L_q$.

How to proceed with this idea? Is there another path? Thank you.

Answer 1

Let us assume that $\mu$ is a finite measure, to simplify exposition. See the bottom of this post for the general, $\sigma$-finite case.

To prove that $f\mapsto \int fg$ is a bounded linear functional on $L^p$, a standard trick is the use of the uniform boundedness principle. Indeed, let $$ g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n, \\ g(x), & \lvert g(x)\rvert <n, \end{cases}$$ (this is known as a truncation of $g$). Define a linear functional on $L^p$ as $$ T_n f:=\int g_n f\, d\mu.$$ This is a bounded functional, because the Hölder inequality gives $$ \lvert T_n f\rvert \le \lVert g_n\rVert_{p'}\lVert f\rVert_p, $$ and $\lVert g_n\rVert_{p'}<\infty$ since $g_n$ is bounded and $\mu$ is finite.

Now, define another functional as $$Tf:=\int gf\, d\mu.$$ So far, we do not know whether $T$ is bounded or not. However, we know that, by dominated convergence, for each $f\in L^p$ we have $T_n f\to Tf$. Moreover, using that $$ \lvert g_n(x)\rvert \le \lvert g(x)\rvert, $$ we can estimate $$ \lvert T_n f\rvert \le \int \lvert gf\rvert\, d\mu <\infty.$$ This last inequality, and the fact that each $T_n$ is a bounded functional, imply by the uniform boundedness principle that there exists $C>0$ such that $$ \lVert T_n\rVert_{(L^p)^\ast}\le C, \quad \text{ for all }n,$$ where, as usual, $\lVert T_n\rVert_{(L^p)^\ast}:=\sup\{ \lvert T_n h\rvert\ |\ h\in L^p, \|h\|_p=1\}$. And since $T_nf\to Tf$, this implies that $\lVert T\rVert_{(L^p)^\ast}\le C$, which is what you wanted to prove.

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The $\sigma$-finite case is essentially the same, with a small additional technical detail. Notice that we used finiteness only to ensure that $g_n\in L^{p'}$, given that $g_n$ is bounded; this is not true if $\mu$ is not a finite measure. To circumvent this, let $\Omega_n$ be a sequence of sets such that $\mu(\Omega_n)<\infty$ and $\bigcup \Omega_n =\Omega$. Redefine $g_n$ so that it reads $$ g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n\ \text{and }x\in \Omega_n, \\ g(x), & \lvert g(x)\rvert <n\ \text{and }x\in \Omega_n,\\ 0,& x\notin \Omega_n. \end{cases}$$ Now, $g_n$ is bounded and supported in a set of finite measure. Thus, $g_n\in L^{p'}$.

The proof goes on from this point exactly as before.