$\phi f \in L^p(\mu)$ whenever $f\in L^p(\mu)$, then $\phi \in L^\infty(\mu)$

Question

Taken from Conway's A Course in Functional Analysis Chapter 3 Section 2 Problem 3

Problem statement: If $(X, \Omega, \mu)$ is a $\sigma$-finite measure space, $\phi: X \rightarrow \mathbb{F}$ is an $\Omega$-measureable function, $1\leq p \leq \infty$, and $\phi f \in L^p(\mu)$ whenever $f\in L^p(\mu)$, then show that $\phi \in L^\infty(\mu)$.

Below is my folioing attempt at a solution.

Suppose that $\phi \notin L^\infty (\mu)$. Then for any $N>0$, there exists a subset $A_N\subset \Omega$ of nonzero measure such that $\phi(x) > N$ for $x \in A_N$. We can define an $f \in L^p(\mu)$ so that $f|A_N = $Id. Thus \begin{align} \left(\int_\Omega |\phi f|^p d\mu\right)^{1/p} &\geq \left(\int_{A_n} |\phi f|^p d\mu\right)^{1/p} \\ &= \left(\int_{A_n} |\phi|^p d\mu\right)^{1/p}\\ &> \left(\int_{A_n} N^p d\mu\right)^{1/p}\\ &= \mu(A_n)^{1/p}N \end{align}

I hope to show that the last equation goes to infinity. I can let $N \rightarrow \infty$ but if I do then $\mu(A_n)$ could also tends towards $0$ even though it must be strictly greater than $0$ by assumption.

Answer 1

First, observe that if $p=\infty$, the conclusion of the problem follows immediately by taking $f\equiv1$.

For the remainder, I will assume that $\phi$ is such that $\phi f\in L_p(\mu)$ for all $f\in L_p(\mu)$, where $(X,\mathcal{F},\mu)$ is $\sigma$-finite, and $1\leq p<\infty$.

To complete the argument of the OP it seems to be unavoidable that, under the conditions of the problem, one has to show first the product operator $M_\phi:f\mapsto \phi f$ on $L_p(\mu)$ is bounded. This may be done by appealing to the power of Baire's category theorem, for example the open map theorem, the Banach-Steinhause theorem (a.k.a. uniform boundedness principle), etc. Here -as suggested by @JochenGlueck- I present one approach based on the closed graph theorem (a corollary to the open map theorem). Towards the end, I will present a different solution to the problem which makes no use of Baire's category theorem.

Completion of the OP's argument: Assume first that the multiplication operator $M_\phi$ is bounded. Then, for some positive constant $c$, $\|M_\phi f\|_p\leq c\|f\|_p$ for all $f\in L_p(\mu)$. If $\phi\notin L_\infty(\mu)$, then for each $N\in\mathbb{N}$, the set $A_N=\{|\phi|>N\}$ has positive measure, and $\sigma$-finiteness provides a set $E_N\subset A_N$ with $0<\mu(E_N)<\infty$. Define $$ f_N=\frac{1}{(\mu(E_N))^{1/p}}\mathbb{1}_{E_N}$$ We observe that $f_N\in L_p(\mu)$ and $\|f_N\|_p=1$. However $$c^p=c^p\|f_N\|^p_P\geq\|M_\phi f_N\|^p_P=\int_X|\phi f_N|^p\,d\mu\geq N^p,\quad N\in\mathbb{N}$$ that is, $c>\sqrt[p]{N}$ for all $N\in\mathbb{N}$. This yields a contradiction.

Now we prove that

Lemma: Under the conditions in the problem, $M_\phi$ is bounded.

Proof: We appeal here to the closed graph theorem. Suppose $\{f_n:m\in\mathbb{N}\}\subset L_p(\mu)$ is such that $$(f_n,M_\phi f_n)\xrightarrow{n\rightarrow\infty}(0,y)$$ in $L_p(\mu)\times L_p(\mu)$. Then $f_n\xrightarrow{n\rightarrow\infty}0$ in $L_p(\mu)$, and $\phi f_n\xrightarrow{n\rightarrow\infty}y$ in $L_p(\mu)$. It follows that along a subsequence $n_k$, $f_{n_k}\xrightarrow{k\rightarrow\infty}0$, and $M_\phi f_{n_k}=\phi f_{n_k}\xrightarrow{k\rightarrow\infty}y$ point wise $\mu$-a.s. Consequently, $y=\phi\cdot 0=0$; therefore, $\operatorname{Graph}(M_\phi)$ is closed, and $M_\phi$ is bounded by the closed graph theorem.

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Alternative solution to the problem: For each $n\in\mathbb{N}$, define the sets $$A_n=\{x\in X: n<|\phi(x)|\leq n+1\}$$ If $\phi\notin L_\infty(\mu)$, then there are infinitely many $A_{n}$'s with $\mu(A_n)>0$. We may choose a strictly increasing subsequence $n_k$ such that $\mu(A_{n_k})>0$.

Since $\mu$ is $\sigma$-finite, there are subsets $E_{n_k}\subset A_{n_k}$ such that $0<\mu(E_{n_k})<\infty$. Define $$ f=\sum^\infty_{k=1}\frac{1}{n_k(\mu(E_{n_k}))^{1/p}}\mathbb{1}_{E_{n_k}}$$ where $\mathbb{1}_A$ stands for the indicator function of set $A$. Observe that $f\in L_p$ since $\int_X|f|^p\,d\mu=\sum^\infty_{k=1}\frac{1}{n^p_k}<\infty$. However $$\begin{align} \int_X|\phi f|^p\,d\mu =\sum^\infty_{k=1} \frac{1}{n^p_k \mu(E_{n_k})}\int_{E_{n_k}}|\phi|^p\,d\mu\geq \sum^\infty_{k=1}1=\infty \end{align} $$ This contradicts the assumption that $\phi f\in L_p(\mu)$ for all $f\in L_p(\mu)$.

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Remark 1. $\sigma$-finiteness may be replaced by semi-finiteness ($\mu$ is semi-finite if any $A\in\mathcal{F}$ with $\mu(A)>0$ contains a subset $B\in\mathcal{F}$ with $0<\mu(B)<\infty$).

Remark 2. In the alternative solution, once it has been established that $\phi\in L_\infty(\mu)$), we obtain that the multiplication operator $M_\phi:f\mapsto \phi f$ on $L_p(\mu)$ is bounded. With some extra effort (semi finiteness), one can even show that $$\sup_{\|f\|_p=1}\|M_\phi f\|_p=\|\phi\|_\infty$$

Answer 2

This exercise in Conway's Functional Analysis is exactly a kind of reciprocal of example 2.2, and it can be easily (and elegantly) solved by using the example 2.2 itself and the Uniform Boundedness Principle.

If $(X, \Omega, \mu)$ is a $\sigma$-finite measure space, $\phi: X \rightarrow \mathbb{F}$ is an $\Omega$-measureable function, $1\leq p \leq \infty$, and $\phi f \in L^p(\mu)$ whenever $f\in L^p(\mu)$, then show that $\phi \in L^\infty(\mu)$.

Proof: Since $\phi: X \rightarrow \mathbb{F}$ is an $\Omega$-measureable function, define, for each $n \in \Bbb N$, $\phi_n= \chi_{\{|\phi|\leq n\}}\phi$ and define $T_n: L^{p} (\mu)\to L^{p} (\mu)$ by $T_n(f)=\phi_n f$.

Clearly $\phi_n \in L^\infty(\mu)$, $T_n$ is a bounded linear operator. Since $(X, \Omega, \mu)$ is a $\sigma$-finite measure space, we have $\|T_n\| = \|\phi_n\|_\infty$ (see example 2.2 in chapter 3, section 2 in Conway's Functional Analysis). Moreover,

$$\sup_n\|(T_nf)\|_p = \sup_n\|\phi_n f\|_p \leq \|\phi f\|_p <\infty$$ for each fixed $f$, because $\phi f \in L^p(\mu)$. Hence, by Uniform Boundedness Principle, $\sup_n \|T_n\| <\infty$. It means that $\sup_n \|\phi_n\|_\infty =a <\infty$.

It is easy to see that $\|\phi\|_\infty \leq a<\infty$. In fact, if $\|\phi\|_\infty > a $, then there is $k \in \Bbb N$ such that $k > a$ and $\mu(\{x \in X: a < \phi(x) \leq k\}) >0$. But the $\|\phi_k\|_\infty >a$. Contradiction, because $\sup_n \|\phi_n\|_\infty =a $.

So $\phi \in L^\infty(\mu)$. $\square$

Remark: It is easy to see that example 2.2 in fact remains true if $(X,\Omega,\mu)$ is just a semi-finite measure space. Then the proof above immediately shows the exercise 3 above also remains true if $(X,\Omega,\mu)$ is just a semi-finite measure space.