How to show following function is in $L^{\infty}(\mu)$

Question

If (X, $\Omega, \mu$) is a $\sigma$-finite measure space, $\phi:X \rightarrow \mathbb{C}$ is an $\Omega$-measurable function, $1 \leq p \leq \infty$, and $\phi f \in L^{p}(\mu)$ whenever $f \in L^{p}(\mu)$, then show that $\phi \in L^{\infty}(\mu)$. I am trying this by using closed graph theorem. Any help.

Answer 1

By contradiction, suppose that $\phi$ is not essentially bounded.

Define the sets $$B_n=\{x\in X~:~|\phi(x)|\in[n,\,n+1)\}\in \Omega.$$ There exists a sequence $\{n_i\}_{i=1}^\infty$ s.t. $\mu(B_{n_i})>0$ for any $i$. Indeed, if such sequence does not exists then for some integer $N>0$ we have $\mu(B_n)=0$ for all $n\geq N$, i.e. $|\phi|\leq N$ $\mu$-a.e.

Define $f\in L^p(\mu)$, $1\leq p <\infty$, as stepwise function: $$ f(x) = \sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{1+1/p}}\frac{1}{\left(\mu(B_{n_i})\right)^{1/p}}\mathbb 1_{\{x\in B_{n_i}\}} $$

Check that $f\in L^p(\mu)$, $1\leq p <\infty$. $$ \int\limits_X |f|^p \, d\mu = \int\limits_X \biggl(\sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{1+1/p}\left(\mu(B_{n_i})\right)^{1/p}}\mathbb 1_{\{x\in B_{n_i}\}}\biggr)^p \, d\mu =\sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{p+1}\mu(B_{n_i})} \int\limits_X\mathbb 1_{\{x\in B_{n_i}\}} \, d\mu =\sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{p+1}}<\infty $$ Look at $\int_X |\phi f|^p \, d\mu $: $$ \int\limits_X |\phi f|^p \, d\mu = \sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{p+1}\mu(B_{n_i})} \int\limits_X|\phi(x)|^p\mathbb 1_{\{x\in B_{n_i}\}} \, d\mu \geq \sum_{i=1}^\infty \frac{1}{{i\vphantom{)}}^{p+1}\mu(B_{n_i})} \int\limits_X n_i^p\mathbb 1_{\{x\in B_{n_i}\}} \, d\mu = \sum_{i=1}^\infty \frac{n_i^p}{{i\vphantom{)}}^{p+1}} \geq \sum_{i=1}^\infty \frac{i^p}{{i\vphantom{)}}^{p+1}}=\sum_{i=1}^\infty \frac{1}{i}=\infty. $$

For $p=\infty$ simply take $f\equiv 1\in L^\infty(\mu)$. By a condition, $\phi f=\phi\in L^\infty(\mu)$.