Let $\mu$ be a finite positive measure on $X$ and let $1\leq p<\infty$. Suppose that $g:X\to \Bbb R$ satisfies $fg\in L^p$ whenever $f\in L^p(\mu)$. I want to show that $||g||_\infty =\sup \{||fg||_p:f\in L^p(\mu)$ such that $||f||_p=1\}<\infty$.
The inequality $||g||_\infty \geq \sup \{||fg||_p:f\in L^p(\mu)$ such that $||f||_p=1\}$ is obvious. So it is left to show the converse inequality and that the value is finite, but I have no idea here. Any hints?
Hint: If $\frac 1{\mu (E)} \int_E |g| d\mu\leq M$ for every set $E$ of positive measure then $|g| \leq M$ a.e. so $\|g\|_{\infty} \leq M$.
Now let $M=\sup \{\|fg\|_p: \|f\|_p=1\}$. Take $f=\frac 1 {\mu (E)^{1/p}} \chi_E$. Then $\|f\|_p=1$. By Holder's inequality we have $\frac 1{\mu (E)} \int_E |g| d\mu\leq \frac 1{\mu (E)} \|f(\mu (E)^{1/p}) g\|_p \|I_E\|^{q}$ where $\frac 1 p+\frac 1 q=1$. The desired equality now follows.
To show that $\|g\|_{\infty} <\infty$ consider $f =\sum a_n\chi_{n \leq |g| <n+1}$ where $a_n$ are positive numbers. Let $c_n=\mu \{x: n \leq |g(x)| <n+1\}$. The hypothesis implies that $\sum a_n^{p} n^{p}c_n <\infty$ whenever $\sum a_n^{p} c_n <\infty$. I leave it to you to show that this can happen only when $c_n=0$ for $n$ sufficiently large. Of course this would then show that $g$ is essentially bounded.