Let $\mu $ be a $\sigma $-finite measure. Show that for any $f\in L_p(\mu)$, $\|f\|_1=\sup\{ \int fg\, d\mu :\|g\|_\infty \leq 1\}$

Question

Let $\mu $ be a $\sigma $-finite measure. Show that for any $f\in L_p(\mu)$, $\|f\|_1=\sup\{ \int fg \, d\mu :\|g\|_\infty \leq 1\}$

I know that Holders inequality implies $\int fg \, d\mu \leq \|f\|_1 \|g\|_\infty$ and for all $\|g\|_\infty \leq 1$, $\int fg \, d\mu \leq \|f\|_1$. But how do I show it's the supremum? My intial idea was to show that there exists a $g$ so that it attains the supremum i.e. $fg=|f|$ but I can't seem to find one.

Any hints please?

Answer 1

Let $g=\operatorname{sign}(f)$ then $\|g\|_\infty \leq 1$ and $\int fg d\mu = \int |f| \, d\mu =\|f\|_1$ which implies $\|f\|_1$ is the supremum.