# Equality for $\lvert\lvert f\rvert\rvert_1$

I am trying to solve the following exercise: Show that for $f\in L_1(\mu)$,

$$\lvert\lvert f\rvert\rvert_1=\sup\Bigg\{\int fg d\mu : \lvert\lvert g\rvert\rvert_\infty\leq 1\Bigg\}$$

I know that as $\lvert\lvert g\rvert\rvert_\infty=\sup\lvert g\rvert$, we have the inequality

$$\int fg d\mu\leq\lvert\lvert g\rvert\rvert_\infty\int f d\mu\leq\int\lvert f\rvert d\mu=\lvert\lvert f\rvert\rvert_1$$

How to prove now that the $1-$norm is in fact the supremum of such integrals?

Thank you

Take $g=1_{\{f\geq0\}}-1_{\{f<0\}}$ to get $\int fgd\mu=\int|f|d\mu=\Vert f\Vert_1$. Note that $g$ is measurable if $f$ is.
(Also you might want to change $\sup$ to $\operatorname{esssup}$ in your use of $\Vert\cdot\Vert_\infty$)
• @par : this is obviously the perfect solution but it needs adjustment if $g$ is required to be in $L_1$ – pmichel31415 Apr 27 '16 at 22:18
• @Mandrathax: was $g$ required to be in $L_1$? I interpreted it as $g\in L_\infty$. – parsiad Apr 27 '16 at 22:20