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 $\Vert\cdot\Vert_\infty$)

  • $\begingroup$ I don't understand how that proves the result $\endgroup$ – user194469 Apr 27 '16 at 22:16
  • $\begingroup$ Can you provide more detail, please? $\endgroup$ – user194469 Apr 27 '16 at 22:16
  • $\begingroup$ @par : this is obviously the perfect solution but it needs adjustment if $g$ is required to be in $L_1$ $\endgroup$ – pmichel31415 Apr 27 '16 at 22:18
  • $\begingroup$ @Mandrathax: was $g$ required to be in $L_1$? I interpreted it as $g\in L_\infty$. $\endgroup$ – parsiad Apr 27 '16 at 22:20
  • $\begingroup$ Well it would have made sense $\endgroup$ – pmichel31415 Apr 27 '16 at 22:27

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