2
$\begingroup$

I know that for $p \in [1,\infty]$ if $X$ is $\sigma$-finite (for the $p=\infty$ case) we have $ \Vert f \Vert_p = \sup_{\substack{g \in L^q\\\Vert g \Vert = 1}} \int_X fg d\mu. $ I always see it stated assuming $f \in L^p$, eg on wikipedia, but I would like to apply it in a situation where this is possibly not the case. Does anyone know if it is still true in the case $\Vert f \Vert_p = \infty$?

  • 0
    See [this thread](http://math.stackexchange.com/q/90303) and [this thread](http://math.stackexchange.com/q/61458) for example.2012-11-08

1 Answers 1

2

If the space is $\sigma$-finite, it works. Indeed, let $\{A_n\}$ an increasing sequence of measurable sets of finite measure, and $g_n:=\chi_{A_n}\operatorname{sgn}(f_n)\chi_{\{|f|. It's an increasing sequence, and by Fatou's lemma, $+\infty\leqslant\liminf_{n\to+\infty}\int_X fg_n,$ giving what we want.

When the space is not $\sigma$-finite, we can cheat, taking $X=\{a\}$ with the measure $\mu(\{a\})=+\infty$.