Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Vert f \Vert_p = \infty$?

share|improve this question
    
If $X$ is $\sigma$-finite, you could choose $g_n = \text{sgn} (f) 1_{X_n}$ where $X_n$ is a nested collection of finite measure sets whose union is $X$. Then $\int fg_n = \int_{X_n} |f|$ and Fatou's lemma shows the desired equality. If $X$ is not $\sigma$-finite, I have no clue at present. –  copper.hat Nov 8 '12 at 18:09
    
You mean $g_n = sgn(f) |f|^{p-1} 1_{X_n}$ I think. –  Sanchez Nov 8 '12 at 18:15
    
@copper.hat In case $p\neq \infty$ I meant $\int|f|^p = \infty$ and in case $p=\infty$ I meant $\mathrm{esssup}|f| = \infty$ and in both cases assuming $f$ is measurable. –  nullUser Nov 8 '12 at 18:15
    
@Sanchez: Yes, you are correct! I missed the exponent on all $f$s there... –  copper.hat Nov 8 '12 at 18:17
    
See this thread and this thread for example. –  user48692 Nov 8 '12 at 18:40
add comment

1 Answer

up vote 2 down vote accepted

If the space is $\sigma$-finite, it works. Indeed, let $\{A_n\}$ an increasing sequence of measurable sets of finite measure, and $g_n:=\chi_{A_n}\operatorname{sgn}(f_n)\chi_{\{|f|<n\}}|f|^{p-1}$. 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 $\mu(\{a\})=+\infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.