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.

Can we find a function such that $f\in L^1([0,1])$ and for any $0\leq a<b\leq 1$ we have that $||f||_{L^{\infty}([a,b])}=\infty$?

share|improve this question
2  
Welcome to Math.SE! Is this homework? If so, please mark it with the homework tag, so that people know to give hints rather than just giving you the solution and depriving you of the opportunity to learn from solving it. –  Jonathan Christensen Jan 24 '13 at 22:50
2  
Try constructing appropriate nonnegative functions $f_n$ so that for each $n$, the support of $f_n$ is contained in the $n$'th dyadic interval $D_n$ and $f_n\in B(L_1)\setminus L_\infty$. Then consider $f=\sum 2^{-n} f_n$. –  David Mitra Jan 24 '13 at 22:59
    
@DavidMitra: I am slightly confused - do you mean $f_{n}\in L_{\infty}-B(L_{1})$? For example in Davide's answer all $f_{n}$ is in $L^{\infty}$. –  Bombyx mori Jan 24 '13 at 23:13
    
@user32240 I think what I wrote is right. We want each $f_n$ to not be in $L_\infty$. This is the case below, with $f_j(x)={1\over\sqrt{|x-r_j|}}$. –  David Mitra Jan 24 '13 at 23:21
    
@user32240, they are not in $L^\infty$, they are in $L^1$ –  Tomás Jan 24 '13 at 23:21

1 Answer 1

Yes, we can. Consider $\{r_j,j\in\Bbb N\}$ an enumeration of rational numbers of $[0,1]$ and $$f(x):=\sum_{j=1}^{+\infty}\frac{2^{—j}}{\sqrt{|x-r_j|}}.$$

share|improve this answer
    
+1 Very nice Answer... –  Tomás Jan 24 '13 at 23:08
    
And $f^2$ is an example of a function that is finite a.e., but nowhere locally integrable, as was pointed out here. –  Martin Jan 25 '13 at 8:54

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.