6

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 $||f||_{L^{\infty}([a,b])}=\infty$?

Davide Giraudo
  • 172,925
Vlad Matei
  • 1,403
  • 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
  • @DavidMitra: I think the essential superum for $f_{n}$ is $\infty$ right? They are in $L^{1}$ but not in $L^{\infty}$..I see. – Bombyx mori Jan 24 '13 at 23:39

1 Answers1

5

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

Davide Giraudo
  • 172,925
  • 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 08:54