# $\int_0^1|f_n|^3\leq 1\Rightarrow \int_E |f_n|<\varepsilon$ when $|E|$ is small

Let $f_n\colon [0,1]\to\mathbb{R}$ be Lebesgue measurable with $$\int_0^1|f_n|^3\leq 1 \mbox{ for all } n.$$ Show that for all $\varepsilon>0$ there exists $\delta>0$ so that if $E\subset[0,1]$ is Lebesgue measurable with $|E|<\delta$ then $$\int_E |f_n|<\varepsilon\mbox{ for all } n$$

• Do you know Hölder's inequality? – 23rd Aug 4 '13 at 23:54
• Yes, I do, why? – cmath Aug 5 '13 at 0:08
• Then try to apply it to get an upper bound of your integral in terms of $|E|$ and $L^3$-norm of $f_n$ – 23rd Aug 5 '13 at 0:09
• Its true, thank you very much. How do I have an example that the conclusion is false if to replace $\|f\|_3\leq 1$ to $\|f_n\|_1\leq 1$? – cmath Aug 5 '13 at 0:19
• You are welcome. What about $f_n=n\cdot\chi_{E_n}$, where $|E_n|=\frac{1}{n}$? – 23rd Aug 5 '13 at 0:29

I don't want to give away the whole answer, but use that $\displaystyle \int_E |f_n|=\int_{[0,1]}\chi_E|f_n|$ and try to use the Holder's inequality hint on the right side (what should $q$ be?).
• Thank you very much. Do you know an example that the conclusion is false if to replace $\|f\|_3\leq 1$ to $\|f_n\|_1\leq 1$? – cmath Aug 5 '13 at 0:21
I would complete proof of JLA. Suppose $p=3$ and then Using the equality $\frac{1}{p} + \frac{1}{q} =1$ we have $q= \frac{3}{2}$. We can write $$\int_E |f_n|=\int_{[0,1]}\chi_E|f_n| \leq \left(\int_{[0,1]}{|f_n|^3}\right)^\frac{1}{3} \left(\int_{[0,1]}{\chi_E}^{\frac{3}{2}}\right)^{\frac{2}{3}}$$