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Let $M>0$, $\{f_n\}\subset L^2([0,1])$ such that $\int_0^1 |f_n|^2 dm\leq M$ and $f_n(x)\to 0$ as $n\to\infty$ almost everywhere, $m$ is Lebesgue measure. Show that for all $0<p<2$, $$\lim_{n\to\infty}\int_0^1|f_n|^pdm =0.$$

Please help me solving this problem, I have a prelim in analysis soon. Any ideas will be appreciated.

Things I believe we should use:

Since $m([0,1])<\infty$, $f_n$ are measurable (since $f_n$ in $L^2$) and $f_n\to 0$ (point-wisely) a.e., we can apply Egorov's Theorem: For $\epsilon>0$, there exists a measurable set $A$ with $m(A)<\epsilon$ and $f_n\to 0$ uniformly on $[0,1]\setminus A$. Now for all $x \in [0,1]\setminus A$, there exists an $N$ natural number such that for all $n\geq N$ we have $|f_n-0|<(\epsilon/2)^{1/p}$. Then, for $n\geq N$: $$\int_0^1|f_n|^pdm =\int_{[0,1]\setminus A}|f_n|^pdm+\int_A |f_n|^pdm \\ <(\epsilon/2)\,\, m([0,1]\setminus A)+\int_A |f_n|^pdm \\ \leq (\epsilon/2) \,\,m([0,1])+\int_A |f_n|^pdm\\ = \frac{\epsilon}{2}+\int_A |f_n|^pdm\\ $$ Also note that $|f_n|^p\in L^{2/p}([0,1])$, that is because $f_n^p$ is measurable, and hence $|f_n|^p$, and $$\|\,|f_n|^p\|_{2/p}^{p/2}=\int_0^1 \left(|f_n|^p\right)^{2/p}\,dm=\int_0^1 |f_n|^2 dm\leq M<\infty $$.

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You are on the right track. By Egoroff's, it is enough to show that for an arbitrary small set $A$ we have $$\int_A |f_n|^p dm \leq \epsilon.$$

By Holders, for $1/q+1/q'=1$, we get $$ \int_A |f_n|^p \cdot 1 \leq \left( \int_A |f_n|^{pq} dm \right)^{1/q} \left( \int_A 1^{q'} dm \right)^{1/q'}.$$

Choose $pq=2$ so that $1/q'=1-1/q=1-p/2=(2-p)/2$ or $q'=2/(2-p)$. Then $$\int_A |f_n|^p dm \leq M^{1/q} m(A)^{1/q'}.$$

The rest follows easily.

NOTES: The idea is to remember you can use Holder with the function $g=1$ and that somehow you want to convert the power of $p$ to a power of $2$. Holder requires the exponents to be slightly greater than 1, so $p<2$ makes sense. In addition, note that $q' \leq 0$ for $p \geq 2$, which does not allow for the same bound.

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  • $\begingroup$ No problem. I edited my answer quite a bit after I realized I was using Holder in a more convoluted way. I forgot $g=1$ is good enough. Also, I was using $|f_n| \geq 1$, which is an assumption we can make because the dominated convergence theorem takes care of $|f_n| \leq 1$. $\endgroup$ – abnry Dec 15 '14 at 5:13

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