3
$\begingroup$

Let $f_n : [0,1] \rightarrow \mathbf{R}$ be a sequence of measurable functions such that

$\bullet$ $f_n \rightarrow 0$ a.e. on $[0,1]$.

$\bullet$ $\int_{[0,1]} |f_n|^2 dm \leq 1$ for all $n \geq 0$.

Then I want to show that $\int_{[0,1]} |f_n| dm \rightarrow 0$ as $n \rightarrow \infty$.

I tried to combine Egorov's Theorem and Dominated Convergence Theorem but I could not find a dominating function for $|f_n|$.

$\endgroup$
6
$\begingroup$

Egoroff's theorem is a good idea. Fix $\varepsilon>0$, $S_\varepsilon$ a set where $\sup_{x\in S_\varepsilon}|f_n(x)|\to 0$ and $\mu([0,1]\setminus S_\varepsilon)<\varepsilon$. We have \begin{align} \int_{[0,1]}|f_n|d\mu&\leqslant\int_{S_\varepsilon}|f_n|d\mu+\int_{[0,1]\setminus S_\varepsilon}|f_n|d\mu\\ &\leqslant\mu(S_{\varepsilon})\sup_{x\in S_\varepsilon}|f_n(x)|+\sqrt{\mu([0,1]\setminus S_\varepsilon)}\sqrt{\int_{[0,1]}|f_n|^2d\mu}\\ &\leqslant\sup_{x\in S_\varepsilon}|f_n(x)|+\varepsilon. \end{align} This gives $\limsup_{n\to +\infty}\int_{[0,1]}|f_n|d\mu\leqslant\varepsilon$ and we conclude as $\varepsilon$ is arbitrary.

$\endgroup$
  • 1
    $\begingroup$ Where does the second inequality come from? Is it Cauchy-Schwarz? $\endgroup$ – Karatug Ozan Bircan Jan 6 '13 at 13:35
  • 1
    $\begingroup$ Yes, exactly, Cauchy-Schwarz. $\endgroup$ – Davide Giraudo Jan 6 '13 at 13:38
0
$\begingroup$

Since the sequence is $L^2$-bounded, it is uniformly integrable and therefore your result holds.

$\endgroup$
  • $\begingroup$ A well-known weakening of the $L^2$-boundedness here is $L \log L$-boundedness, that is $$\int_{[0,1]} |f_n| \log^+ |f_n| dm \le M$$ where $\log^+ x$ is the maximum of $\log x$ and $0$. $\endgroup$ – GEdgar Jan 6 '13 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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