2
$\begingroup$

We are given a sequence of functions $\{f_n\}$ on $L^1[0,1]$ so that for each $g \in L^\infty[0,1]$, we have $\int f_ng$ converges. We are also given that there exists $u \in L^1$ such that $\lambda(|f_n|>M)\leq \lambda(|u| >M)$ for all $M>0$. How to prove that $f_n$ converges weakly to some $f \in L^1$?

My approach goes like this: first consider $|f_n|$ being bounded: $|f_n|\leq M$ for some $M$, and consider $T: L^1 \to (L^\infty)^*$ by $T(f_n)(g) := \int f_ng$ for $g \in L^\infty$, so that $T(f_n)$ converges in $(L^\infty)^*$. By boundedness of the sequence and Alaoglu theorem, we prove that $f_n \rightharpoonup f$. Does it makes sense?

Now for unbounded sequence, consider $\tilde{f}_n := f_n \chi_{|f_n|\leq M}$, so that $\tilde{f}_n$ is bounded and weak limit exists. How should I proceed?

$\endgroup$
1
  • $\begingroup$ I have made a change. $\endgroup$ Commented Mar 10, 2021 at 17:01

1 Answer 1

Reset to default
1
$\begingroup$

Let $\mu (A) =\lim \int_A f_n =\lim \int f_n g$ where $g=\chi_A$. Then $\mu$ is a real measure and $\mu << m$ where $m$ is the Lebesgue measure. Let $f=\frac {d\mu} {dm}$. Now $\int |f_n| =\int_0^{\infty} m\{x:|f_n(x)| >M\} dM \le \int_0^{\infty} m\{x:u(x) >M\} dM=\int u $. Hence, $\sup_n \int |f_n| <\infty$. We now have $\int f_n g =\int fg$ for any simple function $g$. Any $g \in L^{\infty}$ is a uniform limit of simple functions. Now you can use triangle inequality to show that $\int f_n g =\int fg$ for any $g \in L^{\infty}$.

$\endgroup$
2
  • $\begingroup$ Do we have to first decompose $f = f_+ - f_-$ and apply your Radon Nikodym argument for each $f_+$ and $f_-$? And is it clear that $\mu$ defines a (signed) measure? $\endgroup$ Commented Mar 10, 2021 at 13:07
  • $\begingroup$ The fact that $\mu$ is real mesure is proved in Royden's book. Also Radon Nikodym Theorem holds for real measure too, so there is no need to split into positive and negative parts. @MLSo $\endgroup$ Commented Mar 10, 2021 at 23:13

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.