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Suppose that $1<p<\infty,\ f,\ f_n\in L^p([0,1]),\ n\in\mathbb{N},\ ||f_n||_{L^p}\leq 1$ for all $n$. $f_n\to f$ a.e.
Show that $f_n\to f$ weakly and $||f||_{L^p}\leq 1$.

Attempt:
If $f_n\not\to f$ weakly, then there is some subsequence $\{f_{n_k}\}$, $\epsilon> 0$, $g\in L^q$ such that $|\int (f_{n_k}-f)g|\geq \epsilon$

By Alaoglu's theorem, the closed ball in $L^p$ is weakly compact, so there is some subsequence $\{f_{n_{k\ell}}\}$ that converges to $f$ weakly, which is a contradiction.
However, how would I justify that the sequence converges to $f$ and not something else? Is a.e. convergence sufficient?

For the second part, $1\geq \lim_n|\int f_n g| = |\int fg|$, and taking the $\sup$ over $g\in L^q,\ ||g||_q=1$ gives us $||f||_p\leq 1$

I think I have solved most of the problem, but there's are parts that I am unconvinced about.

RaiRsXOT6L
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  • What does $\text{a.e.}$ mean? Sorry if I am amateur at this. I know that i.e. $=$ Id est but are they related? – Mr Pie May 25 '18 at 05:22
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    a.e. means 'almost everywhere'. It means the property holds except in a set of measure $0$. – Fimpellizzeri May 25 '18 at 05:23
  • @Fimpellizieri thanks for telling me :) – Mr Pie May 25 '18 at 05:24
  • I have used uniform integrability in my answer. Reference: Theorem 16.13 of Billinglsey's Probability and measure. If you want to know how to prove uniform integrability using boundedness of $L^{p}$ norms please let me know. – Kavi Rama Murthy May 25 '18 at 05:32
  • Can you give a proof sketch? This is something I have not encountered before. – RaiRsXOT6L May 25 '18 at 05:32
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    let $m$ denote Lebesgue measure. $m{|f_n| >T} \leq \frac 1 {T^{p}} \int_{{|f_n|>T}} |f_n|^{p} dm \leq \frac 1 {T^{p}} \int |f_n|^{p} dm \leq \frac 1 {T^{p}} <\epsilon$ if $T$ is large enough. Next, $\int_{{|f_n|>T}} |f_n|^{p} \leq (\int |f_n|^{p})^{1/p} (m{|f_n| >T})^{1/q}$ by Holder's inequality, $q$ being $\frac p {p-1}$. Hence $\int_{{|f_n|>T}} |f_n|^{p} < \epsilon$ for all $n$. This is the definition of uniform integrability. Now just apply the theorem from Billingsley's book. – Kavi Rama Murthy May 25 '18 at 07:47
  • Related, related and related. – A.Γ. May 25 '18 at 09:16

1 Answers1

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$\{f_n\}$ is uniformly integrable and $f_n \to f$ a.e. so $f_n \to f$ in $L^{1}$. Hence $\int_A f_n \to \int_A f$. If $f_n \to g$ weakly where $g \in L^{q}$ then $\int_A f_n \to \int_A g$. This gives $\int_A f=\int_A g$ for all $A$, so $f=g$ a.e.. For the second part you can simply use Fatou's Lemma: $\int |f|^{p} \leq \liminf \int |f_n|^{p} \leq 1$.

Kavi Rama Murthy
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