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Let $1 < p < \infty$. If $(f_n)$ is a sequence in $L^p(\Omega)$ satisfying

  1. $f_n(x) \to f(x)$ a.e.,
  2. $\|f_n\|_p \to \|f\|_p$,

then does it follow that $\|f_n - f\|_p \to 0$?

Edit. Here is my solution.

For $A \subseteq \Omega$,\begin{align*}\|f_n - f\|_q & \le \|f_n - f\|_{u,\,\Omega - A}|\Omega - A|^{1\over q} + |A|^{{1\over q} - {1\over p}}\|f_n - f\|_p \\ & \le \|f_n - f\|_{u,\,\Omega - A}|\Omega - A|^{1\over q} + 2|A|^{{1\over q} - {1\over p}} \sup \|f_n\|_p \\ & \le \|f_n - f\|_{u,\,\Omega - A} |\Omega - A|^{1\over q} + 2M|A|^{{1\over q} - {1\over p}},\end{align*}where the second line follows by Fatou's lemma. By Egorov's theorem, we can choose $A$ arbitrarily small such that $f_n \rightrightarrows f$ on $\Omega - A$, so we are done.

I was wondering if anyone had any alternative solutions to this problem?

Student
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  • On a finite measure space, it certainly does. Can you share what you've tried? –  Jan 13 '16 at 23:16
  • Try looking at the function $f_n+f-|f_n-f|$, and using Fatous lemma. It is the commonly given hint for this problem. – siegehalver Jan 13 '16 at 23:18
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    @ChristopherHalverson Or probably $|f_n|+|f|-|f_n-f|$; that gives the result for $p=1$. – David C. Ullrich Jan 14 '16 at 00:12
  • @user In fact on a finite measure space it seems to me the argument I suspect you have in mind works (for $p>1$) assuming just that $||f_n||p$ is bounded (and $||f||_p<\infty$). Then _if $||f_n||_p\to||f||_p$, one more epsilon extends that to the case of an infinite measure space. (Take $E$ of finite measure so $\int_E|f|^p>||f||_p^p-\epsilon$. Fatou's Lemma shows that the integral of $|f_n|^p$ over the complement of $E$ is less than $2\epsilon$ for large $n$.) – David C. Ullrich Jan 14 '16 at 00:24
  • @user Never mind what I said - dropped a $p$. – David C. Ullrich Jan 14 '16 at 00:48
  • There's a lot of that solution I don't follow... ? – David C. Ullrich Jan 14 '16 at 03:39
  • Like, you seem to be assuming the space has finite measure. It looks like a proof that $||f_n-f||_q\to0$. You seem to be using the fact that the $L^q$ norm is dominated by the $L^p$ norm, which is not true unless $q\le p$. And you seem to be using the fact that $|A|^{1/q-1/p}\to0$ as $|A|\to0$< which is not true unless $q<p$. ??? – David C. Ullrich Jan 14 '16 at 04:22
  • See this. – David Mitra Jan 14 '16 at 07:52

2 Answers2

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Use the fact that $$ 0 \leq 2^p (|f_n|^p +|f|^p) -|f_n-f|^p. $$ Hence by Fatou's lemma, $$ 2^{p+1} \int_\Omega |f|^p \leq \liminf_{n \to +\infty} \int_\Omega \left( 2^p (|f_n|^p +|f|^p) -|f_n-f|^p \right) = 2^{p+1} \int_\Omega |f|^p - \limsup_{n \to +\infty} \int_\Omega |f_n-f|^p. $$ Remark that this holds true also for $p=1$.

Siminore
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Fatou's Lemma shows that $$\int_A|f|^p\le\liminf\int_A|f_n|^p$$for any $A$. Applying this to $A$ and to $X\setminus A$ and noting that $||f_n||_p\to||f||_p$ we see that in fact $$\int_A|f|^p=\lim\int_A|f_n|^p.$$

Say $\epsilon>0$. Choose a set $E$ of finite measure so $$\int_{X\setminus E}|f|^p<\epsilon.$$Choose $\delta>0$ so $\mu(A)<\delta$ implies $\int_A|f|^p<\epsilon$, and use Egoroff to get $F\subset E$ with $\mu(E\setminus F)<\delta$ and such that $f_n\to f$ uniformly on $F$. Now the thing at the start shows that $$\int_{X\setminus F}|f_n-f|^p<c\epsilon$$for large $n$, and $$\int_{F}|f_n-f|^p\to0.$$

David C. Ullrich
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