$(f_n)$ in $L^p(\Omega)$ satisfying $f_n(x) \to f(x)$ a.e. and $\|f_n\|_p \to \|f\|_p$, then $\|f_n - f\|_p \to 0$?

Question

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?

Answer 1

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 $$\int_{F}|f_n-f|^p\to0.$$

Answer 2

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 $p=1$.