# Showing that for $\{u_n,u\}_{n \geq 1} \subseteq L^p(\Omega)$ it is $u_n \to u$ in $L^p(\Omega)$.

Exercise :

Let $$\Omega \subseteq \mathbb R^n$$ be open and bounded, $$\{u_n, u\}_{n \geq 1} \subseteq L^p(\Omega)$$ with $$1 and we assume that $$\|u_n\|_p \to \|u\|_p, \; u_n \xrightarrow{a.e.} u$$. Show that $$u_n \to u$$ in $$L^p(\Omega)$$.

Attempt :

Since $$\|u_n\|_p \to \|u\|_p$$ then it is :

$$\|u_n\|_p^p \to \|u\|_p^p \Leftrightarrow \int_\Omega |u_n|^p \mathrm{d}x = \int_\Omega |u|^p\mathrm{d}x$$

To show that $$u_n \to u$$ in $$L^p(\Omega)$$, we need to show that $$\|u_n - u\|_p \to 0$$. But :

\begin{align*} \|u_n-u\|_p^p &= \int_\Omega |u_n-u|^p\mathrm{d}x \\ &\leq \int_\Omega \left(|u_n| + |u|\right)^p\mathrm{d}x \end{align*}

But from then on, I don't see how to continue in order to derive an expression of the form $$\|u_n-u\|_p^p < \varepsilon$$ where $$\varepsilon > 0$$.

Any hints will be greatly appreciated.

Since we know that $$u_n\to u$$ almost everywhere we can conclude that $$|u_n-u|^p\to 0$$ almost everywhere. Hence: (we will use Fatou lemma)
$$2\int_{\Omega} |u|^pdx=\int_{\Omega}\lim_{n\to\infty}(|u|^p+|u_n|^p-|u-u_n|^p)dx\leq$$
$$\leq\liminf_{n\to\infty}\int_{\Omega}(|u|^p+|u_n|^p-|u-u_n|^p) dx=2\int_{\Omega} |u|dx-\limsup_{n\to\infty}\int_{\Omega} |u-u_n|^p dx$$
In the last equality we used the fact that $$\int_{\Omega} |u_n|^p\to\int_{\Omega} |u|^p$$.
So what we got is that $$\limsup_{n\to\infty}\int_{\Omega} |u-u_n|^pdx\leq 0$$. Since this is a sequence of non negative real numbers this of course implies $$\lim_{n\to\infty}\int_{\Omega} |u-u_n|^pdx=0$$.