Let $u_n \to u$ in $L^1(\Omega)$. Does $u_n^p \to u^p$ in $L^1(\Omega)$ if we know $u_n^p \in L^1(\Omega)$?

Question

Suppose $u_n \to u$ in $L^1(\Omega)$ where $\Omega$ is a bounded domain. Suppose that $u_n^p \in L^1(\Omega)$ (actually $L^\infty(\Omega)$ for each $n$).

Fix $p \in [1,\infty)$.

So $u_n(x) \to u(x)$ a.e. By continuity, $u_n^p(x) \to u^p(x)$ pointwise a.e. But there is no dominating function so I cannot use DCT. Also $g(r) = r^p$ is not Lipschitz, it is only locally Lipschitz so that does not help. I also cannot use the Nemytskii theory. Is there any way to prove this if it is true?

Answer 1

This is not true if the sequence $(u_n)$ is not uniformly bounded in $L^\infty(\Omega)$:

Take $\Omega=(0,1)$, $$ u_n(x) = n^{1/2} \chi_{[0,1/n]}(x). $$ Then $u_n\to 0$ in $L^1(0,1)$, $u_n\in L^\infty(0,1)$, but $\|u_n\|_{L^2(0,1)}=1$, $\|u_n\|_{L^p(0,1)}\to\infty$ for $p>2$.

If in addition, the sequence is uniformly bounded in $L^\infty(0,1)$, then $u_n\to u$ in every $L^p(\Omega)$, $1<p<\infty$: Since we have pointwise a.e. convergence of a subsequence, it follows that the limit $u$ is in $L^\infty(\Omega)$, too. Then we can estimate $$ \begin{split} \|u_n-u\|_{L^p(\Omega)}^p &= \int_\Omega |u_n-u|^p = \int_\Omega |u_n-u|\cdot|u_n-u|^{p-1}\\ &\le \|u-u_n\|_{L^1(\Omega)} \|u-u_n\|_{L^\infty(\Omega)}^{p-1} \end{split} $$ by Hoelder inequality.

Answer 2

Certainly not if $u^p\not\in L^1(\Omega)$. Take for instance $\Omega=(0,1)$, $u(x)=x^{-1/2}$, $p=2$ and $u_n=\min(u(x),n)$.