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Let $\Omega\subset \mathbb{R}^n$ be a bounded domain, $p\in (1,\infty)$. Suppose that $u_n\in L^p(\Omega)$. By using the fact that $L^p(\Omega)$ is uniformly convex, we know that if $u_n\rightharpoonup u$ and $\|u_n\|_p\rightarrow \|u\|_p$, then $u_n\rightarrow u$.

Now, if $u_n\rightharpoonup u$ and $\|\sqrt{f^2+u_n^2}\|_p\rightarrow\|\sqrt{f^2+u^2}\|_p$ with $f\in L^p(\Omega)$, can we conclude the same thing, i.e. $u_n\rightarrow u$?


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up vote 1 down vote accepted

Yes. Instead of $u_n$ and $u$, consider $\mathbb R^2$-valued functions $(u_n,f)$ and $(u,f)$. The space $L^p(\Omega;\mathbb R^2)$ is uniformly convex; one can either prove this directly or use the fact that the Bochner space $L^p(\Omega;X)$ is uniformly convex whenever $1<p<\infty$ and $X$ is uniformly convex (see here for references). Since $(u_n,f)$ converge weakly to $(u,f)$, and their $L^p(\Omega;\mathbb R^2)$ norms converge, the claim follows.

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