Let $\Omega\subset \mathbb{R}^3$ be a bounded domain. Assume that $(f_n)_{n\in \mathbb{N}}\subset W^{1,2}(\Omega)$ is weakly convergent to $f\in W^{1,2}(\Omega)$.
Consider now $(f_n^3)_{n\in \mathbb{N}}$ which is bounded in $L^2(\Omega)$ by the embedding $W^{1,2}(\Omega)\hookrightarrow L^6(\Omega)$, so that in converges weakly in $L^2(\Omega)$. Is it possible to show that the weak limit is acutally $f^3$?
Weak convergence in $L^2$ of powers of a sequence which is weakly convergent in $W^{1,2}$
functional-analysislp-spacessobolev-spacesweak-convergence
Best Answer
By compact embeddings of $W^{1,2}$ into $L^p$, $p<6$, we have $f_{n_k}\to f$ weakly in $H^1$, strongly in $L^p$ for $p<6$, and pointwise almost everywhere. This implies $f_{n_k}^3 \to f^3$ pointwise almost everywhere.
Since $(f_n^3)$ is bounded in $L^2$ it has a weakly converging subsequnec, so we could have chosen $f_{n_k}$ from the beginning such that $f_{n_k}^3 \rightharpoonup g$ in $L^2$ for some $g$. Pointwise a.e. and weak limits coincide, so $f^3=g$.
By a subsequence-subsequence argument, the whole sequence $(f_n^3)$ converges weakly to $f^3$.