Let $\Omega\subset \mathbb{R}^n$ be bounded and let $(f_n)_{n\in \mathbb{N}}\subset H^1(\Omega)$ be weak convergent to $f$ in $H^1(\Omega)$. Is it true that $(f_n)_{n\in \mathbb{N}}$ admits a subsequence which converges weakly to $f$ in $L^6(\Omega)$ ?
By the compact embedding of $H^1(\Omega)\hookrightarrow L^p(\Omega)$ for $p\in [1,6)$ we have for example a subsequence $(f_{n_k})_{k\in \mathbb{N}}$ so that $f_{n_k}\to f$ in $L^5(\Omega)$. On the other hand we know that there exists $h\in L^6(\Omega)$ so that $f_n\rightharpoonup h$ in $L^6(\Omega)$. This implies $f_n \rightharpoonup h$ in $L^5(\Omega)$ as $L^{\frac{5}{4}}(\Omega)\subset L^{\frac{6}{5}}(\Omega)$. By the strong convergence we conclude $f=h$ but only in $L^5(\Omega)$. Is there a way to extend my argument or maybe a completely different approach to show $f_n\rightharpoonup f$ in $L^6(\Omega)$?
I would be grateful for any advice.
Best Answer
If $n\le 3$ then the embedding $H^1\hookrightarrow L^6$ is continuous, this implies weak convergence in $L^6$ directly, without using compact embeddings.