0

Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain and $1<p<\infty$. Suppose $$ W^{1,p}(\Omega):=\{u:\Omega\to\mathbb{R}\text{ measurable such that }\|u\|_{W^{1,p}(\Omega)}<\infty\}, $$ where $$ \|u\|_{W^{1,p}(\Omega)}=\left(\|u\|_{L^p(\Omega)}+\|\nabla u\|_{L^p(\Omega)}\right)^\frac{1}{p}. $$ Let $X\subset W^{1,p}(\Omega)$ be such that $\int_{\Omega}v\,dx =0$. Then by the Sobolev embedding theorem, $X$ is a normed linear space under the norm $$ \|u\|_{X}=\left(\int_{\Omega}|\nabla u|^p\,dx\right)^\frac{1}{p}. $$ I want to know if $X$ is a reflexive and separable Banach space under the $\|\cdot\|_X$ norm?

If so, can we obtain it from the reflexive and separable Banach space property of $W^{1,p}(\Omega)$?

Kindly help.

Thanks.

Mathguide
  • 421
  • A closed subspace of a reflexive Banach space is reflexive (https://math.stackexchange.com/questions/1527170/a-closed-subspace-of-a-reflexive-banach-space-is-reflexive) and a subset of a separable metric space is separable (https://math.stackexchange.com/questions/516886/prove-that-a-subset-of-a-separable-set-is-itself-separable). – MaoWao Jul 02 '21 at 16:37
  • Plus the fact that the two norms of $X$ and of $W^{1,p}$ are equivalent on $X$, thus $X$ is a closed subspace of $W^{1,p}$ with codimension 1. – Lorenzo Pompili Jul 02 '21 at 16:40
  • Thank you very much. – Mathguide Jul 03 '21 at 11:17

0 Answers0