1

There are Sobolev/Rellich-Kondrachov theorem imbedding :

Let $\Omega\subset \mathbb R^n $ open bounded with Lipschitz boundary.

1) If $1\leq p<n$ then $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ for every $q\in [1,p^*]$ where $p^*=\frac{np}{n-p}$.

2) If $q\in [1,q^*)$ the imbedding is compact.

I'm not exactly sure what it exactly mean. Can I interpret it as :

1) If $(u_n)$ is a bounded sequence of $W^{1,p}$, then there is a subsequence that converge weakly to $u$ in $W^{1,p}$ ?

2) If $(u_n)$ is a bounded sequence of $W^{1,p}$, then there is a subsequence that converge strongly to $u$ in $W^{1,p}$ ?

If yes, how can it be shown consider the hypothesis of the theorem ?

user330587
  • 1,624

2 Answers2

2

Part (1) is a direct consequence of the Sobolev inequality, which says that there exists a positive $C$, specific to your choice of $p$, $q$ and $\Omega$, such that $|| u ||_{L^q(\Omega)} \leq C || u ||_{W^{1,p}(\Omega)}$ for all $u \in W^{1,p}(\Omega)$. This tells us that elements of $W^{1,p}(\Omega)$ have finite $L^q$-norm, so they can be considered as legitimate elements of $L^q(\Omega)$. By considering elements of $W^{1,p}(\Omega)$ as elements of $L^q(\Omega)$, we get a natural map $W^{1,p} (\Omega)\hookrightarrow L^q(\Omega)$, and this map is a bounded linear map, by virtue of the above inequality.

Part (2) is the Rellich compactness theorem. This says that our map $W^{1,p}(\Omega) \hookrightarrow L^q(\Omega)$ is compact. Remember, saying that our map $W^{1,p} (\Omega) \hookrightarrow L^q(\Omega)$ is a compact map is the same as saying that the closure of the image of the unit ball (and hence, any ball) in $W^{1,p}(\Omega)$ is a compact subset of $L^q(\Omega)$. And a subset of a metric space has compact closure iff every sequence in this subset has a convergent subsequence. Therefore if $u_n$ is a bounded sequence in $W^{1,p}(\Omega)$, then there is a subsequence $u_{n_k}$ that is strongly convergent in $L^q(\Omega)$. (Note that $u_{n_k}$ does not necessarily converge strongly in $W^{1,p}(\Omega)$ itself, so the final sentence in your post is not true.)

Kenny Wong
  • 32,192
1

First, $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ can be interpreted as $W^{1,p}(\Omega )\subset L^q(\Omega )$ what mean that $$\|u\|_{L^q}\leq \gamma \|u\|_{W^{1,p}},$$ for a certain constant $\gamma $. Now if $p\in ]1,\infty [$, if $(u_n)\subset W^{1,p}(\Omega )$ is a bounded sequence, since $W^{1,p}(\Omega )$ is reflexive, there is a subsequence (still denoted $u_n$) that converge weakly. Let denote $u$ the limit. The fact that you have a compact embedding just tells you that $(u_n)$ converge strongly to $u$ in $L^q$.

Surb
  • 55,662