I have a question about Sobolev space.

Let $\Omega$ be an open subset of $\mathbb{R}^{d}$,

we consider the Sobolev space

$H^{1}(\Omega):=\left\{ u \in L^{2}(\Omega) : D_{j}u \in L^{2}(\Omega), j=1,\ldots,n \right\}$

with norm


where $D_{j}u=\partial u/ \partial x_{j}$ is the distributional deriavtive. Moreover, we let

$X:=\text{closure of }\left\{ u|{}_{\Omega} \in H^{1}(\Omega): u \in C_{c}(\bar{\Omega})\right\} \text{in } H^{1}(\Omega) ,$

$Y:=\text{closure of }\left\{ u|{}_{\Omega} \in H^{1}(\Omega): u \in C(\bar{\Omega})\right\} \text{in } H^{1}(\Omega),$

where $C_{c}(\bar{\Omega})$ denotes the space of all continuous real valued functions on $\bar{\Omega}$ with compact support and $C(\bar{\Omega})$ denotes the space of all continuous real valued functions on $\bar{\Omega}$.

My question

Can we show $X=Y$ ?

My attempt

It is clear that $X \subset Y$. If $\Omega$ is bounded, $X=Y$ (since $C_{c}(\bar{\Omega})=C(\bar{\Omega})$ holds).

Can we show $X=Y$ when $\Omega$ is an arbitary open subset?

If you know, please tell me.

Thank you in advance.

  • $\begingroup$ Isn't $X=H_0^1(\Omega)$? $\endgroup$ – Siminore Apr 28 '16 at 9:09
  • $\begingroup$ I think $H_{0}^{1}(\Omega) \subset X$ is true. Can you show $X \subset H_{0}^{1}(\Omega)$? $\endgroup$ – sharpe Apr 28 '16 at 9:32

If you have $y \in H^1(\Omega) \cap C(\bar\Omega)$ you can use a truncation argument to obtain $x \in H^1(\Omega) \cap C_c(\bar\Omega)$ such that $\|x-y\|_{H^1}$ is arbitrarily small.

  • $\begingroup$ Thank you for your reply. But I don't know truncation argument. What is truncation argument? $\endgroup$ – sharpe Apr 29 '16 at 5:10
  • $\begingroup$ You take $\varphi \in C_0^\infty(\mathbb R^n)$ such that $\varphi = 1$ on some $B_r(0)$ and $\varphi = 0$ on $B_R(0)$ with $0 < r < R$. Then, you set $x = y \, \varphi$. $\endgroup$ – gerw Apr 29 '16 at 5:14

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