0
$\begingroup$

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

$\|u\|^{2}_{H^{1}(\Omega)}:=\|u\|^{2}_{L^{2}(\Omega)}+\sum_{j=1}^{n}\|D_{j}u\|_{L^{2}(\Omega)}^{2}$,

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.