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

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

For any $\epsilon>0$, $f \in Y$, there exists $\varphi \in X$ s.t. $\|f-\varphi\|_{H^{1}(\Omega)}<\epsilon$ ?

My attempt 1

By Tietze extension theorem, we can find $F \in C(\mathbb{R}^{d})$ such that $f=F$ on $\bar{\Omega}$. Let $J_{\delta}F=\int_{\mathbb{R}^{d}}j_{\delta}(x-y)F(y)\,dy$, where $j_{\delta}$ is mollifier. Then $F_{\delta} \in C_{c}^{\infty}(\mathbb{R}^{d})$ and $J_{\delta}F \to f$ in $L^{2}(\Omega)$ as $\delta \to 0$. Futhermore, since $f \in H^{1}(\Omega)$, \begin{align*} \frac{\partial}{\partial x_{i}}J_{\delta}F-\frac{\partial f}{\partial x_{i}}&=J_{\delta}\frac{\partial F}{\partial x_{i}}-\frac{\partial F}{\partial x_{i}}\quad \text{on } \Omega \\ & \to 0 \quad \text{in } L^{2}(\Omega). \end{align*} Therefore $\|J_{\delta}F -f\|_{H^{1}(\Omega)} \to 0$ as $\delta \to 0$. But I don't know $J_{\delta}F \left| \right. _{\Omega} \in C_{c}(\bar{\Omega})$. Can we show $J_{\delta}F \left| \right. _{\Omega} \in C_{c}(\bar{\Omega})$ ?

My attempt 2

If there exist open subset $O$ and compact subset $K$ of $\mathbb{R}^{d}$ s.t. $O \subset K \subset \Omega$ and $\|f\|_{H^{1}(\Omega \setminus K)}<\epsilon$, we take $\psi \in C_{c}^{\infty}(\mathbb{R}^{d})$ s.t. $\psi=1$ on $O$ and $\psi=0$ on $\mathbb{R}^{d} \setminus K$.

Then \begin{align*} \|f \psi -f\|_{H^{1}(\Omega)} &\leq \|f \psi -f\|_{H^{1}(O)}+\|f \psi -f\|_{H^{1}(K \setminus O)}+\|f \psi -f\|_{H^{1}(\Omega \setminus K)} \\ &=\|f -f\|_{H^{1}(O)}+\|f \psi -f\|_{H^{1}(K \setminus O)}+\|0 -f\|_{H^{1}(\Omega \setminus K)} \\ &\leq 0+\|f \psi -f\|_{H^{1}(K \setminus O)}+\epsilon. \end{align*}

But I don't know how to deal with $\|f \psi -f\|_{H^{1}(K \setminus O)}$ and existence of such $O,K$.

If you know, please tell me. Thank you in advance.

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closed as off-topic by Michael Renardy, Franz Lemmermeyer, Stefan Kohl, Alex Degtyarev, Christian Remling Apr 29 '16 at 17:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Michael Renardy, Franz Lemmermeyer, Alex Degtyarev
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Waht do you mean by $C_c(\overline{\Omega})$? Continuous functions on $\mathbb R^d$ such that the support is a compact subset of $\overline{\Omega}$? For "nice" $\Omega$ this would be the space of continuous functions on $\Omega$ vanishing at the boundary. $\endgroup$ – Jochen Wengenroth Apr 29 '16 at 12:50
  • $\begingroup$ Thank you for your comment. $C_{c}(\bar{\Omega})$ denotes the space of all continuous real valued functions on $\bar{\Omega}$ with compact support. $\endgroup$ – sharpe Apr 29 '16 at 13:54
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Use cutoff functions as in attempt 2, but construct $\psi$ more carefully. This is a very standard construction.

Start with a function $\psi \in C^\infty_c(\mathbb{R}^d)$ having $\psi = 1$ on the ball $B(0,1)$, supported inside $B(0,2)$, and with $0 \le \psi \le 1$ everywhere. Let $M = \max_i \sup_{B(0,2)} |D_i \psi|$ which is finite. Set $\psi_n(x) = \psi(x/n)$. If $f \in Y$ then clearly $\psi_n f \in X$ for every $n$. Now $$\|f - \psi_n f\|_{H^1}^2 = \|f - \psi_n f\|_{L^2}^2 + \sum_i \|D_i(f - \psi_n f)\|_{L^2}^2.$$ Note that $\psi_n \to 1$ pointwise and boundedly. Thus $f - \psi_n f \to 0$ pointwise, and $|f - \psi_n f|^2 \le |f|^2 \in L^1$ because $f \in L^2$. So by dominated convergence we have $\|f - \psi_n f\|_{L^2}^2 = \int |f - \psi_n f|^2 \to 0$, taking care of the first term.

Next, by the product rule (which is valid for weak derivatives) we have $D_i(f - \psi_n f) = (1 - \psi_n) D_i f - f D_i \psi_n$. As above, since $\psi_n \to 1$ pointwise and boundedly, and $D_i f \in L^2$, we get by dominated convergence that $(1 - \psi_n) D_i f \to 0$ in $L^2$. Also, $|D_i \psi_n| \le M/n$ so $D_i \psi_n \to 0$ pointwise and boundedly as well, and $f \in L^2$, so $f D_i \psi_n \to 0$ in $L^2$.

We have now shown that $\psi_n f \to f$ in $H^1$, as desired.

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  • $\begingroup$ Thanks for your reply! Your proof is very easy to understand! $\endgroup$ – sharpe Apr 29 '16 at 14:27

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