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Is the space $C^\infty_0(\Omega)$ of smooth functions with compact support, dense in the set of smooth functions $C^\infty(\Omega)$ with respect to the norms $\|\cdot\|_{L^p}$ and the Sobolev-Norm $\|\cdot\|_{W^{1,p}}$ for $\Omega \subset \mathbb{R}^n$ and $1 \leq p < \infty$? I.e. does there exist for every $u \in C^\infty(\Omega)$ a sequence $(u_k)_{k \in \mathbb{N}} \subset C^\infty_0(\Omega)$ such that $$\|u_k-u\|_{L^p} \rightarrow 0 \text{ and/or } \|u_k-u\|_{W^{1,p}} \rightarrow 0?$$

I suspect that in general the answer is yes in the case of the $L^p$-norm and no in the case of the $W^{1,p}$-norm. However for $\Omega = \mathbb{R}^n$ it seems to be true in both cases:

If $\Omega = \mathbb{R}^n$ and $u \in C^\infty(\mathbb{R}^n)$ we can choose a function $\eta \in C^\infty_0(B_2(0))$ with $\eta = 1$ inside $B_1(0)$ (where $B_R(0)$ is the ball with radius $R$ inside $\mathbb{R}^n$) and define for $k \in \mathbb{N}$ the functions $\eta_k(x) = \eta(\frac{x}{k}) \in C^\infty_0(B_{2k}(0))$. Then set $u_k := \eta_k u \in C^\infty_0(\mathbb{R}^n)$. By dominated convergence, we then have $$\|u_k-u\|_{L^p} \rightarrow 0$$ right?

Similarly, it follows that $$\|\nabla u_k - \nabla u\|_{L^p} \rightarrow 0$$ because $(\nabla \eta_k) u \rightarrow 0$ pointwise almost everywhere. Hence also $\|u_k-u\|_{W^{1,p}} \rightarrow 0.$

Are my thoughts correct? And what about general $\Omega \subset \mathbb{R}^n$?

EDIT: I forgot to mention that we always have to consider $u \in C^\infty(\Omega) \cap L^p(\Omega)$ or $u \in C^\infty(\Omega) \cap W^{1,p}(\Omega)$ respectively, for the above.

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  • $\begingroup$ If we take, $f(x)= \sin(x), x\in \mathbb R,$ then $f\in C^{\infty}(\mathbb R)$ but $f\notin L^{1}(\mathbb R);$ now can we expect to approximate, $f$ by $C_{c}^{\infty}(\mathbb R)$ in $L^{1}-$ norm ? (To me it seems no, but I may be wong) $\endgroup$ – Inquisitive Aug 21 '14 at 13:20
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By one hand, if $\Omega\subset \mathbb{R}^N$ is any open set then, $C_0^\infty(\Omega)$ is dense in $L^p(\Omega)$, as you can see, for example, in Brezis chapter 4. Once $$C_0^\infty(\Omega)\subset C^\infty(\Omega)\cap L^p(\Omega)\subset L^p(\Omega),$$ the result follows.

On the other hand, if $-\infty<a<b<\infty$, you can see in chapter 8 of Brezis book that $$W_0^{1,p}(a,b)=\{u\in W^{1,p}(a,b):\ u(a)=u(b)=0\}.$$

As $W_0^{1,p}(a,b)$ is the closure of $C_0^\infty(a,b)$ in the $W^{1,p}$ norm, we must conclude that $C_0^\infty(a,b)$ is not dense in $C^\infty(a,b)$ in the $W^{1,p}$ norm.

An analogous example will work in every dimension $N$, for example, if you consider $p>N$.

Remark 1: If $\Omega=\mathbb{R}^N$ then $W_0^{1,p}(\Omega)=W^{1,p}(\Omega)$.

Remark 2: Sometime I read in some book, which I fail to remember what book is, that if $C\subset \mathbb{R}^N$ is a closed set with null capacity then, $$W^{1,p}(\mathbb{R}^N\setminus C)=W_0^{1,p}(\mathbb{R}^n\setminus C)$$

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  • $\begingroup$ Thanks a lot for your answer! So, my reasoning above for $C^{\infty}_0(\mathbb{R}^n)$ being dense in $C^{\infty}(\mathbb{R}^n)$ w.r.t. the $W^{1,p}$-norm is correct? $\endgroup$ – Tom Bombadil Aug 25 '14 at 10:25
  • $\begingroup$ Yes, they are good. You have to change bounded by compact in your question. $\endgroup$ – Tomás Aug 25 '14 at 11:12

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