Let $p \geq 1$. I know that there exists a continuous and linear extension operator $$ E : W^{1,p}(\mathbb{R}_+^n) \to W^{1,p}(\mathbb{R}^n) .$$

I read that from the existence of such an extension one can deduce that $C_c^{\infty}(\bar{\mathbb{R}^n_+})$ is dense in $W^{1,p}(\mathbb{R}_+^n)$, where for $C_c^{\infty}(\bar{\mathbb{R}^n_+})$ I mean smooth functions on $\mathbb{R}_+^n$ whose support is contained in $\bar{\mathbb{R}_+^n}$(I don't know whether this is a standard notation or not).

I tried to use convolution and use that $C_c^{\infty}(\mathbb{R}^n)$ is dense in $W^{1,p}(\mathbb{R}^n)$ but I wasn't able to get anything .

  • $\begingroup$ for $C_c^{\infty}(\bar{\mathbb{R}^n_+})$ the support of the functions can contain elements of the boundary of $\mathbb R^n_+$, correct? $\endgroup$ – supinf Dec 10 '18 at 9:52
  • $\begingroup$ Sorry it was not clear I edited $\endgroup$ – Tommaso Scognamiglio Dec 10 '18 at 10:14

$C_c^{\infty}(\bar{\mathbb{R}^n_+})$ is dense in $W^{1,p}(\mathbb{R}_+^n)$,

This is not correct. We choose $n=1,p=1$. Lets assume that we can approximate the constant function $1\in W^{1,p}(\mathbb R^n_+)$ with functions from $C_c^\infty (\overline{\mathbb R_+^n})$.

Let $\phi\in C_c^\infty (\overline{\mathbb R_+^n})$. Then $\phi(0)=0$.

If $\phi(x) \leq \frac12$ for all $x\in (0,1)$, then we have $$ \| 1- \phi |_{W^{1,1}} \geq \|1-\phi|_{L^1} \geq \frac12.$$ So we can assume that $\phi(x)>\frac12$ for some $x\in (0,1)$. Then $$ \frac12 = \int_0^x \phi'(y) \mathrm dy \leq \int_0^1 | \phi'(y) |\mathrm dy \leq \|\phi'\|_{L^1} \leq \|1-\phi\|_{W^{1,1}}. $$ So in any case, we cannot approximate $1$ with $\phi$.

  • $\begingroup$ You misunderstood the definition of $C_c^\infty(\overline{\mathbb{R}^n})$ the op is using. It's the space of all smooth functions with compact support in $\mathbb{R}^{n-1}\times [0,\infty)$. In the one-dimensional case this means that they have support in $[0,R]$ for some $R>0$, but not necessarily satisfy $\phi(0)=0$. $\endgroup$ – MaoWao Dec 10 '18 at 10:46

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