Approximating Sobolev functions in $W^{1,p}(mathbb{R}_+^n)$












0












$begingroup$


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 .






sobolev-spaces convolution smooth-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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 .






sobolev-spaces convolution smooth-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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 .






sobolev-spaces convolution smooth-functions







share|cite|improve this question











$endgroup$




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 .






sobolev-spaces convolution smooth-functions




sobolev-spaces convolution smooth-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 10:14







Tommaso Scognamiglio

















asked Dec 10 '18 at 9:08









Tommaso ScognamiglioTommaso Scognamiglio

492312




492312












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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




$begingroup$
Sorry it was not clear I edited
$endgroup$
– Tommaso Scognamiglio
Dec 10 '18 at 10:14










1 Answer
1






active

oldest

votes


















-1












$begingroup$


$C_c^{infty}(bar{mathbb{R}^n_+})$ 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 $1in W^{1,p}(mathbb R^n_+)$ with functions from $C_c^infty (overline{mathbb R_+^n})$.



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



If $phi(x) leq frac12$ for all $xin (0,1)$, 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 $xin (0,1)$.
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 $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$


$C_c^{infty}(bar{mathbb{R}^n_+})$ 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 $1in W^{1,p}(mathbb R^n_+)$ with functions from $C_c^infty (overline{mathbb R_+^n})$.



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



If $phi(x) leq frac12$ for all $xin (0,1)$, 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 $xin (0,1)$.
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 $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















-1












$begingroup$


$C_c^{infty}(bar{mathbb{R}^n_+})$ 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 $1in W^{1,p}(mathbb R^n_+)$ with functions from $C_c^infty (overline{mathbb R_+^n})$.



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



If $phi(x) leq frac12$ for all $xin (0,1)$, 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 $xin (0,1)$.
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 $phi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














-1












-1








-1





$begingroup$


$C_c^{infty}(bar{mathbb{R}^n_+})$ 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 $1in W^{1,p}(mathbb R^n_+)$ with functions from $C_c^infty (overline{mathbb R_+^n})$.



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



If $phi(x) leq frac12$ for all $xin (0,1)$, 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 $xin (0,1)$.
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 $phi$.






share|cite|improve this answer









$endgroup$




$C_c^{infty}(bar{mathbb{R}^n_+})$ 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 $1in W^{1,p}(mathbb R^n_+)$ with functions from $C_c^infty (overline{mathbb R_+^n})$.



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



If $phi(x) leq frac12$ for all $xin (0,1)$, 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 $xin (0,1)$.
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 $phi$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 10:27









supinfsupinf

6,3561028




6,3561028












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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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