Truncation of a Sobolev function is still Sobolev












2












$begingroup$


Let $Omega subset mathbb{R}^d$ a regular domain (compact boundary $C^1$) e let $u in H^1(Omega)$. Let $tr(u) equiv alpha$.



I would like to prove that the truncated function $$bar{u}=u chi_{{u < alpha }}+alpha chi_{{u geq alpha }}$$ is still a Sobolev function with $dfrac{partial u}{partial x_i}=dfrac{partial u}{partial x_i}chi_{{u <1}}$ and $tr(bar{u})=1$.



I tried to apply the direct definition with the existence of weak derivative but I wasn't able to do that.






sobolev-spaces weak-derivatives







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  • $begingroup$
    L'esame di istituzioni si avvicina...
    $endgroup$
    – tommy1996q
    Jan 4 at 10:35
















2












$begingroup$


Let $Omega subset mathbb{R}^d$ a regular domain (compact boundary $C^1$) e let $u in H^1(Omega)$. Let $tr(u) equiv alpha$.



I would like to prove that the truncated function $$bar{u}=u chi_{{u < alpha }}+alpha chi_{{u geq alpha }}$$ is still a Sobolev function with $dfrac{partial u}{partial x_i}=dfrac{partial u}{partial x_i}chi_{{u <1}}$ and $tr(bar{u})=1$.



I tried to apply the direct definition with the existence of weak derivative but I wasn't able to do that.






sobolev-spaces weak-derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    L'esame di istituzioni si avvicina...
    $endgroup$
    – tommy1996q
    Jan 4 at 10:35














2












2








2





$begingroup$


Let $Omega subset mathbb{R}^d$ a regular domain (compact boundary $C^1$) e let $u in H^1(Omega)$. Let $tr(u) equiv alpha$.



I would like to prove that the truncated function $$bar{u}=u chi_{{u < alpha }}+alpha chi_{{u geq alpha }}$$ is still a Sobolev function with $dfrac{partial u}{partial x_i}=dfrac{partial u}{partial x_i}chi_{{u <1}}$ and $tr(bar{u})=1$.



I tried to apply the direct definition with the existence of weak derivative but I wasn't able to do that.






sobolev-spaces weak-derivatives







share|cite|improve this question











$endgroup$




Let $Omega subset mathbb{R}^d$ a regular domain (compact boundary $C^1$) e let $u in H^1(Omega)$. Let $tr(u) equiv alpha$.



I would like to prove that the truncated function $$bar{u}=u chi_{{u < alpha }}+alpha chi_{{u geq alpha }}$$ is still a Sobolev function with $dfrac{partial u}{partial x_i}=dfrac{partial u}{partial x_i}chi_{{u <1}}$ and $tr(bar{u})=1$.



I tried to apply the direct definition with the existence of weak derivative but I wasn't able to do that.






sobolev-spaces weak-derivatives




sobolev-spaces weak-derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:38







Tommaso Scognamiglio

















asked Jan 4 at 9:44









Tommaso ScognamiglioTommaso Scognamiglio

490312




490312












  • $begingroup$
    L'esame di istituzioni si avvicina...
    $endgroup$
    – tommy1996q
    Jan 4 at 10:35


















  • $begingroup$
    L'esame di istituzioni si avvicina...
    $endgroup$
    – tommy1996q
    Jan 4 at 10:35
















$begingroup$
L'esame di istituzioni si avvicina...
$endgroup$
– tommy1996q
Jan 4 at 10:35




$begingroup$
L'esame di istituzioni si avvicina...
$endgroup$
– tommy1996q
Jan 4 at 10:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

Yes, this is the case.



Note that $u-alphain H_0^1(Omega)$.
It is a known result that $max(v,0)in H_0^1(Omega)$ if $vin H_0^1(Omega)$,
see here
for example (using $G(x)=max(x,0)$), or here.



Using
$$
bar u = max(u-alpha,0)+alpha
$$
it follows that $bar uin H^1(Omega)$.



old answer (before an important correction was made in the original post):



No, this is not the case.



We choose $d=1$, $Omega=(0,2pi)$ and the function
$$
u(x) = 2+sin(x).
$$
Clearly, $uin H^1(Omega)$ and $alpha=2$.
However, $bar u$ is not continuous and has a jump at $x=pi$:
$$
bar u(x) =
begin{cases}
1 & xin (0,pi)
\
2+sin(x) & xin [pi,2pi)
end{cases}
$$
Therefore it is not a Sobolev function.
(Note that for $d=1$ the functions in $H^1(Omega)$ are continuous.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
    $endgroup$
    – Tommaso Scognamiglio
    Jan 4 at 11:39










  • $begingroup$
    ok I answered the new question as well.
    $endgroup$
    – supinf
    Jan 4 at 12:10











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Yes, this is the case.



Note that $u-alphain H_0^1(Omega)$.
It is a known result that $max(v,0)in H_0^1(Omega)$ if $vin H_0^1(Omega)$,
see here
for example (using $G(x)=max(x,0)$), or here.



Using
$$
bar u = max(u-alpha,0)+alpha
$$
it follows that $bar uin H^1(Omega)$.



old answer (before an important correction was made in the original post):



No, this is not the case.



We choose $d=1$, $Omega=(0,2pi)$ and the function
$$
u(x) = 2+sin(x).
$$
Clearly, $uin H^1(Omega)$ and $alpha=2$.
However, $bar u$ is not continuous and has a jump at $x=pi$:
$$
bar u(x) =
begin{cases}
1 & xin (0,pi)
\
2+sin(x) & xin [pi,2pi)
end{cases}
$$
Therefore it is not a Sobolev function.
(Note that for $d=1$ the functions in $H^1(Omega)$ are continuous.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
    $endgroup$
    – Tommaso Scognamiglio
    Jan 4 at 11:39










  • $begingroup$
    ok I answered the new question as well.
    $endgroup$
    – supinf
    Jan 4 at 12:10
















1












$begingroup$

Yes, this is the case.



Note that $u-alphain H_0^1(Omega)$.
It is a known result that $max(v,0)in H_0^1(Omega)$ if $vin H_0^1(Omega)$,
see here
for example (using $G(x)=max(x,0)$), or here.



Using
$$
bar u = max(u-alpha,0)+alpha
$$
it follows that $bar uin H^1(Omega)$.



old answer (before an important correction was made in the original post):



No, this is not the case.



We choose $d=1$, $Omega=(0,2pi)$ and the function
$$
u(x) = 2+sin(x).
$$
Clearly, $uin H^1(Omega)$ and $alpha=2$.
However, $bar u$ is not continuous and has a jump at $x=pi$:
$$
bar u(x) =
begin{cases}
1 & xin (0,pi)
\
2+sin(x) & xin [pi,2pi)
end{cases}
$$
Therefore it is not a Sobolev function.
(Note that for $d=1$ the functions in $H^1(Omega)$ are continuous.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
    $endgroup$
    – Tommaso Scognamiglio
    Jan 4 at 11:39










  • $begingroup$
    ok I answered the new question as well.
    $endgroup$
    – supinf
    Jan 4 at 12:10














1












1








1





$begingroup$

Yes, this is the case.



Note that $u-alphain H_0^1(Omega)$.
It is a known result that $max(v,0)in H_0^1(Omega)$ if $vin H_0^1(Omega)$,
see here
for example (using $G(x)=max(x,0)$), or here.



Using
$$
bar u = max(u-alpha,0)+alpha
$$
it follows that $bar uin H^1(Omega)$.



old answer (before an important correction was made in the original post):



No, this is not the case.



We choose $d=1$, $Omega=(0,2pi)$ and the function
$$
u(x) = 2+sin(x).
$$
Clearly, $uin H^1(Omega)$ and $alpha=2$.
However, $bar u$ is not continuous and has a jump at $x=pi$:
$$
bar u(x) =
begin{cases}
1 & xin (0,pi)
\
2+sin(x) & xin [pi,2pi)
end{cases}
$$
Therefore it is not a Sobolev function.
(Note that for $d=1$ the functions in $H^1(Omega)$ are continuous.)






share|cite|improve this answer











$endgroup$



Yes, this is the case.



Note that $u-alphain H_0^1(Omega)$.
It is a known result that $max(v,0)in H_0^1(Omega)$ if $vin H_0^1(Omega)$,
see here
for example (using $G(x)=max(x,0)$), or here.



Using
$$
bar u = max(u-alpha,0)+alpha
$$
it follows that $bar uin H^1(Omega)$.



old answer (before an important correction was made in the original post):



No, this is not the case.



We choose $d=1$, $Omega=(0,2pi)$ and the function
$$
u(x) = 2+sin(x).
$$
Clearly, $uin H^1(Omega)$ and $alpha=2$.
However, $bar u$ is not continuous and has a jump at $x=pi$:
$$
bar u(x) =
begin{cases}
1 & xin (0,pi)
\
2+sin(x) & xin [pi,2pi)
end{cases}
$$
Therefore it is not a Sobolev function.
(Note that for $d=1$ the functions in $H^1(Omega)$ are continuous.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 12:10

























answered Jan 4 at 10:40









supinfsupinf

6,4351028




6,4351028












  • $begingroup$
    Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
    $endgroup$
    – Tommaso Scognamiglio
    Jan 4 at 11:39










  • $begingroup$
    ok I answered the new question as well.
    $endgroup$
    – supinf
    Jan 4 at 12:10


















  • $begingroup$
    Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
    $endgroup$
    – Tommaso Scognamiglio
    Jan 4 at 11:39










  • $begingroup$
    ok I answered the new question as well.
    $endgroup$
    – supinf
    Jan 4 at 12:10
















$begingroup$
Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
$endgroup$
– Tommaso Scognamiglio
Jan 4 at 11:39




$begingroup$
Sorry I made a stupid mistake:now the truncated function should be continous in the 1 dimensional case everytime.
$endgroup$
– Tommaso Scognamiglio
Jan 4 at 11:39












$begingroup$
ok I answered the new question as well.
$endgroup$
– supinf
Jan 4 at 12:10




$begingroup$
ok I answered the new question as well.
$endgroup$
– supinf
Jan 4 at 12:10


















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