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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061466%2ftruncation-of-a-sobolev-function-is-still-sobolev%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061466%2ftruncation-of-a-sobolev-function-is-still-sobolev%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more