Mixing
Is the functional $w mapsto int_0^1 | | w(t) | - 1 | dt$ $C^1$ or even smooth?
$begingroup$
Let $H:= H(I;mathbb{R}^3)$ be the space of $L^2$ + absolutely continuous functions with $L^2$ derivative.
For $w in H$ consider the functional
$$psi(w) = int_0^1 | | w(t) |^2 - 1 | dt$$
where $| cdot |$ denotes the standard norm in $mathbb {R^3}$ and $| cdot |$ the absolute value in $mathbb{R}$.
I want to show that $Omega := H(I;S^2) = psi^{-1}(0)$ is a $C^infty$-submanifold of codimension $1$ of $H$.
I think I'm at the point where I only need to make sure that the functional $psi$ is smooth.
To compute the first derivative, I calculated the first variation which is
$$left.frac{d}{dvarepsilon}right|_{varepsilon=0} psi(w + varepsilon eta) = int_0^1 text{sgn}(| w |^2 - 1) 2 weta dt text{ for all } eta in H$$
(which is a linear and bounded functional in $eta$ and makes it the Frechét differential $D_w psi$ at $w$ that is unequal $0$ making the implicit function theorem usable for the proof of the highlighted statement above.)
In turn $Omega$ is at least a $C^1$-submanifold. Now calculating further derivatives / variations of $psi$ to show smoothness, the second variation yields
$$left.frac{d}{dvarepsilon^2}right|_{varepsilon=0} psi(w + varepsilon eta) = left.frac{d}{dvarepsilon}right|_{varepsilon=0} int_0^1 text{sgn}(| w |^2 + 2varepsilon w eta + varepsilon^2|eta |^2 - 1) 2(w + varepsiloneta) dt $$
The derivative of the sgn-function is $0$ almost everywhere, continuing above terms with product rule yields (? not sure here)
$$ int_0^1 text{sgn}(| w |^2 - 1) 2 eta dt $$
Is this correct?
proof-verification smooth-manifolds calculus-of-variations frechet-derivative global-analysis
asked Jan 27 at 14:56
NhatNhat
1,0261017
$endgroup$
add a comment |
$begingroup$
Let $H:= H(I;mathbb{R}^3)$ be the space of $L^2$ + absolutely continuous functions with $L^2$ derivative.
For $w in H$ consider the functional
$$psi(w) = int_0^1 | | w(t) |^2 - 1 | dt$$
where $| cdot |$ denotes the standard norm in $mathbb {R^3}$ and $| cdot |$ the absolute value in $mathbb{R}$.
I want to show that $Omega := H(I;S^2) = psi^{-1}(0)$ is a $C^infty$-submanifold of codimension $1$ of $H$.
I think I'm at the point where I only need to make sure that the functional $psi$ is smooth.
To compute the first derivative, I calculated the first variation which is
$$left.frac{d}{dvarepsilon}right|_{varepsilon=0} psi(w + varepsilon eta) = int_0^1 text{sgn}(| w |^2 - 1) 2 weta dt text{ for all } eta in H$$
(which is a linear and bounded functional in $eta$ and makes it the Frechét differential $D_w psi$ at $w$ that is unequal $0$ making the implicit function theorem usable for the proof of the highlighted statement above.)
In turn $Omega$ is at least a $C^1$-submanifold. Now calculating further derivatives / variations of $psi$ to show smoothness, the second variation yields
$$left.frac{d}{dvarepsilon^2}right|_{varepsilon=0} psi(w + varepsilon eta) = left.frac{d}{dvarepsilon}right|_{varepsilon=0} int_0^1 text{sgn}(| w |^2 + 2varepsilon w eta + varepsilon^2|eta |^2 - 1) 2(w + varepsiloneta) dt $$
The derivative of the sgn-function is $0$ almost everywhere, continuing above terms with product rule yields (? not sure here)
$$ int_0^1 text{sgn}(| w |^2 - 1) 2 eta dt $$
Is this correct?
proof-verification smooth-manifolds calculus-of-variations frechet-derivative global-analysis
asked Jan 27 at 14:56
NhatNhat
1,0261017
$endgroup$
1
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52
add a comment |
$begingroup$
Let $H:= H(I;mathbb{R}^3)$ be the space of $L^2$ + absolutely continuous functions with $L^2$ derivative.
For $w in H$ consider the functional
$$psi(w) = int_0^1 | | w(t) |^2 - 1 | dt$$
where $| cdot |$ denotes the standard norm in $mathbb {R^3}$ and $| cdot |$ the absolute value in $mathbb{R}$.
I want to show that $Omega := H(I;S^2) = psi^{-1}(0)$ is a $C^infty$-submanifold of codimension $1$ of $H$.
I think I'm at the point where I only need to make sure that the functional $psi$ is smooth.
To compute the first derivative, I calculated the first variation which is
$$left.frac{d}{dvarepsilon}right|_{varepsilon=0} psi(w + varepsilon eta) = int_0^1 text{sgn}(| w |^2 - 1) 2 weta dt text{ for all } eta in H$$
(which is a linear and bounded functional in $eta$ and makes it the Frechét differential $D_w psi$ at $w$ that is unequal $0$ making the implicit function theorem usable for the proof of the highlighted statement above.)
In turn $Omega$ is at least a $C^1$-submanifold. Now calculating further derivatives / variations of $psi$ to show smoothness, the second variation yields
$$left.frac{d}{dvarepsilon^2}right|_{varepsilon=0} psi(w + varepsilon eta) = left.frac{d}{dvarepsilon}right|_{varepsilon=0} int_0^1 text{sgn}(| w |^2 + 2varepsilon w eta + varepsilon^2|eta |^2 - 1) 2(w + varepsiloneta) dt $$
The derivative of the sgn-function is $0$ almost everywhere, continuing above terms with product rule yields (? not sure here)
$$ int_0^1 text{sgn}(| w |^2 - 1) 2 eta dt $$
Is this correct?
proof-verification smooth-manifolds calculus-of-variations frechet-derivative global-analysis
asked Jan 27 at 14:56
NhatNhat
1,0261017
$endgroup$
Let $H:= H(I;mathbb{R}^3)$ be the space of $L^2$ + absolutely continuous functions with $L^2$ derivative.
For $w in H$ consider the functional
$$psi(w) = int_0^1 | | w(t) |^2 - 1 | dt$$
where $| cdot |$ denotes the standard norm in $mathbb {R^3}$ and $| cdot |$ the absolute value in $mathbb{R}$.
I want to show that $Omega := H(I;S^2) = psi^{-1}(0)$ is a $C^infty$-submanifold of codimension $1$ of $H$.
I think I'm at the point where I only need to make sure that the functional $psi$ is smooth.
To compute the first derivative, I calculated the first variation which is
$$left.frac{d}{dvarepsilon}right|_{varepsilon=0} psi(w + varepsilon eta) = int_0^1 text{sgn}(| w |^2 - 1) 2 weta dt text{ for all } eta in H$$
(which is a linear and bounded functional in $eta$ and makes it the Frechét differential $D_w psi$ at $w$ that is unequal $0$ making the implicit function theorem usable for the proof of the highlighted statement above.)
In turn $Omega$ is at least a $C^1$-submanifold. Now calculating further derivatives / variations of $psi$ to show smoothness, the second variation yields
$$left.frac{d}{dvarepsilon^2}right|_{varepsilon=0} psi(w + varepsilon eta) = left.frac{d}{dvarepsilon}right|_{varepsilon=0} int_0^1 text{sgn}(| w |^2 + 2varepsilon w eta + varepsilon^2|eta |^2 - 1) 2(w + varepsiloneta) dt $$
The derivative of the sgn-function is $0$ almost everywhere, continuing above terms with product rule yields (? not sure here)
$$ int_0^1 text{sgn}(| w |^2 - 1) 2 eta dt $$
Is this correct?
proof-verification smooth-manifolds calculus-of-variations frechet-derivative global-analysis
proof-verification smooth-manifolds calculus-of-variations frechet-derivative global-analysis
asked Jan 27 at 14:56
NhatNhat
1,0261017
asked Jan 27 at 14:56
NhatNhat
1,0261017
edited Jan 29 at 17:50
Nhat
asked Jan 27 at 14:56
NhatNhat
1,0261017
asked Jan 27 at 14:56
NhatNhat
1,0261017
asked Jan 27 at 14:56
NhatNhat
1,0261017
1,0261017
1
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52
add a comment |
1
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52
1
1
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52
add a comment |
0
active
oldest
votes
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089658%2fis-the-functional-w-mapsto-int-01-wt-1-dt-c1-or-even-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089658%2fis-the-functional-w-mapsto-int-01-wt-1-dt-c1-or-even-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
$psi(w)=0$ means that $|w|equiv1$, and your differentials can't be right (because $psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth.
$endgroup$
– Lukas Geyer
Jan 27 at 20:59
$begingroup$
@LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $| w | = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0?
$endgroup$
– Nhat
Jan 27 at 21:12
$begingroup$
Yes, it does yield exactly the same preimage, mappings into the unit sphere.
$endgroup$
– Lukas Geyer
Jan 28 at 2:24
$begingroup$
While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question.
$endgroup$
– Nhat
Jan 28 at 11:52