The composition of a L^2 function and a diffeomorphism is well-defined?
0
$begingroup$
Let $f in L^2((0,1);mathbb{R})$ with respect to the Lebesgue measure.
Let $g$ be a $C^1$-diffeomorphism from $(0,2)$ into its image $g((0,2)) subset (0,1)$.
Can I define the composition function $h:(0,2) rightarrow mathbb{R}$ by
$$h(x)=f(g(x)),$$
for almost every $x in (0,2)$ ?
Is it enough to say that the measure of $g((0,2))$ is not zero ?
real-analysis measure-theory functions
asked Jan 11 at 12:15
perturbationperturbation
878
$endgroup$
|
show 1 more comment
0
$begingroup$
Let $f in L^2((0,1);mathbb{R})$ with respect to the Lebesgue measure.
Let $g$ be a $C^1$-diffeomorphism from $(0,2)$ into its image $g((0,2)) subset (0,1)$.
Can I define the composition function $h:(0,2) rightarrow mathbb{R}$ by
$$h(x)=f(g(x)),$$
for almost every $x in (0,2)$ ?
Is it enough to say that the measure of $g((0,2))$ is not zero ?
real-analysis measure-theory functions
asked Jan 11 at 12:15
perturbationperturbation
878
$endgroup$
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
1
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
1
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32
|
show 1 more comment
0
0
0
$begingroup$
Let $f in L^2((0,1);mathbb{R})$ with respect to the Lebesgue measure.
Let $g$ be a $C^1$-diffeomorphism from $(0,2)$ into its image $g((0,2)) subset (0,1)$.
Can I define the composition function $h:(0,2) rightarrow mathbb{R}$ by
$$h(x)=f(g(x)),$$
for almost every $x in (0,2)$ ?
Is it enough to say that the measure of $g((0,2))$ is not zero ?
real-analysis measure-theory functions
asked Jan 11 at 12:15
perturbationperturbation
878
$endgroup$
Let $f in L^2((0,1);mathbb{R})$ with respect to the Lebesgue measure.
Let $g$ be a $C^1$-diffeomorphism from $(0,2)$ into its image $g((0,2)) subset (0,1)$.
Can I define the composition function $h:(0,2) rightarrow mathbb{R}$ by
$$h(x)=f(g(x)),$$
for almost every $x in (0,2)$ ?
Is it enough to say that the measure of $g((0,2))$ is not zero ?
real-analysis measure-theory functions
real-analysis measure-theory functions
asked Jan 11 at 12:15
perturbationperturbation
878
asked Jan 11 at 12:15
perturbationperturbation
878
asked Jan 11 at 12:15
perturbationperturbation
878
asked Jan 11 at 12:15
perturbationperturbation
878
asked Jan 11 at 12:15
perturbationperturbation
878
878
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
1
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
1
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32
|
show 1 more comment
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
1
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
1
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
1
1
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
1
1
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32
|
show 1 more 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
});
}
});
draft saved
draft discarded
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%2f3069773%2fthe-composition-of-a-l2-function-and-a-diffeomorphism-is-well-defined%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
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
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%2f3069773%2fthe-composition-of-a-l2-function-and-a-diffeomorphism-is-well-defined%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
$begingroup$
Where is the problem? As the image of $g$ is in $(0, 1)$, the composition is well-defined. It has nothing to do with measure theory.
$endgroup$
– Stockfish
Jan 11 at 12:40
$begingroup$
Assume that $g$ sends $(0,2)$ into a point $x_0$ (in particular in $(0,1)$ as you say). Then we can not define $f(x_0)$ as $f$ is only defined a.e. no ?
$endgroup$
– perturbation
Jan 11 at 12:54
1
$begingroup$
Not exactly, if $g$ was constant for instance, the composition would not be defined, because $f$ is not a function, but an equivalence class of functions. However, if $f_1$ and $f_2$ are actual functions ae equal (in the class of $f$), then since $g$ is a diffeomorphism into its image, the pre-image under $g$ of ${f_1 neq f_2}$ has null measure, as you pointed out. So yes, the composition is well-defined. Not sure if it is in $L^2$ though but it should work provided $g’$ has some proper uniform regularity.
$endgroup$
– Mindlack
Jan 11 at 12:54
$begingroup$
@Mindlack Thank you, this helped me. I didnt remember that the elements in these equivalence classes are in fact functions defined for every $x$. Now, how do you actually prove that the measure of $g^{-1}(N)$ is equal to zero if so is the measure of $N$ ? FInally, I think the composition is indeed $L^2$ by using the change of variable theorem (I don't know how it is properly called in english).
$endgroup$
– perturbation
Jan 11 at 13:25
1
$begingroup$
By the change of variable theorem, if $N$ is any Borel set, the measure of $g(N)=int_N{|g’|}$. Since $g’$ does not vanish because $g$ is a diffeomorphism, $N$ has null measure iff $g(N)$ does. For the $L^2$ part: the integral of $|h|^2$ is the integral on the range of $g$ of $|f|^2/|g’circ g^{-1}|$, and $1/|g’circ g^{-1}|$ can be unbounded.
$endgroup$
– Mindlack
Jan 11 at 13:32