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 ?










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$endgroup$












  • $begingroup$
    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
















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 ?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














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 ?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










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