$L^p$ convergence of composition












3












$begingroup$


Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.



Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?



I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.






sequences-and-series analysis measure-theory convergence lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
    $endgroup$
    – humanStampedist
    Apr 30 '18 at 9:08










  • $begingroup$
    please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
    $endgroup$
    – JustDroppedIn
    Apr 30 '18 at 18:19










  • $begingroup$
    Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
    $endgroup$
    – user490393
    May 1 '18 at 7:16










  • $begingroup$
    Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:54










  • $begingroup$
    Is $U$ of finite measure?
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:55


















3












$begingroup$


Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.



Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?



I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.






sequences-and-series analysis measure-theory convergence lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
    $endgroup$
    – humanStampedist
    Apr 30 '18 at 9:08










  • $begingroup$
    please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
    $endgroup$
    – JustDroppedIn
    Apr 30 '18 at 18:19










  • $begingroup$
    Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
    $endgroup$
    – user490393
    May 1 '18 at 7:16










  • $begingroup$
    Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:54










  • $begingroup$
    Is $U$ of finite measure?
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:55
















3












3








3


1



$begingroup$


Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.



Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?



I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.






sequences-and-series analysis measure-theory convergence lp-spaces







share|cite|improve this question











$endgroup$




Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.



Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?



I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.






sequences-and-series analysis measure-theory convergence lp-spaces




sequences-and-series analysis measure-theory convergence lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 20:52









Davide Giraudo

128k17156268




128k17156268










asked Apr 30 '18 at 8:33









user490393user490393

265




265












  • $begingroup$
    Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
    $endgroup$
    – humanStampedist
    Apr 30 '18 at 9:08










  • $begingroup$
    please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
    $endgroup$
    – JustDroppedIn
    Apr 30 '18 at 18:19










  • $begingroup$
    Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
    $endgroup$
    – user490393
    May 1 '18 at 7:16










  • $begingroup$
    Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:54










  • $begingroup$
    Is $U$ of finite measure?
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:55




















  • $begingroup$
    Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
    $endgroup$
    – humanStampedist
    Apr 30 '18 at 9:08










  • $begingroup$
    please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
    $endgroup$
    – JustDroppedIn
    Apr 30 '18 at 18:19










  • $begingroup$
    Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
    $endgroup$
    – user490393
    May 1 '18 at 7:16










  • $begingroup$
    Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:54










  • $begingroup$
    Is $U$ of finite measure?
    $endgroup$
    – JustDroppedIn
    May 1 '18 at 14:55


















$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08




$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08












$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19




$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19












$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16




$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16












$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54




$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54












$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55






$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55












1 Answer
1






active

oldest

votes


















1












$begingroup$

For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.



Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.






share|cite|improve this answer











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    active

    oldest

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    1












    $begingroup$

    For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
    $$
    t_n(x) = frac{k}{2^n}
    $$     for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
    $$
    fmapsto fcirc t_n
    $$     is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
    $$
    1_mathbb{Q}circ t_n (x) = 1,
    $$     while
    $$
    0circ t_n(x) = 0.
    $$     Therefore, we see that composition may not be well-defined in general.



    Instead of $L^p$ of equivalence classes, we may adopt the space
    $$
    mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
    $$     of actual functions. However, we can see that the convergence can also fail in this case by the example
    $$
    1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
    $$     Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
      $$
      t_n(x) = frac{k}{2^n}
      $$       for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
      $$
      fmapsto fcirc t_n
      $$       is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
      $$
      1_mathbb{Q}circ t_n (x) = 1,
      $$       while
      $$
      0circ t_n(x) = 0.
      $$       Therefore, we see that composition may not be well-defined in general.



      Instead of $L^p$ of equivalence classes, we may adopt the space
      $$
      mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
      $$       of actual functions. However, we can see that the convergence can also fail in this case by the example
      $$
      1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
      $$       Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
        $$
        t_n(x) = frac{k}{2^n}
        $$         for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
        $$
        fmapsto fcirc t_n
        $$         is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
        $$
        1_mathbb{Q}circ t_n (x) = 1,
        $$         while
        $$
        0circ t_n(x) = 0.
        $$         Therefore, we see that composition may not be well-defined in general.



        Instead of $L^p$ of equivalence classes, we may adopt the space
        $$
        mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
        $$         of actual functions. However, we can see that the convergence can also fail in this case by the example
        $$
        1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
        $$         Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.






        share|cite|improve this answer











        $endgroup$



        For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
        $$
        t_n(x) = frac{k}{2^n}
        $$         for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
        $$
        fmapsto fcirc t_n
        $$         is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
        $$
        1_mathbb{Q}circ t_n (x) = 1,
        $$         while
        $$
        0circ t_n(x) = 0.
        $$         Therefore, we see that composition may not be well-defined in general.



        Instead of $L^p$ of equivalence classes, we may adopt the space
        $$
        mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
        $$         of actual functions. However, we can see that the convergence can also fail in this case by the example
        $$
        1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
        $$         Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 6:12

























        answered Dec 11 '18 at 21:34









        SongSong

        18.5k21651




        18.5k21651






























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