What does mean : element of $L^p$ are equivalence class rather than function.












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What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $fin L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent class" ?










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  • 1




    $begingroup$
    If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:32
















0












$begingroup$


What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $fin L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent class" ?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:32














0












0








0





$begingroup$


What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $fin L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent class" ?










share|cite|improve this question









$endgroup$




What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $fin L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent class" ?







lp-spaces






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asked Dec 11 '18 at 12:28









NewMathNewMath

4059




4059








  • 1




    $begingroup$
    If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:32














  • 1




    $begingroup$
    If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:32








1




1




$begingroup$
If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
$endgroup$
– Surb
Dec 11 '18 at 12:32




$begingroup$
If you define a function in $L^p$, it's defined up to a set of measure $0$. I.e. if $f(x)=e^{-x^2}$ in $L^p$, this mean that $f(x)=e^{-x^2}$ a.e. The problem of this definition, it's if you fix a $x$, you don't really know what is $f(x)$. You cannot really evaluate the function at a given point.
$endgroup$
– Surb
Dec 11 '18 at 12:32










3 Answers
3






active

oldest

votes


















2












$begingroup$

This kind of thing occurs frequently in mathematics.



You may say a "rational number" $frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $frac{7}{12}=frac{-7}{-12}=frac{21}{36}=dots$.



You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.overline{9} = 1$ or not.



You may say that $mathbb C = mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
    $endgroup$
    – NewMath
    Dec 11 '18 at 13:32










  • $begingroup$
    There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
    $endgroup$
    – nicomezi
    Dec 11 '18 at 14:15



















0












$begingroup$

Let $X$ be a measure space. IF $f,gcolon Xlongrightarrowmathbb R$ are measurable functions, we usually identify $f$ with $g$ whe the set ${xin X,|,f(x)neq g(x)}$ has null measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$lVert f-grVert_p=0iff fsim g,$$where $sim$ is the equivalence relation that I have described. Otherwise, we could have distinct function such that the distance between them would be $0$. That is, $bigl(L^p(X),lVertcdotrVert_0bigr)$ would not be a metric space.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
    $endgroup$
    – NewMath
    Dec 11 '18 at 12:34








  • 1




    $begingroup$
    @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:37



















0












$begingroup$

Let's say you have a measurable function $f colon X rightarrow mathbb{R}$ on $(X,mu,Sigma)$. You can say two distinct things about $f$:




  1. The function $f$ is $p$-integrable meaning $int_X |f|^p dmu < infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,mu, Sigma)$.

  2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes
    $$ mathcal{L}^p(X,mu,Sigma) := L^p(X,mu,Sigma) / sim $$
    where $f sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $mathcal{L}^p$, you actually mean that $[f] in mathcal{L}^p(X,musigma)$.


The reason the $mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm



$$ | f |_p = int_X |f|^p , du. $$



This is not a norm on $L^p$ but only a semi-norm since it is possible that $| f |_p = 0$ even though $f neq 0$. By taking the quotient, the semi-norm descends to an honest to god norm on the quotient space $mathcal{L}^p$.





Thus, an element $f in L^p(X,mu,Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x in X$. However, an element $[f] in mathcal{L}^p(X,mu,Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) neq g(x)$ so the operation of evaluating an equivalence class at a point $x in X$ is not well-defined).






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    3 Answers
    3






    active

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    3 Answers
    3






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    active

    oldest

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    2












    $begingroup$

    This kind of thing occurs frequently in mathematics.



    You may say a "rational number" $frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $frac{7}{12}=frac{-7}{-12}=frac{21}{36}=dots$.



    You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.overline{9} = 1$ or not.



    You may say that $mathbb C = mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
      $endgroup$
      – NewMath
      Dec 11 '18 at 13:32










    • $begingroup$
      There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
      $endgroup$
      – nicomezi
      Dec 11 '18 at 14:15
















    2












    $begingroup$

    This kind of thing occurs frequently in mathematics.



    You may say a "rational number" $frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $frac{7}{12}=frac{-7}{-12}=frac{21}{36}=dots$.



    You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.overline{9} = 1$ or not.



    You may say that $mathbb C = mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
      $endgroup$
      – NewMath
      Dec 11 '18 at 13:32










    • $begingroup$
      There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
      $endgroup$
      – nicomezi
      Dec 11 '18 at 14:15














    2












    2








    2





    $begingroup$

    This kind of thing occurs frequently in mathematics.



    You may say a "rational number" $frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $frac{7}{12}=frac{-7}{-12}=frac{21}{36}=dots$.



    You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.overline{9} = 1$ or not.



    You may say that $mathbb C = mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".






    share|cite|improve this answer









    $endgroup$



    This kind of thing occurs frequently in mathematics.



    You may say a "rational number" $frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $frac{7}{12}=frac{-7}{-12}=frac{21}{36}=dots$.



    You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.overline{9} = 1$ or not.



    You may say that $mathbb C = mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 12:52









    GEdgarGEdgar

    63.3k268172




    63.3k268172












    • $begingroup$
      Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
      $endgroup$
      – NewMath
      Dec 11 '18 at 13:32










    • $begingroup$
      There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
      $endgroup$
      – nicomezi
      Dec 11 '18 at 14:15


















    • $begingroup$
      Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
      $endgroup$
      – NewMath
      Dec 11 '18 at 13:32










    • $begingroup$
      There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
      $endgroup$
      – nicomezi
      Dec 11 '18 at 14:15
















    $begingroup$
    Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
    $endgroup$
    – NewMath
    Dec 11 '18 at 13:32




    $begingroup$
    Sorry, but I don't understand why $7/12$ is an equivalence class, neither why a real number is an equivalence class with Cauchy sequence... neither why $mathbb C={a+ibmid a,binmathbb R}$ is an equivalent class... and this form is only for beginner with $i^2=-1$. It's not true that $mathbb C={a+ibmid a,binmathbb R}$ ? uuggghhh...
    $endgroup$
    – NewMath
    Dec 11 '18 at 13:32












    $begingroup$
    There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
    $endgroup$
    – nicomezi
    Dec 11 '18 at 14:15




    $begingroup$
    There are several ways to construct $mathbb{C}$, you can even represent complex numbers by matrices. But I think it should appear clearly to you that if we define $C$ by a quotient field, the elements of this field are equivalence classes. The other examples may require some hindsight on the question.
    $endgroup$
    – nicomezi
    Dec 11 '18 at 14:15











    0












    $begingroup$

    Let $X$ be a measure space. IF $f,gcolon Xlongrightarrowmathbb R$ are measurable functions, we usually identify $f$ with $g$ whe the set ${xin X,|,f(x)neq g(x)}$ has null measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$lVert f-grVert_p=0iff fsim g,$$where $sim$ is the equivalence relation that I have described. Otherwise, we could have distinct function such that the distance between them would be $0$. That is, $bigl(L^p(X),lVertcdotrVert_0bigr)$ would not be a metric space.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
      $endgroup$
      – NewMath
      Dec 11 '18 at 12:34








    • 1




      $begingroup$
      @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
      $endgroup$
      – Surb
      Dec 11 '18 at 12:37
















    0












    $begingroup$

    Let $X$ be a measure space. IF $f,gcolon Xlongrightarrowmathbb R$ are measurable functions, we usually identify $f$ with $g$ whe the set ${xin X,|,f(x)neq g(x)}$ has null measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$lVert f-grVert_p=0iff fsim g,$$where $sim$ is the equivalence relation that I have described. Otherwise, we could have distinct function such that the distance between them would be $0$. That is, $bigl(L^p(X),lVertcdotrVert_0bigr)$ would not be a metric space.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
      $endgroup$
      – NewMath
      Dec 11 '18 at 12:34








    • 1




      $begingroup$
      @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
      $endgroup$
      – Surb
      Dec 11 '18 at 12:37














    0












    0








    0





    $begingroup$

    Let $X$ be a measure space. IF $f,gcolon Xlongrightarrowmathbb R$ are measurable functions, we usually identify $f$ with $g$ whe the set ${xin X,|,f(x)neq g(x)}$ has null measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$lVert f-grVert_p=0iff fsim g,$$where $sim$ is the equivalence relation that I have described. Otherwise, we could have distinct function such that the distance between them would be $0$. That is, $bigl(L^p(X),lVertcdotrVert_0bigr)$ would not be a metric space.






    share|cite|improve this answer









    $endgroup$



    Let $X$ be a measure space. IF $f,gcolon Xlongrightarrowmathbb R$ are measurable functions, we usually identify $f$ with $g$ whe the set ${xin X,|,f(x)neq g(x)}$ has null measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$lVert f-grVert_p=0iff fsim g,$$where $sim$ is the equivalence relation that I have described. Otherwise, we could have distinct function such that the distance between them would be $0$. That is, $bigl(L^p(X),lVertcdotrVert_0bigr)$ would not be a metric space.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 12:32









    José Carlos SantosJosé Carlos Santos

    171k23132240




    171k23132240












    • $begingroup$
      I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
      $endgroup$
      – NewMath
      Dec 11 '18 at 12:34








    • 1




      $begingroup$
      @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
      $endgroup$
      – Surb
      Dec 11 '18 at 12:37


















    • $begingroup$
      I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
      $endgroup$
      – NewMath
      Dec 11 '18 at 12:34








    • 1




      $begingroup$
      @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
      $endgroup$
      – Surb
      Dec 11 '18 at 12:37
















    $begingroup$
    I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
    $endgroup$
    – NewMath
    Dec 11 '18 at 12:34






    $begingroup$
    I understand the "reason" of the relation $f=g$ a.e., what I don't understand it's element in $L^p$ are not really functions... For me there are functions, and I don't really understand why they would not be functions... (thank you for your answer btw)
    $endgroup$
    – NewMath
    Dec 11 '18 at 12:34






    1




    1




    $begingroup$
    @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:37




    $begingroup$
    @NewMath: As I said in my comment below your question : If $f$ is defined in $L^p$ (i.e. up to a set of measure 0), you can't evaluate $f$ at a given point.
    $endgroup$
    – Surb
    Dec 11 '18 at 12:37











    0












    $begingroup$

    Let's say you have a measurable function $f colon X rightarrow mathbb{R}$ on $(X,mu,Sigma)$. You can say two distinct things about $f$:




    1. The function $f$ is $p$-integrable meaning $int_X |f|^p dmu < infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,mu, Sigma)$.

    2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes
      $$ mathcal{L}^p(X,mu,Sigma) := L^p(X,mu,Sigma) / sim $$
      where $f sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $mathcal{L}^p$, you actually mean that $[f] in mathcal{L}^p(X,musigma)$.


    The reason the $mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm



    $$ | f |_p = int_X |f|^p , du. $$



    This is not a norm on $L^p$ but only a semi-norm since it is possible that $| f |_p = 0$ even though $f neq 0$. By taking the quotient, the semi-norm descends to an honest to god norm on the quotient space $mathcal{L}^p$.





    Thus, an element $f in L^p(X,mu,Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x in X$. However, an element $[f] in mathcal{L}^p(X,mu,Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) neq g(x)$ so the operation of evaluating an equivalence class at a point $x in X$ is not well-defined).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let's say you have a measurable function $f colon X rightarrow mathbb{R}$ on $(X,mu,Sigma)$. You can say two distinct things about $f$:




      1. The function $f$ is $p$-integrable meaning $int_X |f|^p dmu < infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,mu, Sigma)$.

      2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes
        $$ mathcal{L}^p(X,mu,Sigma) := L^p(X,mu,Sigma) / sim $$
        where $f sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $mathcal{L}^p$, you actually mean that $[f] in mathcal{L}^p(X,musigma)$.


      The reason the $mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm



      $$ | f |_p = int_X |f|^p , du. $$



      This is not a norm on $L^p$ but only a semi-norm since it is possible that $| f |_p = 0$ even though $f neq 0$. By taking the quotient, the semi-norm descends to an honest to god norm on the quotient space $mathcal{L}^p$.





      Thus, an element $f in L^p(X,mu,Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x in X$. However, an element $[f] in mathcal{L}^p(X,mu,Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) neq g(x)$ so the operation of evaluating an equivalence class at a point $x in X$ is not well-defined).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let's say you have a measurable function $f colon X rightarrow mathbb{R}$ on $(X,mu,Sigma)$. You can say two distinct things about $f$:




        1. The function $f$ is $p$-integrable meaning $int_X |f|^p dmu < infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,mu, Sigma)$.

        2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes
          $$ mathcal{L}^p(X,mu,Sigma) := L^p(X,mu,Sigma) / sim $$
          where $f sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $mathcal{L}^p$, you actually mean that $[f] in mathcal{L}^p(X,musigma)$.


        The reason the $mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm



        $$ | f |_p = int_X |f|^p , du. $$



        This is not a norm on $L^p$ but only a semi-norm since it is possible that $| f |_p = 0$ even though $f neq 0$. By taking the quotient, the semi-norm descends to an honest to god norm on the quotient space $mathcal{L}^p$.





        Thus, an element $f in L^p(X,mu,Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x in X$. However, an element $[f] in mathcal{L}^p(X,mu,Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) neq g(x)$ so the operation of evaluating an equivalence class at a point $x in X$ is not well-defined).






        share|cite|improve this answer









        $endgroup$



        Let's say you have a measurable function $f colon X rightarrow mathbb{R}$ on $(X,mu,Sigma)$. You can say two distinct things about $f$:




        1. The function $f$ is $p$-integrable meaning $int_X |f|^p dmu < infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,mu, Sigma)$.

        2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes
          $$ mathcal{L}^p(X,mu,Sigma) := L^p(X,mu,Sigma) / sim $$
          where $f sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $mathcal{L}^p$, you actually mean that $[f] in mathcal{L}^p(X,musigma)$.


        The reason the $mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm



        $$ | f |_p = int_X |f|^p , du. $$



        This is not a norm on $L^p$ but only a semi-norm since it is possible that $| f |_p = 0$ even though $f neq 0$. By taking the quotient, the semi-norm descends to an honest to god norm on the quotient space $mathcal{L}^p$.





        Thus, an element $f in L^p(X,mu,Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x in X$. However, an element $[f] in mathcal{L}^p(X,mu,Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) neq g(x)$ so the operation of evaluating an equivalence class at a point $x in X$ is not well-defined).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 12:50









        levaplevap

        47.9k33274




        47.9k33274






























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