What's the meaning of denoting $ell^p$ as $ell^p(mathbb{N})$?












2












$begingroup$


What's the meaning of denoting $ell^p$ as $ell^p(mathbb{N})$?



I read that it's like $ell^p$ "over $mathbb{N}$". But $l^p$ is sequences indexed by $mathbb{N}$.



So it seems weird to treat the index set as an input to $ell^p$? Since if $x_n$ is a function, then it could have much more as input or domain than $mathbb{N}$.






notation lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
    $endgroup$
    – the_candyman
    Jan 27 at 20:48












  • $begingroup$
    @the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
    $endgroup$
    – mavavilj
    Jan 27 at 20:49
















2












$begingroup$


What's the meaning of denoting $ell^p$ as $ell^p(mathbb{N})$?



I read that it's like $ell^p$ "over $mathbb{N}$". But $l^p$ is sequences indexed by $mathbb{N}$.



So it seems weird to treat the index set as an input to $ell^p$? Since if $x_n$ is a function, then it could have much more as input or domain than $mathbb{N}$.






notation lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
    $endgroup$
    – the_candyman
    Jan 27 at 20:48












  • $begingroup$
    @the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
    $endgroup$
    – mavavilj
    Jan 27 at 20:49














2












2








2





$begingroup$


What's the meaning of denoting $ell^p$ as $ell^p(mathbb{N})$?



I read that it's like $ell^p$ "over $mathbb{N}$". But $l^p$ is sequences indexed by $mathbb{N}$.



So it seems weird to treat the index set as an input to $ell^p$? Since if $x_n$ is a function, then it could have much more as input or domain than $mathbb{N}$.






notation lp-spaces







share|cite|improve this question











$endgroup$




What's the meaning of denoting $ell^p$ as $ell^p(mathbb{N})$?



I read that it's like $ell^p$ "over $mathbb{N}$". But $l^p$ is sequences indexed by $mathbb{N}$.



So it seems weird to treat the index set as an input to $ell^p$? Since if $x_n$ is a function, then it could have much more as input or domain than $mathbb{N}$.






notation lp-spaces




notation lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 20:54









José Carlos Santos

170k23132238




170k23132238










asked Jan 27 at 20:39









mavaviljmavavilj

2,85611137




2,85611137












  • $begingroup$
    You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
    $endgroup$
    – the_candyman
    Jan 27 at 20:48












  • $begingroup$
    @the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
    $endgroup$
    – mavavilj
    Jan 27 at 20:49


















  • $begingroup$
    You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
    $endgroup$
    – the_candyman
    Jan 27 at 20:48












  • $begingroup$
    @the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
    $endgroup$
    – mavavilj
    Jan 27 at 20:49
















$begingroup$
You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
$endgroup$
– the_candyman
Jan 27 at 20:48






$begingroup$
You could also have $ell^p(mathbb{Z})$, or $ell^p(I)$, where $I subset mathbb{N}$, no?
$endgroup$
– the_candyman
Jan 27 at 20:48














$begingroup$
@the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
$endgroup$
– mavavilj
Jan 27 at 20:49




$begingroup$
@the_candyman Sure but I was confused at first when seeing $ell^p(mathbb{N})$. Because I thought that it's different than $ell^p$.
$endgroup$
– mavavilj
Jan 27 at 20:49










2 Answers
2






active

oldest

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2












$begingroup$

Because, given a set $X$, $ell^p(X)=L^p(X,mathcal{P}(X),Omega)$, where $Omega$ is the counting measure (that is, $Omega(A)=#A$).






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can define $ell ^p$ to be a space of sequences with any set of indices (usually it is $mathbb{N}$), so the space is depended on the set of indices.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But it is dependent on the function of $x_n(x)$ as well?
      $endgroup$
      – mavavilj
      Jan 27 at 20:50










    • $begingroup$
      No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
      $endgroup$
      – MOMO
      Jan 27 at 21:12













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Because, given a set $X$, $ell^p(X)=L^p(X,mathcal{P}(X),Omega)$, where $Omega$ is the counting measure (that is, $Omega(A)=#A$).






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Because, given a set $X$, $ell^p(X)=L^p(X,mathcal{P}(X),Omega)$, where $Omega$ is the counting measure (that is, $Omega(A)=#A$).






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Because, given a set $X$, $ell^p(X)=L^p(X,mathcal{P}(X),Omega)$, where $Omega$ is the counting measure (that is, $Omega(A)=#A$).






        share|cite|improve this answer









        $endgroup$



        Because, given a set $X$, $ell^p(X)=L^p(X,mathcal{P}(X),Omega)$, where $Omega$ is the counting measure (that is, $Omega(A)=#A$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 20:50









        José Carlos SantosJosé Carlos Santos

        170k23132238




        170k23132238























            0












            $begingroup$

            You can define $ell ^p$ to be a space of sequences with any set of indices (usually it is $mathbb{N}$), so the space is depended on the set of indices.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But it is dependent on the function of $x_n(x)$ as well?
              $endgroup$
              – mavavilj
              Jan 27 at 20:50










            • $begingroup$
              No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
              $endgroup$
              – MOMO
              Jan 27 at 21:12


















            0












            $begingroup$

            You can define $ell ^p$ to be a space of sequences with any set of indices (usually it is $mathbb{N}$), so the space is depended on the set of indices.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But it is dependent on the function of $x_n(x)$ as well?
              $endgroup$
              – mavavilj
              Jan 27 at 20:50










            • $begingroup$
              No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
              $endgroup$
              – MOMO
              Jan 27 at 21:12
















            0












            0








            0





            $begingroup$

            You can define $ell ^p$ to be a space of sequences with any set of indices (usually it is $mathbb{N}$), so the space is depended on the set of indices.






            share|cite|improve this answer









            $endgroup$



            You can define $ell ^p$ to be a space of sequences with any set of indices (usually it is $mathbb{N}$), so the space is depended on the set of indices.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 27 at 20:49









            MOMOMOMO

            717312




            717312












            • $begingroup$
              But it is dependent on the function of $x_n(x)$ as well?
              $endgroup$
              – mavavilj
              Jan 27 at 20:50










            • $begingroup$
              No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
              $endgroup$
              – MOMO
              Jan 27 at 21:12




















            • $begingroup$
              But it is dependent on the function of $x_n(x)$ as well?
              $endgroup$
              – mavavilj
              Jan 27 at 20:50










            • $begingroup$
              No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
              $endgroup$
              – MOMO
              Jan 27 at 21:12


















            $begingroup$
            But it is dependent on the function of $x_n(x)$ as well?
            $endgroup$
            – mavavilj
            Jan 27 at 20:50




            $begingroup$
            But it is dependent on the function of $x_n(x)$ as well?
            $endgroup$
            – mavavilj
            Jan 27 at 20:50












            $begingroup$
            No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
            $endgroup$
            – MOMO
            Jan 27 at 21:12






            $begingroup$
            No. It is depended only on the set of indices. Given a set of indices $I$, the space $ell^p(I)$ is well-defined as the space of all sequences $(a_i)_{iin I}$ satisfying $sum_{iin I}|a_i|^p<infty$
            $endgroup$
            – MOMO
            Jan 27 at 21:12




















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