Which convergence when discussing density in $L^p$?












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For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?






measure-theory lp-spaces







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  • 2




    $begingroup$
    $L^p$ convergence
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:13






  • 2




    $begingroup$
    When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
    $endgroup$
    – Jakobian
    Dec 28 '18 at 21:16










  • $begingroup$
    All right, thanks a lot to both.
    $endgroup$
    – Bermudes
    Dec 28 '18 at 21:27
















1












$begingroup$


For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?






measure-theory lp-spaces







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $L^p$ convergence
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:13






  • 2




    $begingroup$
    When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
    $endgroup$
    – Jakobian
    Dec 28 '18 at 21:16










  • $begingroup$
    All right, thanks a lot to both.
    $endgroup$
    – Bermudes
    Dec 28 '18 at 21:27














1












1








1





$begingroup$


For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?






measure-theory lp-spaces







share|cite|improve this question









$endgroup$




For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?






measure-theory lp-spaces




measure-theory lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 21:11









BermudesBermudes

18713




18713








  • 2




    $begingroup$
    $L^p$ convergence
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:13






  • 2




    $begingroup$
    When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
    $endgroup$
    – Jakobian
    Dec 28 '18 at 21:16










  • $begingroup$
    All right, thanks a lot to both.
    $endgroup$
    – Bermudes
    Dec 28 '18 at 21:27














  • 2




    $begingroup$
    $L^p$ convergence
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:13






  • 2




    $begingroup$
    When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
    $endgroup$
    – Jakobian
    Dec 28 '18 at 21:16










  • $begingroup$
    All right, thanks a lot to both.
    $endgroup$
    – Bermudes
    Dec 28 '18 at 21:27








2




2




$begingroup$
$L^p$ convergence
$endgroup$
– mathworker21
Dec 28 '18 at 21:13




$begingroup$
$L^p$ convergence
$endgroup$
– mathworker21
Dec 28 '18 at 21:13




2




2




$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16




$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16












$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27




$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27










1 Answer
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A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.






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    $begingroup$

    A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.






        share|cite|improve this answer









        $endgroup$



        A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 21 at 10:07









        Davide GiraudoDavide Giraudo

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        127k16151264






























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