For which $p in [1,infty]$ does $g, g_{A}$ and $g_{A^{c}} in mathcal{L}^p$ hold?












1












$begingroup$


Let $s in mathbb R, d in mathbb N$, $g:mathbb R^{d}to bar{mathbb R}, x mapsto |x|^s$
.



Set $E_{d}:={xin mathbb R^{d}:|x|leq 1}$



Determine $p in [1,infty]$ for which $g, gchi_{E_{d}}$ and $gchi_{E_{d}^{c}} in mathcal{L}^{p}$



Idea: Assume $g geq 0$ without loss of generality otherwise: $g:=g_{+}-g_{-}$



for $p =1$



$int_{mathbb R^{d}}gdlambda^{d}(x)=int_{mathbb R^{d}}x^sdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}x^sdlambda)...dlambda=int_{mathbb R}...frac{x^{s+1}}{s+1}vert_{-infty}^{infty}...dlambda$ so $notin mathcal{L}^{1} $



And if $gnotin mathcal{L}^{1}$ I want to say $g notin mathcal{L}^{p},forall pin]1,infty]$ but this does not seem possible given that fact that $lambda^{d}(mathbb R^{d})=infty$, so this shortcut is not possible.



Next $int_{E^{d}}gdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}gchi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}x^{s}chi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}frac{x^{s+1}}{s+1}|^{1}_{-1}dlambda)...dlambda$



and $frac{x^{s+1}}{s+1}|^{1}_{-1}=frac{1}{s+1}-frac{(-1)^{s+1}}{s+1}$ and if $s $ us an odd number then we get $0$, and therefore $int_{E^{d}}gdlambda^{d}(x)=0$, I for all other cases of $s in mathbb R$, we get $gchi_{E_{d}}in mathcal{L}^1$ but I am unsure on how to reason it precisely.



On $int_{(E^{d})^{c}}gdlambda^{d}(x)$, I am lost and for all cases $p in ]1,infty]$ I am unsure on what to do for all three function, particulary given:



$(int_{mathbb R^{d}}x^{sp}dlambda^{d}(x))^{frac{1}{p}}$, where $p in ]1,infty]$



Any guidance on this problem is greatly appreciated.






real-analysis integration measure-theory lp-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $s in mathbb R, d in mathbb N$, $g:mathbb R^{d}to bar{mathbb R}, x mapsto |x|^s$
    .



    Set $E_{d}:={xin mathbb R^{d}:|x|leq 1}$



    Determine $p in [1,infty]$ for which $g, gchi_{E_{d}}$ and $gchi_{E_{d}^{c}} in mathcal{L}^{p}$



    Idea: Assume $g geq 0$ without loss of generality otherwise: $g:=g_{+}-g_{-}$



    for $p =1$



    $int_{mathbb R^{d}}gdlambda^{d}(x)=int_{mathbb R^{d}}x^sdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}x^sdlambda)...dlambda=int_{mathbb R}...frac{x^{s+1}}{s+1}vert_{-infty}^{infty}...dlambda$ so $notin mathcal{L}^{1} $



    And if $gnotin mathcal{L}^{1}$ I want to say $g notin mathcal{L}^{p},forall pin]1,infty]$ but this does not seem possible given that fact that $lambda^{d}(mathbb R^{d})=infty$, so this shortcut is not possible.



    Next $int_{E^{d}}gdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}gchi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}x^{s}chi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}frac{x^{s+1}}{s+1}|^{1}_{-1}dlambda)...dlambda$



    and $frac{x^{s+1}}{s+1}|^{1}_{-1}=frac{1}{s+1}-frac{(-1)^{s+1}}{s+1}$ and if $s $ us an odd number then we get $0$, and therefore $int_{E^{d}}gdlambda^{d}(x)=0$, I for all other cases of $s in mathbb R$, we get $gchi_{E_{d}}in mathcal{L}^1$ but I am unsure on how to reason it precisely.



    On $int_{(E^{d})^{c}}gdlambda^{d}(x)$, I am lost and for all cases $p in ]1,infty]$ I am unsure on what to do for all three function, particulary given:



    $(int_{mathbb R^{d}}x^{sp}dlambda^{d}(x))^{frac{1}{p}}$, where $p in ]1,infty]$



    Any guidance on this problem is greatly appreciated.






    real-analysis integration measure-theory lp-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $s in mathbb R, d in mathbb N$, $g:mathbb R^{d}to bar{mathbb R}, x mapsto |x|^s$
      .



      Set $E_{d}:={xin mathbb R^{d}:|x|leq 1}$



      Determine $p in [1,infty]$ for which $g, gchi_{E_{d}}$ and $gchi_{E_{d}^{c}} in mathcal{L}^{p}$



      Idea: Assume $g geq 0$ without loss of generality otherwise: $g:=g_{+}-g_{-}$



      for $p =1$



      $int_{mathbb R^{d}}gdlambda^{d}(x)=int_{mathbb R^{d}}x^sdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}x^sdlambda)...dlambda=int_{mathbb R}...frac{x^{s+1}}{s+1}vert_{-infty}^{infty}...dlambda$ so $notin mathcal{L}^{1} $



      And if $gnotin mathcal{L}^{1}$ I want to say $g notin mathcal{L}^{p},forall pin]1,infty]$ but this does not seem possible given that fact that $lambda^{d}(mathbb R^{d})=infty$, so this shortcut is not possible.



      Next $int_{E^{d}}gdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}gchi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}x^{s}chi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}frac{x^{s+1}}{s+1}|^{1}_{-1}dlambda)...dlambda$



      and $frac{x^{s+1}}{s+1}|^{1}_{-1}=frac{1}{s+1}-frac{(-1)^{s+1}}{s+1}$ and if $s $ us an odd number then we get $0$, and therefore $int_{E^{d}}gdlambda^{d}(x)=0$, I for all other cases of $s in mathbb R$, we get $gchi_{E_{d}}in mathcal{L}^1$ but I am unsure on how to reason it precisely.



      On $int_{(E^{d})^{c}}gdlambda^{d}(x)$, I am lost and for all cases $p in ]1,infty]$ I am unsure on what to do for all three function, particulary given:



      $(int_{mathbb R^{d}}x^{sp}dlambda^{d}(x))^{frac{1}{p}}$, where $p in ]1,infty]$



      Any guidance on this problem is greatly appreciated.






      real-analysis integration measure-theory lp-spaces







      share|cite|improve this question









      $endgroup$




      Let $s in mathbb R, d in mathbb N$, $g:mathbb R^{d}to bar{mathbb R}, x mapsto |x|^s$
      .



      Set $E_{d}:={xin mathbb R^{d}:|x|leq 1}$



      Determine $p in [1,infty]$ for which $g, gchi_{E_{d}}$ and $gchi_{E_{d}^{c}} in mathcal{L}^{p}$



      Idea: Assume $g geq 0$ without loss of generality otherwise: $g:=g_{+}-g_{-}$



      for $p =1$



      $int_{mathbb R^{d}}gdlambda^{d}(x)=int_{mathbb R^{d}}x^sdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}x^sdlambda)...dlambda=int_{mathbb R}...frac{x^{s+1}}{s+1}vert_{-infty}^{infty}...dlambda$ so $notin mathcal{L}^{1} $



      And if $gnotin mathcal{L}^{1}$ I want to say $g notin mathcal{L}^{p},forall pin]1,infty]$ but this does not seem possible given that fact that $lambda^{d}(mathbb R^{d})=infty$, so this shortcut is not possible.



      Next $int_{E^{d}}gdlambda^{d}(x)=int_{mathbb R}...(int_{mathbb R}gchi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}x^{s}chi_{[-1,1]}dlambda)...dlambda=int_{mathbb R}...(int_{mathbb R}frac{x^{s+1}}{s+1}|^{1}_{-1}dlambda)...dlambda$



      and $frac{x^{s+1}}{s+1}|^{1}_{-1}=frac{1}{s+1}-frac{(-1)^{s+1}}{s+1}$ and if $s $ us an odd number then we get $0$, and therefore $int_{E^{d}}gdlambda^{d}(x)=0$, I for all other cases of $s in mathbb R$, we get $gchi_{E_{d}}in mathcal{L}^1$ but I am unsure on how to reason it precisely.



      On $int_{(E^{d})^{c}}gdlambda^{d}(x)$, I am lost and for all cases $p in ]1,infty]$ I am unsure on what to do for all three function, particulary given:



      $(int_{mathbb R^{d}}x^{sp}dlambda^{d}(x))^{frac{1}{p}}$, where $p in ]1,infty]$



      Any guidance on this problem is greatly appreciated.






      real-analysis integration measure-theory lp-spaces




      real-analysis integration 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 Jan 8 at 14:13









      MinaThumaMinaThuma

      1968




      1968






















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          $begingroup$

          A problem in your approach is that you wrote $x^s$ for an element of $mathbb R^d$, while we want to work to the norm of $x$ to the power $s$.



          Here are some steps to solve the problem:




          1. First treat completely the one dimensional case, which consists in studying the convergence of $int_0^1 x^{ps}mathrm dx$ and $int_1^{+infty} x^{ps}mathrm dx$.

          2. The function $xmapsto leftlvert xrightrvert^{ps}$ is radial, in sense that it depends only on the norm of $x$ hence (by a change in polar coordinates for example) $$int_{E_d}leftlvert xrightrvert^{ps}=c_d int_{0}^1t^{ps}t^{n-1}mathrm dt;$$
            $$int_{E_d^c}leftlvert xrightrvert^{ps}=c_d int_{1}^{+infty}t^{ps}t^{n-1}mathrm dt.$$






          share|cite|improve this answer











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            $begingroup$

            A problem in your approach is that you wrote $x^s$ for an element of $mathbb R^d$, while we want to work to the norm of $x$ to the power $s$.



            Here are some steps to solve the problem:




            1. First treat completely the one dimensional case, which consists in studying the convergence of $int_0^1 x^{ps}mathrm dx$ and $int_1^{+infty} x^{ps}mathrm dx$.

            2. The function $xmapsto leftlvert xrightrvert^{ps}$ is radial, in sense that it depends only on the norm of $x$ hence (by a change in polar coordinates for example) $$int_{E_d}leftlvert xrightrvert^{ps}=c_d int_{0}^1t^{ps}t^{n-1}mathrm dt;$$
              $$int_{E_d^c}leftlvert xrightrvert^{ps}=c_d int_{1}^{+infty}t^{ps}t^{n-1}mathrm dt.$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              A problem in your approach is that you wrote $x^s$ for an element of $mathbb R^d$, while we want to work to the norm of $x$ to the power $s$.



              Here are some steps to solve the problem:




              1. First treat completely the one dimensional case, which consists in studying the convergence of $int_0^1 x^{ps}mathrm dx$ and $int_1^{+infty} x^{ps}mathrm dx$.

              2. The function $xmapsto leftlvert xrightrvert^{ps}$ is radial, in sense that it depends only on the norm of $x$ hence (by a change in polar coordinates for example) $$int_{E_d}leftlvert xrightrvert^{ps}=c_d int_{0}^1t^{ps}t^{n-1}mathrm dt;$$
                $$int_{E_d^c}leftlvert xrightrvert^{ps}=c_d int_{1}^{+infty}t^{ps}t^{n-1}mathrm dt.$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                A problem in your approach is that you wrote $x^s$ for an element of $mathbb R^d$, while we want to work to the norm of $x$ to the power $s$.



                Here are some steps to solve the problem:




                1. First treat completely the one dimensional case, which consists in studying the convergence of $int_0^1 x^{ps}mathrm dx$ and $int_1^{+infty} x^{ps}mathrm dx$.

                2. The function $xmapsto leftlvert xrightrvert^{ps}$ is radial, in sense that it depends only on the norm of $x$ hence (by a change in polar coordinates for example) $$int_{E_d}leftlvert xrightrvert^{ps}=c_d int_{0}^1t^{ps}t^{n-1}mathrm dt;$$
                  $$int_{E_d^c}leftlvert xrightrvert^{ps}=c_d int_{1}^{+infty}t^{ps}t^{n-1}mathrm dt.$$






                share|cite|improve this answer











                $endgroup$



                A problem in your approach is that you wrote $x^s$ for an element of $mathbb R^d$, while we want to work to the norm of $x$ to the power $s$.



                Here are some steps to solve the problem:




                1. First treat completely the one dimensional case, which consists in studying the convergence of $int_0^1 x^{ps}mathrm dx$ and $int_1^{+infty} x^{ps}mathrm dx$.

                2. The function $xmapsto leftlvert xrightrvert^{ps}$ is radial, in sense that it depends only on the norm of $x$ hence (by a change in polar coordinates for example) $$int_{E_d}leftlvert xrightrvert^{ps}=c_d int_{0}^1t^{ps}t^{n-1}mathrm dt;$$
                  $$int_{E_d^c}leftlvert xrightrvert^{ps}=c_d int_{1}^{+infty}t^{ps}t^{n-1}mathrm dt.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 21:30

























                answered Jan 8 at 21:34









                Davide GiraudoDavide Giraudo

                128k17156268




                128k17156268






























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