neitheir $L^p((0, infty),mathbb{C}) subset L^q((0, infty),mathbb{C})$ nor $L^q((0, infty),mathbb{C}) subset...











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I have to show that for $1le p < q <infty$

neitheir
$L^p((0, infty),mathbb{C}) subset L^q((0, infty),mathbb{C})$

nor
$L^q((0, infty),mathbb{C}) subset L^p((0, infty),mathbb{C})$.



I previously had to show, that there is no $a in mathbb{R}$ with which the function f:$(0, infty) to mathbb{R}$ f(x)=x^a

would be lebesgue integrable and there is a hint that i should use this information, but i have no idea how to do so.



I would be very thankfull for any kind of help.










share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I have to show that for $1le p < q <infty$

    neitheir
    $L^p((0, infty),mathbb{C}) subset L^q((0, infty),mathbb{C})$

    nor
    $L^q((0, infty),mathbb{C}) subset L^p((0, infty),mathbb{C})$.



    I previously had to show, that there is no $a in mathbb{R}$ with which the function f:$(0, infty) to mathbb{R}$ f(x)=x^a

    would be lebesgue integrable and there is a hint that i should use this information, but i have no idea how to do so.



    I would be very thankfull for any kind of help.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have to show that for $1le p < q <infty$

      neitheir
      $L^p((0, infty),mathbb{C}) subset L^q((0, infty),mathbb{C})$

      nor
      $L^q((0, infty),mathbb{C}) subset L^p((0, infty),mathbb{C})$.



      I previously had to show, that there is no $a in mathbb{R}$ with which the function f:$(0, infty) to mathbb{R}$ f(x)=x^a

      would be lebesgue integrable and there is a hint that i should use this information, but i have no idea how to do so.



      I would be very thankfull for any kind of help.










      share|cite|improve this question













      I have to show that for $1le p < q <infty$

      neitheir
      $L^p((0, infty),mathbb{C}) subset L^q((0, infty),mathbb{C})$

      nor
      $L^q((0, infty),mathbb{C}) subset L^p((0, infty),mathbb{C})$.



      I previously had to show, that there is no $a in mathbb{R}$ with which the function f:$(0, infty) to mathbb{R}$ f(x)=x^a

      would be lebesgue integrable and there is a hint that i should use this information, but i have no idea how to do so.



      I would be very thankfull for any kind of help.







      lp-spaces






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      asked Nov 18 at 10:25









      Zweistein

      11




      11






















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          Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+infty)tomathbb{R}$ given by
          $$
          f(x)=x^{-frac{1}{r}}chi_{(0,1)}(x)
          $$

          it's easy to show that $fin L^p(0,+infty)$ but $fnotin L^q(0,+infty)$. For the second part you might just take
          $$
          g(x)=x^{-frac{1}{r}}chi_{(1,+infty)}(x)
          $$

          Here $fin L^q(0,+infty)$ but $fnotin L^p(0,+infty)$. This example should show you that in general for $p<q$ you have $L^qsubset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $infty$", for example with $mathbb{N}$ and the counting measure, then is reversed: $ell^psubsetell^q$.






          share|cite|improve this answer























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            active

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            up vote
            0
            down vote













            Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+infty)tomathbb{R}$ given by
            $$
            f(x)=x^{-frac{1}{r}}chi_{(0,1)}(x)
            $$

            it's easy to show that $fin L^p(0,+infty)$ but $fnotin L^q(0,+infty)$. For the second part you might just take
            $$
            g(x)=x^{-frac{1}{r}}chi_{(1,+infty)}(x)
            $$

            Here $fin L^q(0,+infty)$ but $fnotin L^p(0,+infty)$. This example should show you that in general for $p<q$ you have $L^qsubset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $infty$", for example with $mathbb{N}$ and the counting measure, then is reversed: $ell^psubsetell^q$.






            share|cite|improve this answer



























              up vote
              0
              down vote













              Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+infty)tomathbb{R}$ given by
              $$
              f(x)=x^{-frac{1}{r}}chi_{(0,1)}(x)
              $$

              it's easy to show that $fin L^p(0,+infty)$ but $fnotin L^q(0,+infty)$. For the second part you might just take
              $$
              g(x)=x^{-frac{1}{r}}chi_{(1,+infty)}(x)
              $$

              Here $fin L^q(0,+infty)$ but $fnotin L^p(0,+infty)$. This example should show you that in general for $p<q$ you have $L^qsubset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $infty$", for example with $mathbb{N}$ and the counting measure, then is reversed: $ell^psubsetell^q$.






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+infty)tomathbb{R}$ given by
                $$
                f(x)=x^{-frac{1}{r}}chi_{(0,1)}(x)
                $$

                it's easy to show that $fin L^p(0,+infty)$ but $fnotin L^q(0,+infty)$. For the second part you might just take
                $$
                g(x)=x^{-frac{1}{r}}chi_{(1,+infty)}(x)
                $$

                Here $fin L^q(0,+infty)$ but $fnotin L^p(0,+infty)$. This example should show you that in general for $p<q$ you have $L^qsubset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $infty$", for example with $mathbb{N}$ and the counting measure, then is reversed: $ell^psubsetell^q$.






                share|cite|improve this answer














                Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+infty)tomathbb{R}$ given by
                $$
                f(x)=x^{-frac{1}{r}}chi_{(0,1)}(x)
                $$

                it's easy to show that $fin L^p(0,+infty)$ but $fnotin L^q(0,+infty)$. For the second part you might just take
                $$
                g(x)=x^{-frac{1}{r}}chi_{(1,+infty)}(x)
                $$

                Here $fin L^q(0,+infty)$ but $fnotin L^p(0,+infty)$. This example should show you that in general for $p<q$ you have $L^qsubset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $infty$", for example with $mathbb{N}$ and the counting measure, then is reversed: $ell^psubsetell^q$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 18 at 10:56

























                answered Nov 18 at 10:50









                Marco

                1909




                1909






























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