$f, 1/f$ integrable implies that $f^2$, $1/f^2$ integrable?












1














on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.



I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?






integration measure-theory borel-measures







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    Nov 27 '18 at 14:50










  • Why do you suspect your statement is true?
    – Umberto P.
    Nov 27 '18 at 14:58










  • I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:02
















1














on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.



I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?






integration measure-theory borel-measures







share|cite|improve this question
























  • MathJax is required for questions
    – Akash Roy
    Nov 27 '18 at 14:50










  • Why do you suspect your statement is true?
    – Umberto P.
    Nov 27 '18 at 14:58










  • I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:02














1












1








1







on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.



I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?






integration measure-theory borel-measures







share|cite|improve this question















on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.



I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?






integration measure-theory borel-measures




integration measure-theory borel-measures






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 14:54

























asked Nov 27 '18 at 14:49









mathsisrad

62




62












  • MathJax is required for questions
    – Akash Roy
    Nov 27 '18 at 14:50










  • Why do you suspect your statement is true?
    – Umberto P.
    Nov 27 '18 at 14:58










  • I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:02


















  • MathJax is required for questions
    – Akash Roy
    Nov 27 '18 at 14:50










  • Why do you suspect your statement is true?
    – Umberto P.
    Nov 27 '18 at 14:58










  • I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:02
















MathJax is required for questions
– Akash Roy
Nov 27 '18 at 14:50




MathJax is required for questions
– Akash Roy
Nov 27 '18 at 14:50












Why do you suspect your statement is true?
– Umberto P.
Nov 27 '18 at 14:58




Why do you suspect your statement is true?
– Umberto P.
Nov 27 '18 at 14:58












I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 '18 at 15:02




I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 '18 at 15:02










2 Answers
2






active

oldest

votes


















2














$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.






share|cite|improve this answer





















  • yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
    – mathsisrad
    Nov 27 '18 at 15:04






  • 1




    $0$ is not part of the measure space; we don't need to define it there.
    – Sambo
    Nov 27 '18 at 15:06






  • 1




    @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
    – Ian
    Nov 27 '18 at 15:07










  • hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:09










  • Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
    – Robert Israel
    Nov 27 '18 at 15:25





















0














You might prefer trying to prove something true.



Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.






    share|cite|improve this answer





















    • yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
      – mathsisrad
      Nov 27 '18 at 15:04






    • 1




      $0$ is not part of the measure space; we don't need to define it there.
      – Sambo
      Nov 27 '18 at 15:06






    • 1




      @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
      – Ian
      Nov 27 '18 at 15:07










    • hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
      – mathsisrad
      Nov 27 '18 at 15:09










    • Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
      – Robert Israel
      Nov 27 '18 at 15:25


















    2














    $sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.






    share|cite|improve this answer





















    • yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
      – mathsisrad
      Nov 27 '18 at 15:04






    • 1




      $0$ is not part of the measure space; we don't need to define it there.
      – Sambo
      Nov 27 '18 at 15:06






    • 1




      @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
      – Ian
      Nov 27 '18 at 15:07










    • hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
      – mathsisrad
      Nov 27 '18 at 15:09










    • Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
      – Robert Israel
      Nov 27 '18 at 15:25
















    2












    2








    2






    $sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.






    share|cite|improve this answer












    $sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 15:03









    Robert Israel

    318k23208457




    318k23208457












    • yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
      – mathsisrad
      Nov 27 '18 at 15:04






    • 1




      $0$ is not part of the measure space; we don't need to define it there.
      – Sambo
      Nov 27 '18 at 15:06






    • 1




      @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
      – Ian
      Nov 27 '18 at 15:07










    • hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
      – mathsisrad
      Nov 27 '18 at 15:09










    • Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
      – Robert Israel
      Nov 27 '18 at 15:25




















    • yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
      – mathsisrad
      Nov 27 '18 at 15:04






    • 1




      $0$ is not part of the measure space; we don't need to define it there.
      – Sambo
      Nov 27 '18 at 15:06






    • 1




      @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
      – Ian
      Nov 27 '18 at 15:07










    • hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
      – mathsisrad
      Nov 27 '18 at 15:09










    • Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
      – Robert Israel
      Nov 27 '18 at 15:25


















    yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
    – mathsisrad
    Nov 27 '18 at 15:04




    yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
    – mathsisrad
    Nov 27 '18 at 15:04




    1




    1




    $0$ is not part of the measure space; we don't need to define it there.
    – Sambo
    Nov 27 '18 at 15:06




    $0$ is not part of the measure space; we don't need to define it there.
    – Sambo
    Nov 27 '18 at 15:06




    1




    1




    @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
    – Ian
    Nov 27 '18 at 15:07




    @mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
    – Ian
    Nov 27 '18 at 15:07












    hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:09




    hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
    – mathsisrad
    Nov 27 '18 at 15:09












    Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
    – Robert Israel
    Nov 27 '18 at 15:25






    Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
    – Robert Israel
    Nov 27 '18 at 15:25













    0














    You might prefer trying to prove something true.



    Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
    Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.






    share|cite|improve this answer


























      0














      You might prefer trying to prove something true.



      Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
      Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.






      share|cite|improve this answer
























        0












        0








        0






        You might prefer trying to prove something true.



        Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
        Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.






        share|cite|improve this answer












        You might prefer trying to prove something true.



        Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
        Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 15:05









        Federico

        4,709514




        4,709514






























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