$left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p = sum_{k=1}^{infty} left| frac{(-1)^k}{k}...












0












$begingroup$


Here's the problem statement, which is the context for my question,



Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by



$$
u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
$$



For which $p geq 1$ does $int |u|^p dm < infty$?



The solution manual says that: Consider



$$
u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
$$



$u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that



$$
|u|^p=sum_{k=1}^{infty}
left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$
...



so, I don't get the above equality. How do I get from



$$
|u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
quad text{to} quad
sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
$$



It seems that I at least have to use the triangle inequality, but that would ruin the equality.






real-analysis lebesgue-measure







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Here's the problem statement, which is the context for my question,



    Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by



    $$
    u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
    $$



    For which $p geq 1$ does $int |u|^p dm < infty$?



    The solution manual says that: Consider



    $$
    u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
    $$



    $u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that



    $$
    |u|^p=sum_{k=1}^{infty}
    left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
    $$
    ...



    so, I don't get the above equality. How do I get from



    $$
    |u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
    quad text{to} quad
    sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
    $$



    It seems that I at least have to use the triangle inequality, but that would ruin the equality.






    real-analysis lebesgue-measure







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Here's the problem statement, which is the context for my question,



      Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by



      $$
      u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
      $$



      For which $p geq 1$ does $int |u|^p dm < infty$?



      The solution manual says that: Consider



      $$
      u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
      $$



      $u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that



      $$
      |u|^p=sum_{k=1}^{infty}
      left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
      $$
      ...



      so, I don't get the above equality. How do I get from



      $$
      |u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
      quad text{to} quad
      sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
      $$



      It seems that I at least have to use the triangle inequality, but that would ruin the equality.






      real-analysis lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Here's the problem statement, which is the context for my question,



      Consider the measure space $(mathbb{R}, B(mathbb{R}), m)$ where $m$ is the Lebesgue measure, and let $u:mathbb{R} rightarrow mathbb{R}$ be given by



      $$
      u=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}
      $$



      For which $p geq 1$ does $int |u|^p dm < infty$?



      The solution manual says that: Consider



      $$
      u_n=sum_{k=1}^{n} frac{(-1)^k}{k} 1_{[k, k+1)}
      $$



      $u_n rightarrow u$, (actually $u_n(x)=u(x)$ for all $n > x-1$) and since the sets $[k,k+1)$ are disjoint for all $k geq 1$, we have that



      $$
      |u|^p=sum_{k=1}^{infty}
      left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
      $$
      ...



      so, I don't get the above equality. How do I get from



      $$
      |u|^p= left| sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)} right|^p
      quad text{to} quad
      sum_{k=1}^{infty} left| frac{(-1)^k}{k} right|^p 1_{[k, k+1)}
      $$



      It seems that I at least have to use the triangle inequality, but that would ruin the equality.






      real-analysis lebesgue-measure




      real-analysis lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 28 at 20:31









      TomTom

      285




      285






















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          $begingroup$

          If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
          $$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
          where $p=lfloor x rfloor$.






          share|cite|improve this answer









          $endgroup$














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            0












            $begingroup$

            If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
            $$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
            where $p=lfloor x rfloor$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
              $$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
              where $p=lfloor x rfloor$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
                $$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
                where $p=lfloor x rfloor$.






                share|cite|improve this answer









                $endgroup$



                If you evaluate the sum that defines $u$ at a given point $x$, you'll see that at most one of the terms is non-zero. That's because
                $$u(x)=sum_{k=1}^{infty} frac{(-1)^k}{k} 1_{[k, k+1)}(x) = frac{(-1)^p}{p} $$
                where $p=lfloor x rfloor$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 20:59









                Stefan LafonStefan Lafon

                3,005212




                3,005212






























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