Norm of Fourier series












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$begingroup$


I am reading the proof of the statement that no non-zero multiplication operator on $L^2([0,1])$ is compact in this post. And I would like to address it as a seperate post as I am only curious about one of the step.



So We have ${f_n}$ orthonormal basis for $L^2([0,1])$ where $f_n=e^{2pi inx}$. We would like to show that for every nonzero $gin L^infty$, the sequence ${gf_n}$ is not Cauchy.



We start by writing $g=sum a_n f_n$ and $g f_k=sum a_nf_{n+k}$. Now we want to show that $ lVert gf_{k+j}-gf_klVert> c$ for some postive constant c.



now $gf_{k+j}-gf_k=sum c_n f_n$ and by Bessels's inequality we have
$ lVert gf_{k+j}-gf_klVert^2> sum |c_n|^2$. My question is in our case, how do we now any of the $|c_n|$ is non-zero?



Thanks for your help!






real-analysis functional-analysis operator-theory hilbert-spaces orthonormal







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    0












    $begingroup$


    I am reading the proof of the statement that no non-zero multiplication operator on $L^2([0,1])$ is compact in this post. And I would like to address it as a seperate post as I am only curious about one of the step.



    So We have ${f_n}$ orthonormal basis for $L^2([0,1])$ where $f_n=e^{2pi inx}$. We would like to show that for every nonzero $gin L^infty$, the sequence ${gf_n}$ is not Cauchy.



    We start by writing $g=sum a_n f_n$ and $g f_k=sum a_nf_{n+k}$. Now we want to show that $ lVert gf_{k+j}-gf_klVert> c$ for some postive constant c.



    now $gf_{k+j}-gf_k=sum c_n f_n$ and by Bessels's inequality we have
    $ lVert gf_{k+j}-gf_klVert^2> sum |c_n|^2$. My question is in our case, how do we now any of the $|c_n|$ is non-zero?



    Thanks for your help!






    real-analysis functional-analysis operator-theory hilbert-spaces orthonormal







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am reading the proof of the statement that no non-zero multiplication operator on $L^2([0,1])$ is compact in this post. And I would like to address it as a seperate post as I am only curious about one of the step.



      So We have ${f_n}$ orthonormal basis for $L^2([0,1])$ where $f_n=e^{2pi inx}$. We would like to show that for every nonzero $gin L^infty$, the sequence ${gf_n}$ is not Cauchy.



      We start by writing $g=sum a_n f_n$ and $g f_k=sum a_nf_{n+k}$. Now we want to show that $ lVert gf_{k+j}-gf_klVert> c$ for some postive constant c.



      now $gf_{k+j}-gf_k=sum c_n f_n$ and by Bessels's inequality we have
      $ lVert gf_{k+j}-gf_klVert^2> sum |c_n|^2$. My question is in our case, how do we now any of the $|c_n|$ is non-zero?



      Thanks for your help!






      real-analysis functional-analysis operator-theory hilbert-spaces orthonormal







      share|cite|improve this question









      $endgroup$




      I am reading the proof of the statement that no non-zero multiplication operator on $L^2([0,1])$ is compact in this post. And I would like to address it as a seperate post as I am only curious about one of the step.



      So We have ${f_n}$ orthonormal basis for $L^2([0,1])$ where $f_n=e^{2pi inx}$. We would like to show that for every nonzero $gin L^infty$, the sequence ${gf_n}$ is not Cauchy.



      We start by writing $g=sum a_n f_n$ and $g f_k=sum a_nf_{n+k}$. Now we want to show that $ lVert gf_{k+j}-gf_klVert> c$ for some postive constant c.



      now $gf_{k+j}-gf_k=sum c_n f_n$ and by Bessels's inequality we have
      $ lVert gf_{k+j}-gf_klVert^2> sum |c_n|^2$. My question is in our case, how do we now any of the $|c_n|$ is non-zero?



      Thanks for your help!






      real-analysis functional-analysis operator-theory hilbert-spaces orthonormal




      real-analysis functional-analysis operator-theory hilbert-spaces orthonormal






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 22 at 3:20









      user136592user136592

      818511




      818511






















          1 Answer
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          $begingroup$

          We have
          $$begin{split}
          gf_{k+j}-gf_k &=sum a_nf_{n+k+j}-sum a_nf_{n+k}\
          &=sum (a_{n-k-j}-a_{n-k})f_n
          end{split}$$
          Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$|g|^2 = sum |a_n|^2 = +infty$$






          share|cite|improve this answer









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            $begingroup$

            We have
            $$begin{split}
            gf_{k+j}-gf_k &=sum a_nf_{n+k+j}-sum a_nf_{n+k}\
            &=sum (a_{n-k-j}-a_{n-k})f_n
            end{split}$$
            Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$|g|^2 = sum |a_n|^2 = +infty$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We have
              $$begin{split}
              gf_{k+j}-gf_k &=sum a_nf_{n+k+j}-sum a_nf_{n+k}\
              &=sum (a_{n-k-j}-a_{n-k})f_n
              end{split}$$
              Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$|g|^2 = sum |a_n|^2 = +infty$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We have
                $$begin{split}
                gf_{k+j}-gf_k &=sum a_nf_{n+k+j}-sum a_nf_{n+k}\
                &=sum (a_{n-k-j}-a_{n-k})f_n
                end{split}$$
                Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$|g|^2 = sum |a_n|^2 = +infty$$






                share|cite|improve this answer









                $endgroup$



                We have
                $$begin{split}
                gf_{k+j}-gf_k &=sum a_nf_{n+k+j}-sum a_nf_{n+k}\
                &=sum (a_{n-k-j}-a_{n-k})f_n
                end{split}$$
                Therefore, $c_n=a_{n-k-j}-a_{n-k}$. So, there must be at least one $c_n$ that's non-zero, because if there weren't any, you'd have $a_{n-k-j}=a_{n-k}$ for all $n$. That is, you'd have $a_n=a_{n+j}$ for all $n$, which would imply that either $a_n$ is zero for all $n$, or that $$|g|^2 = sum |a_n|^2 = +infty$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 23:36









                Stefan LafonStefan Lafon

                2,83519




                2,83519






























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