Show that for $|a|<1$ the series $sum_{n=1}^{infty}a^nf_n$ converges in $L_2([-1,1])$.












0












$begingroup$


Let $f_n(x)=x^{n+1/2}$ for $ngeq 1$ and $xin [-1,1]$. Show that for $|a|<1$ the series $sum_{n=1}^{infty}a^nf_n$ converges in $L_2([-1,1])$. First, I have shown that $sum_{n=1}^{infty}a^nf_n(x)$ converges to $frac{ax^{3/2}}{1-ax}$ for $xin [-1,1]$. After that, I'm having difficulties to show that
$$
int_{-1}^{1}left | frac{ax^{3/2}}{1-ax} right |^2,mathrm{d}x=|a|^2int_{-1}^{1}frac{|x^{3}|}{left | 1-ax right |^2},mathrm{d}x
$$
is less than $+infty$. The denominator in the integral is troublesome.






functional-analysis convergence hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Isn't there a problem with $x^{n+1/2}$ for $x<0?$
    $endgroup$
    – zhw.
    Jan 10 at 23:35










  • $begingroup$
    @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
    $endgroup$
    – UnknownW
    Jan 10 at 23:40










  • $begingroup$
    @zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
    $endgroup$
    – UnknownW
    Jan 11 at 15:33












  • $begingroup$
    That's unusual, so you should probably mention it in your question.
    $endgroup$
    – zhw.
    Jan 11 at 19:42
















0












$begingroup$


Let $f_n(x)=x^{n+1/2}$ for $ngeq 1$ and $xin [-1,1]$. Show that for $|a|<1$ the series $sum_{n=1}^{infty}a^nf_n$ converges in $L_2([-1,1])$. First, I have shown that $sum_{n=1}^{infty}a^nf_n(x)$ converges to $frac{ax^{3/2}}{1-ax}$ for $xin [-1,1]$. After that, I'm having difficulties to show that
$$
int_{-1}^{1}left | frac{ax^{3/2}}{1-ax} right |^2,mathrm{d}x=|a|^2int_{-1}^{1}frac{|x^{3}|}{left | 1-ax right |^2},mathrm{d}x
$$
is less than $+infty$. The denominator in the integral is troublesome.






functional-analysis convergence hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Isn't there a problem with $x^{n+1/2}$ for $x<0?$
    $endgroup$
    – zhw.
    Jan 10 at 23:35










  • $begingroup$
    @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
    $endgroup$
    – UnknownW
    Jan 10 at 23:40










  • $begingroup$
    @zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
    $endgroup$
    – UnknownW
    Jan 11 at 15:33












  • $begingroup$
    That's unusual, so you should probably mention it in your question.
    $endgroup$
    – zhw.
    Jan 11 at 19:42














0












0








0





$begingroup$


Let $f_n(x)=x^{n+1/2}$ for $ngeq 1$ and $xin [-1,1]$. Show that for $|a|<1$ the series $sum_{n=1}^{infty}a^nf_n$ converges in $L_2([-1,1])$. First, I have shown that $sum_{n=1}^{infty}a^nf_n(x)$ converges to $frac{ax^{3/2}}{1-ax}$ for $xin [-1,1]$. After that, I'm having difficulties to show that
$$
int_{-1}^{1}left | frac{ax^{3/2}}{1-ax} right |^2,mathrm{d}x=|a|^2int_{-1}^{1}frac{|x^{3}|}{left | 1-ax right |^2},mathrm{d}x
$$
is less than $+infty$. The denominator in the integral is troublesome.






functional-analysis convergence hilbert-spaces







share|cite|improve this question









$endgroup$




Let $f_n(x)=x^{n+1/2}$ for $ngeq 1$ and $xin [-1,1]$. Show that for $|a|<1$ the series $sum_{n=1}^{infty}a^nf_n$ converges in $L_2([-1,1])$. First, I have shown that $sum_{n=1}^{infty}a^nf_n(x)$ converges to $frac{ax^{3/2}}{1-ax}$ for $xin [-1,1]$. After that, I'm having difficulties to show that
$$
int_{-1}^{1}left | frac{ax^{3/2}}{1-ax} right |^2,mathrm{d}x=|a|^2int_{-1}^{1}frac{|x^{3}|}{left | 1-ax right |^2},mathrm{d}x
$$
is less than $+infty$. The denominator in the integral is troublesome.






functional-analysis convergence hilbert-spaces




functional-analysis convergence hilbert-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 23:21









UnknownWUnknownW

1,025922




1,025922












  • $begingroup$
    Isn't there a problem with $x^{n+1/2}$ for $x<0?$
    $endgroup$
    – zhw.
    Jan 10 at 23:35










  • $begingroup$
    @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
    $endgroup$
    – UnknownW
    Jan 10 at 23:40










  • $begingroup$
    @zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
    $endgroup$
    – UnknownW
    Jan 11 at 15:33












  • $begingroup$
    That's unusual, so you should probably mention it in your question.
    $endgroup$
    – zhw.
    Jan 11 at 19:42


















  • $begingroup$
    Isn't there a problem with $x^{n+1/2}$ for $x<0?$
    $endgroup$
    – zhw.
    Jan 10 at 23:35










  • $begingroup$
    @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
    $endgroup$
    – UnknownW
    Jan 10 at 23:40










  • $begingroup$
    @zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
    $endgroup$
    – UnknownW
    Jan 11 at 15:33












  • $begingroup$
    That's unusual, so you should probably mention it in your question.
    $endgroup$
    – zhw.
    Jan 11 at 19:42
















$begingroup$
Isn't there a problem with $x^{n+1/2}$ for $x<0?$
$endgroup$
– zhw.
Jan 10 at 23:35




$begingroup$
Isn't there a problem with $x^{n+1/2}$ for $x<0?$
$endgroup$
– zhw.
Jan 10 at 23:35












$begingroup$
@zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
$endgroup$
– UnknownW
Jan 10 at 23:40




$begingroup$
@zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer.
$endgroup$
– UnknownW
Jan 10 at 23:40












$begingroup$
@zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
$endgroup$
– UnknownW
Jan 11 at 15:33






$begingroup$
@zhw. It is a function from $[-1,1]$ into $mathbb{C}$, so it's all right. For $x<0$, we may put $sqrt{x}=isqrt{-x}$.
$endgroup$
– UnknownW
Jan 11 at 15:33














$begingroup$
That's unusual, so you should probably mention it in your question.
$endgroup$
– zhw.
Jan 11 at 19:42




$begingroup$
That's unusual, so you should probably mention it in your question.
$endgroup$
– zhw.
Jan 11 at 19:42










2 Answers
2






active

oldest

votes


















0












$begingroup$

Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so



$$|a|^2int_{-1}^1 frac{|x^3|}{|1-ax|^2}dx = a^2int_0^1 frac{x^3}{(1-ax)^2}dx,-a^2int_{-1}^0frac{x^3}{(1-ax)^2}dx.$$



Substituting $u=1-ax$ should finish the job.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $sum |g_n|_2 <infty,$ then $sum g_n$ converges in $L^2.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What does $||cdot ||_2$ mean?
      $endgroup$
      – UnknownW
      Jan 15 at 22:31












    • $begingroup$
      It's the $L^2$ norm
      $endgroup$
      – zhw.
      Jan 15 at 22:58











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so



    $$|a|^2int_{-1}^1 frac{|x^3|}{|1-ax|^2}dx = a^2int_0^1 frac{x^3}{(1-ax)^2}dx,-a^2int_{-1}^0frac{x^3}{(1-ax)^2}dx.$$



    Substituting $u=1-ax$ should finish the job.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so



      $$|a|^2int_{-1}^1 frac{|x^3|}{|1-ax|^2}dx = a^2int_0^1 frac{x^3}{(1-ax)^2}dx,-a^2int_{-1}^0frac{x^3}{(1-ax)^2}dx.$$



      Substituting $u=1-ax$ should finish the job.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so



        $$|a|^2int_{-1}^1 frac{|x^3|}{|1-ax|^2}dx = a^2int_0^1 frac{x^3}{(1-ax)^2}dx,-a^2int_{-1}^0frac{x^3}{(1-ax)^2}dx.$$



        Substituting $u=1-ax$ should finish the job.






        share|cite|improve this answer









        $endgroup$



        Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so



        $$|a|^2int_{-1}^1 frac{|x^3|}{|1-ax|^2}dx = a^2int_0^1 frac{x^3}{(1-ax)^2}dx,-a^2int_{-1}^0frac{x^3}{(1-ax)^2}dx.$$



        Substituting $u=1-ax$ should finish the job.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 23:42









        Mose WintnerMose Wintner

        958410




        958410























            0












            $begingroup$

            Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $sum |g_n|_2 <infty,$ then $sum g_n$ converges in $L^2.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What does $||cdot ||_2$ mean?
              $endgroup$
              – UnknownW
              Jan 15 at 22:31












            • $begingroup$
              It's the $L^2$ norm
              $endgroup$
              – zhw.
              Jan 15 at 22:58
















            0












            $begingroup$

            Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $sum |g_n|_2 <infty,$ then $sum g_n$ converges in $L^2.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What does $||cdot ||_2$ mean?
              $endgroup$
              – UnknownW
              Jan 15 at 22:31












            • $begingroup$
              It's the $L^2$ norm
              $endgroup$
              – zhw.
              Jan 15 at 22:58














            0












            0








            0





            $begingroup$

            Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $sum |g_n|_2 <infty,$ then $sum g_n$ converges in $L^2.$






            share|cite|improve this answer









            $endgroup$



            Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $sum |g_n|_2 <infty,$ then $sum g_n$ converges in $L^2.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 10 at 23:42









            zhw.zhw.

            73.8k43175




            73.8k43175












            • $begingroup$
              What does $||cdot ||_2$ mean?
              $endgroup$
              – UnknownW
              Jan 15 at 22:31












            • $begingroup$
              It's the $L^2$ norm
              $endgroup$
              – zhw.
              Jan 15 at 22:58


















            • $begingroup$
              What does $||cdot ||_2$ mean?
              $endgroup$
              – UnknownW
              Jan 15 at 22:31












            • $begingroup$
              It's the $L^2$ norm
              $endgroup$
              – zhw.
              Jan 15 at 22:58
















            $begingroup$
            What does $||cdot ||_2$ mean?
            $endgroup$
            – UnknownW
            Jan 15 at 22:31






            $begingroup$
            What does $||cdot ||_2$ mean?
            $endgroup$
            – UnknownW
            Jan 15 at 22:31














            $begingroup$
            It's the $L^2$ norm
            $endgroup$
            – zhw.
            Jan 15 at 22:58




            $begingroup$
            It's the $L^2$ norm
            $endgroup$
            – zhw.
            Jan 15 at 22:58


















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