Power series is locally normally convergent in its convergence radius $B(0,R)$












1












$begingroup$


$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.



What I did so far:




  • Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
    $$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$

  • Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
    So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.


Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.






complex-analysis power-series







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  • $begingroup$
    @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
    $endgroup$
    – dan
    Dec 30 '18 at 17:36










  • $begingroup$
    Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
    $endgroup$
    – Martin R
    Dec 30 '18 at 17:41












  • $begingroup$
    @MartinR ,Right! I'm sorry I miss typed, just corrected it.
    $endgroup$
    – dan
    Dec 30 '18 at 17:43










  • $begingroup$
    Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:37
















1












$begingroup$


$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.



What I did so far:




  • Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
    $$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$

  • Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
    So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.


Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.






complex-analysis power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
    $endgroup$
    – dan
    Dec 30 '18 at 17:36










  • $begingroup$
    Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
    $endgroup$
    – Martin R
    Dec 30 '18 at 17:41












  • $begingroup$
    @MartinR ,Right! I'm sorry I miss typed, just corrected it.
    $endgroup$
    – dan
    Dec 30 '18 at 17:43










  • $begingroup$
    Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:37














1












1








1





$begingroup$


$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.



What I did so far:




  • Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
    $$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$

  • Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
    So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.


Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.






complex-analysis power-series







share|cite|improve this question











$endgroup$




$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.



What I did so far:




  • Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
    $$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$

  • Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
    So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.


Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.






complex-analysis power-series




complex-analysis power-series






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 12:35









LutzL

60.7k42057




60.7k42057










asked Dec 30 '18 at 16:38









dandan

623613




623613












  • $begingroup$
    @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
    $endgroup$
    – dan
    Dec 30 '18 at 17:36










  • $begingroup$
    Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
    $endgroup$
    – Martin R
    Dec 30 '18 at 17:41












  • $begingroup$
    @MartinR ,Right! I'm sorry I miss typed, just corrected it.
    $endgroup$
    – dan
    Dec 30 '18 at 17:43










  • $begingroup$
    Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:37


















  • $begingroup$
    @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
    $endgroup$
    – dan
    Dec 30 '18 at 17:36










  • $begingroup$
    Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
    $endgroup$
    – Martin R
    Dec 30 '18 at 17:41












  • $begingroup$
    @MartinR ,Right! I'm sorry I miss typed, just corrected it.
    $endgroup$
    – dan
    Dec 30 '18 at 17:43










  • $begingroup$
    Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
    $endgroup$
    – LutzL
    Dec 31 '18 at 12:37
















$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36




$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36












$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41






$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41














$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43




$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43












$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37




$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37










1 Answer
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You have done all the necessary work, you just need to put it together:



For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    You have done all the necessary work, you just need to put it together:



    For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
    $U=B(0, r_1)$ is a neighbourhood of $z_0$, and
    $ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
    $$
    sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
    $$
    for some $M > 0$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You have done all the necessary work, you just need to put it together:



      For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
      $U=B(0, r_1)$ is a neighbourhood of $z_0$, and
      $ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
      $$
      sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
      $$
      for some $M > 0$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You have done all the necessary work, you just need to put it together:



        For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
        $U=B(0, r_1)$ is a neighbourhood of $z_0$, and
        $ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
        $$
        sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
        $$
        for some $M > 0$.






        share|cite|improve this answer









        $endgroup$



        You have done all the necessary work, you just need to put it together:



        For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
        $U=B(0, r_1)$ is a neighbourhood of $z_0$, and
        $ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
        $$
        sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
        $$
        for some $M > 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 17:58









        Martin RMartin R

        31k33561




        31k33561






























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