/ Unlabelled / Does convergence of polynomials imply that of its coefficients?

Senin, 18 Maret 2019

Does convergence of polynomials imply that of its coefficients?





















7
























$begingroup$





Let ${p_{n}}$ be a sequence of polynomials and $f$ a continuous function
on $[0,1]$ such that $intlimits_{0}^{1}|p_{n}(x)-f(x)|dxto 0$.
Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude
that $underset{nrightarrow infty }{lim }c_{n,k}$ exists for each $k$?.



What I know so far: if the degrees of $p_{n}^{prime }s$ are bounded then
this is true. In fact we can replace $L^{1}$ convergence by convergence in
any norm on $C[0,1]$; to see this we just have to note that for fixed $N$, $%
sum_{k=0}^{N}c_{i}x^{i}rightarrow (c_{0},c_{1},...,c_{N})$ is a
linear map on a finite dimensional subspace and hence it is continuous. My
guess is that the result fails when there is no restriction on the degrees.
But if $p_{n}(z)$ converges uniformly in some disk around $0$ in the complex
plane then the conclusion holds. To construct a counterexample we have to
avoid this situation. May be there is a very simple example but I haven't been
to find one. Thank you for investing your time on this.










functional-analysis














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$endgroup$
































    7
























    $begingroup$





    Let ${p_{n}}$ be a sequence of polynomials and $f$ a continuous function
    on $[0,1]$ such that $intlimits_{0}^{1}|p_{n}(x)-f(x)|dxto 0$.
    Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude
    that $underset{nrightarrow infty }{lim }c_{n,k}$ exists for each $k$?.



    What I know so far: if the degrees of $p_{n}^{prime }s$ are bounded then
    this is true. In fact we can replace $L^{1}$ convergence by convergence in
    any norm on $C[0,1]$; to see this we just have to note that for fixed $N$, $%
    sum_{k=0}^{N}c_{i}x^{i}rightarrow (c_{0},c_{1},...,c_{N})$     is a
    linear map on a finite dimensional subspace and hence it is continuous. My
    guess is that the result fails when there is no restriction on the degrees.
    But if $p_{n}(z)$ converges uniformly in some disk around $0$ in the complex
    plane then the conclusion holds. To construct a counterexample we have to
    avoid this situation. May be there is a very simple example but I haven't been
    to find one. Thank you for investing your time on this.










    functional-analysis














    share|cite|improve this question






















    $endgroup$




























      7






















      7














      7






      $begingroup$





      Let ${p_{n}}$ be a sequence of polynomials and $f$ a continuous function
      on $[0,1]$ such that $intlimits_{0}^{1}|p_{n}(x)-f(x)|dxto 0$.
      Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude
      that $underset{nrightarrow infty }{lim }c_{n,k}$ exists for each $k$?.



      What I know so far: if the degrees of $p_{n}^{prime }s$ are bounded then
      this is true. In fact we can replace $L^{1}$ convergence by convergence in
      any norm on $C[0,1]$; to see this we just have to note that for fixed $N$, $%
      sum_{k=0}^{N}c_{i}x^{i}rightarrow (c_{0},c_{1},...,c_{N})$       is a
      linear map on a finite dimensional subspace and hence it is continuous. My
      guess is that the result fails when there is no restriction on the degrees.
      But if $p_{n}(z)$ converges uniformly in some disk around $0$ in the complex
      plane then the conclusion holds. To construct a counterexample we have to
      avoid this situation. May be there is a very simple example but I haven't been
      to find one. Thank you for investing your time on this.










      functional-analysis














      share|cite|improve this question






















      $endgroup$







      Let ${p_{n}}$ be a sequence of polynomials and $f$ a continuous function
      on $[0,1]$ such that $intlimits_{0}^{1}|p_{n}(x)-f(x)|dxto 0$.
      Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude
      that $underset{nrightarrow infty }{lim }c_{n,k}$ exists for each $k$?.



      What I know so far: if the degrees of $p_{n}^{prime }s$ are bounded then
      this is true. In fact we can replace $L^{1}$ convergence by convergence in
      any norm on $C[0,1]$; to see this we just have to note that for fixed $N$, $%
      sum_{k=0}^{N}c_{i}x^{i}rightarrow (c_{0},c_{1},...,c_{N})$       is a
      linear map on a finite dimensional subspace and hence it is continuous. My
      guess is that the result fails when there is no restriction on the degrees.
      But if $p_{n}(z)$ converges uniformly in some disk around $0$ in the complex
      plane then the conclusion holds. To construct a counterexample we have to
      avoid this situation. May be there is a very simple example but I haven't been
      to find one. Thank you for investing your time on this.








      functional-analysis







      functional-analysis










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      edited 25 mins ago













      Bernard



      123k741116







      123k741116















      asked 3 hours ago













      Kavi Rama MurthyKavi Rama Murthy



      68k53067







      68k53067



































          1 Answer

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          active



          oldest



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          6
























          $begingroup$



          Consider the sequence of polynomials



          $$
          p_n(x)=left{
          begin{array}{@{}ll@{}}
          (1-x)^n, & text{if} nequiv 0 mod 2 \
          x^n, & text{if} nequiv 1 mod 2
          end{array}right.
          $$



          Then $p_n(x)$ converges to $0$ but the constant term is oscillating.









          share|cite|improve this answer


















          $endgroup$


















          • 2









            $begingroup$

            Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

            $endgroup$

            – Kavi Rama Murthy

            2 hours ago






















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          1 Answer

          1











          active



          oldest



          votes















          1 Answer

          1











          active



          oldest



          votes

















          active



          oldest



          votes











          active



          oldest



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          6
























          $begingroup$



          Consider the sequence of polynomials



          $$
          p_n(x)=left{
          begin{array}{@{}ll@{}}
          (1-x)^n, & text{if} nequiv 0 mod 2 \
          x^n, & text{if} nequiv 1 mod 2
          end{array}right.
          $$



          Then $p_n(x)$ converges to $0$ but the constant term is oscillating.









          share|cite|improve this answer


















          $endgroup$


















          • 2









            $begingroup$

            Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

            $endgroup$

            – Kavi Rama Murthy

            2 hours ago






























          6
























          $begingroup$



          Consider the sequence of polynomials



          $$
          p_n(x)=left{
          begin{array}{@{}ll@{}}
          (1-x)^n, & text{if} nequiv 0 mod 2 \
          x^n, & text{if} nequiv 1 mod 2
          end{array}right.
          $$



          Then $p_n(x)$ converges to $0$ but the constant term is oscillating.









          share|cite|improve this answer


















          $endgroup$


















          • 2









            $begingroup$

            Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

            $endgroup$

            – Kavi Rama Murthy

            2 hours ago


























          6






















          6














          6






          $begingroup$



          Consider the sequence of polynomials



          $$
          p_n(x)=left{
          begin{array}{@{}ll@{}}
          (1-x)^n, & text{if} nequiv 0 mod 2 \
          x^n, & text{if} nequiv 1 mod 2
          end{array}right.
          $$



          Then $p_n(x)$ converges to $0$ but the constant term is oscillating.









          share|cite|improve this answer


















          $endgroup$





          Consider the sequence of polynomials



          $$
          p_n(x)=left{
          begin{array}{@{}ll@{}}
          (1-x)^n, & text{if} nequiv 0 mod 2 \
          x^n, & text{if} nequiv 1 mod 2
          end{array}right.
          $$



          Then $p_n(x)$ converges to $0$ but the constant term is oscillating.









          share|cite|improve this answer





















          share|cite|improve this answer



          share|cite|improve this answer














          answered 2 hours ago













          ChronusZedChronusZed



          811415







          811415














          • 2









            $begingroup$

            Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

            $endgroup$

            – Kavi Rama Murthy

            2 hours ago

























          • 2









            $begingroup$

            Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

            $endgroup$

            – Kavi Rama Murthy

            2 hours ago














          2







          2







          $begingroup$

          Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

          $endgroup$

          – Kavi Rama Murthy

          2 hours ago







          $begingroup$

          Thanks. I just noticed that teh conclusion fails even with uniform convergence. Consider $x(1-x)p_n(x)$ with the $p_n$' you have defined and look at the coefficient of $x$.

          $endgroup$

          – Kavi Rama Murthy

          2 hours ago
































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