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26

2

$begingroup$

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 $undersetnrightarrow infty lim c_n,k$ 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 $N$, $%
sum_k=0^Nc_ix^irightarrow (c_0,c_1,...,c_N)$ 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. Maybe there is a very simple example but I haven't been
to find one. Thank you for investing your time on this.

functional-analysis polynomials convergence

share|cite|improve this question

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

add a comment | 

26

2

$begingroup$

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 $undersetnrightarrow infty lim c_n,k$ 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 $N$, $%
sum_k=0^Nc_ix^irightarrow (c_0,c_1,...,c_N)$ 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. Maybe there is a very simple example but I haven't been
to find one. Thank you for investing your time on this.

functional-analysis polynomials convergence

share|cite|improve this question

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

add a comment | 

$begingroup$

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 $undersetnrightarrow infty lim c_n,k$ 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 $N$, $%
sum_k=0^Nc_ix^irightarrow (c_0,c_1,...,c_N)$ 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. Maybe there is a very simple example but I haven't been
to find one. Thank you for investing your time on this.

functional-analysis polynomials convergence

share|cite|improve this question

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

share|cite|improve this question

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

share|cite|improve this question

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

edited Mar 19 at 17:56

Rodrigo de Azevedo

13.2k41960

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

asked Mar 18 at 8:13

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

1 Answer
1

active

oldest

votes

33

$begingroup$

Consider the sequence of polynomials

$$
p_n(x)=left{
beginarray@ll@
(1-x)^n, & textif nequiv 0 mod 2 \
x^n, & textif nequiv 1 mod 2
endarrayright.
$$

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

share|cite|improve this answer

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 18 at 8:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  Mar 19 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if the polynomial converges to something other than $0$ though?
  $endgroup$
  – Jam
  Mar 23 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jam You can make it work if $f$ is any function that can be approximated by polynomials. Indeed, if $g_n$ is a sequence of polynomials converging to $f$, then we can assume $g_n(0)$ converges to $f(0)$ since otherwise we're done. But then $p_n+g_n$ still converges to $f$ while $p_n(0)+g_n(0)$ has two subsequences which converge to $f(0)$ and $f(0)+1$ respectively.
  $endgroup$
  – Ethan MacBrough
  Mar 23 at 20:47
  

  
  
  
  

add a comment | 

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33

$begingroup$

Consider the sequence of polynomials

$$
p_n(x)=left{
beginarray@ll@
(1-x)^n, & textif nequiv 0 mod 2 \
x^n, & textif nequiv 1 mod 2
endarrayright.
$$

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

share|cite|improve this answer

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 18 at 8:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  Mar 19 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if the polynomial converges to something other than $0$ though?
  $endgroup$
  – Jam
  Mar 23 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jam You can make it work if $f$ is any function that can be approximated by polynomials. Indeed, if $g_n$ is a sequence of polynomials converging to $f$, then we can assume $g_n(0)$ converges to $f(0)$ since otherwise we're done. But then $p_n+g_n$ still converges to $f$ while $p_n(0)+g_n(0)$ has two subsequences which converge to $f(0)$ and $f(0)+1$ respectively.
  $endgroup$
  – Ethan MacBrough
  Mar 23 at 20:47
  

  
  
  
  

add a comment | 

33

$begingroup$

Consider the sequence of polynomials

$$
p_n(x)=left{
beginarray@ll@
(1-x)^n, & textif nequiv 0 mod 2 \
x^n, & textif nequiv 1 mod 2
endarrayright.
$$

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

share|cite|improve this answer

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 18 at 8:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  Mar 19 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if the polynomial converges to something other than $0$ though?
  $endgroup$
  – Jam
  Mar 23 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jam You can make it work if $f$ is any function that can be approximated by polynomials. Indeed, if $g_n$ is a sequence of polynomials converging to $f$, then we can assume $g_n(0)$ converges to $f(0)$ since otherwise we're done. But then $p_n+g_n$ still converges to $f$ while $p_n(0)+g_n(0)$ has two subsequences which converge to $f(0)$ and $f(0)+1$ respectively.
  $endgroup$
  – Ethan MacBrough
  Mar 23 at 20:47
  

  
  
  
  

add a comment | 

$begingroup$

Consider the sequence of polynomials

$$
p_n(x)=left{
beginarray@ll@
(1-x)^n, & textif nequiv 0 mod 2 \
x^n, & textif nequiv 1 mod 2
endarrayright.
$$

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

share|cite|improve this answer

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

$endgroup$

share|cite|improve this answer

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617

answered Mar 18 at 8:32

Ethan MacBroughEthan MacBrough

1,241617