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21

1

$begingroup$

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

share|cite|improve this question

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

$endgroup$

add a comment | 

21

1

$begingroup$

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

share|cite|improve this question

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

$endgroup$

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

share|cite|improve this question

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

$endgroup$

share|cite|improve this question

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

share|cite|improve this question

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

edited 15 hours ago

YuiTo Cheng

2,0592637

edited 15 hours ago

YuiTo Cheng

2,0592637

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

asked 21 hours ago

Kavi Rama MurthyKavi Rama Murthy

68.2k53069

1 Answer
1

active

oldest

votes

26

$begingroup$

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 21 hours ago

Ethan MacBroughEthan MacBrough

981617

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  5 hours ago
  
  

  
  
  
  

add a comment | 

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26

$begingroup$

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 21 hours ago

Ethan MacBroughEthan MacBrough

981617

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  5 hours ago
  
  

  
  
  
  

add a comment | 

26

$begingroup$

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 21 hours ago

Ethan MacBroughEthan MacBrough

981617

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: Or, use Ethan's sequence with the interval $[1/4, 3/4]$, on which it converges uniformly.
  $endgroup$
  – Nate Eldredge
  5 hours ago
  
  

  
  
  
  

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 21 hours ago

Ethan MacBroughEthan MacBrough

981617

$endgroup$

share|cite|improve this answer

answered 21 hours ago

Ethan MacBroughEthan MacBrough

981617

answered 21 hours ago

Ethan MacBroughEthan MacBrough

981617

answered 21 hours ago

Ethan MacBroughEthan MacBrough

981617