Does convergence of polynomials imply that of its coefficients? The Next CEO of Stack OverflowDifferentiation continuous iff domain is finite dimensionalShow convergence of a sequence of continuous functions $f_n$ to a continuous function $f$ does not imply convergence of corresponding integrals.Convergence problem in different normsDominated convergence theorem and uniformly convergenceShowing that $displaystyleundersetnrightarrow inftylimint_0^1 f_n = int_0^1undersetnrightarrow inftylim f_n$Proof that uniform convergence implies convergence in norm of function spaceFor a piecewise function $g_n(t)$, find $g(t)=limlimits_nrightarrow inftyg_n(t)$.Uniform convergence of even polynomials in $(a,b)$Moving limit into sup normShowing that a linear map does not achieve its norm
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Does convergence of polynomials imply that of its coefficients?
The Next CEO of Stack OverflowDifferentiation continuous iff domain is finite dimensionalShow convergence of a sequence of continuous functions $f_n$ to a continuous function $f$ does not imply convergence of corresponding integrals.Convergence problem in different normsDominated convergence theorem and uniformly convergenceShowing that $displaystyleundersetnrightarrow inftylimint_0^1 f_n = int_0^1undersetnrightarrow inftylim f_n$Proof that uniform convergence implies convergence in norm of function spaceFor a piecewise function $g_n(t)$, find $g(t)=limlimits_nrightarrow inftyg_n(t)$.Uniform convergence of even polynomials in $(a,b)$Moving limit into sup normShowing that a linear map does not achieve its norm
$begingroup$
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
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
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
edited Mar 19 at 17:56
Rodrigo de Azevedo
13.2k41960
asked Mar 18 at 8:13
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
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
functional-analysis polynomials convergence
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
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
edited Mar 19 at 17:56
Rodrigo de Azevedo
13.2k41960
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
asked Mar 18 at 8:13
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
70.8k53170
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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 |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered Mar 18 at 8:32
Ethan MacBroughEthan MacBrough
1,241617
$endgroup$
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.
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
answered Mar 18 at 8:32
Ethan MacBroughEthan MacBrough
1,241617
1,241617
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 |
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
7
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
$begingroup$
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
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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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