Show some theorem concerning of the uniform convergence on compacts of a sequence of polynomials of order...
$begingroup$
Let $k,sin mathbb N$,
let $x_0,x_1,...,x_s$ be given pairweise different real numbers,
let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,
and let
$$
P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
$$
be a sequence of real polynomials of order $leq k-1$.
Assume that there exist limits of derivatives:
$$
lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
$$
$$
lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
$$
...........................
$$
lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
$$
I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
$$
lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
$$
Maybe proof or references.
Thanks.
analysis polynomials convergence uniform-convergence
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
$endgroup$
add a comment |
$begingroup$
Let $k,sin mathbb N$,
let $x_0,x_1,...,x_s$ be given pairweise different real numbers,
let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,
and let
$$
P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
$$
be a sequence of real polynomials of order $leq k-1$.
Assume that there exist limits of derivatives:
$$
lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
$$
$$
lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
$$
...........................
$$
lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
$$
I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
$$
lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
$$
Maybe proof or references.
Thanks.
analysis polynomials convergence uniform-convergence
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
$endgroup$
add a comment |
$begingroup$
Let $k,sin mathbb N$,
let $x_0,x_1,...,x_s$ be given pairweise different real numbers,
let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,
and let
$$
P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
$$
be a sequence of real polynomials of order $leq k-1$.
Assume that there exist limits of derivatives:
$$
lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
$$
$$
lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
$$
...........................
$$
lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
$$
I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
$$
lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
$$
Maybe proof or references.
Thanks.
analysis polynomials convergence uniform-convergence
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
$endgroup$
Let $k,sin mathbb N$,
let $x_0,x_1,...,x_s$ be given pairweise different real numbers,
let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,
and let
$$
P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
$$
be a sequence of real polynomials of order $leq k-1$.
Assume that there exist limits of derivatives:
$$
lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
$$
$$
lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
$$
...........................
$$
lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
$$
I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
$$
lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
$$
Maybe proof or references.
Thanks.
analysis polynomials convergence uniform-convergence
analysis polynomials convergence uniform-convergence
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
edited Dec 14 '18 at 2:29
Alex Ravsky
40.7k32282
40.7k32282
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
asked Dec 9 '18 at 15:15
RichardRichard
1,79511946
1,79511946
add a comment |
add a comment |
1 Answer
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$begingroup$
It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
$$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$
Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$
It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by
$$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$
But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.
Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
$endgroup$
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
$$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$
Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$
It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by
$$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$
But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.
Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
$endgroup$
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
add a comment |
$begingroup$
It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
$$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$
Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$
It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by
$$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$
But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.
Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
$endgroup$
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
add a comment |
$begingroup$
It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
$$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$
Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$
It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by
$$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$
But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.
Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
$endgroup$
It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
$$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$
Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$
It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by
$$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$
But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.
Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
answered Dec 14 '18 at 5:22
Alex RavskyAlex Ravsky
40.7k32282
40.7k32282
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
add a comment |
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
1
1
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
$begingroup$
You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
$endgroup$
– Lukas Geyer
Dec 15 '18 at 13:51
add a comment |
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