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Let $k,s\in \mathbb N$,

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 $$ P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (\textrm{ for } n\in \mathbb N, x\in \mathbb R) $$ be a sequence of real polynomials of order $\leq k-1$.

Assume that there exist limits of derivatives:

$$ \lim_{n\rightarrow \infty} P_n^{(j)}(x_0) (\textrm{ for } j=0,1,...,m_0); $$ $$ \lim_{n\rightarrow \infty} P_n^{(j)}(x_1) (\textrm{ for } j=0,1,...,m_1); $$ ........................... $$ \lim_{n\rightarrow \infty} P_n^{(j)}(x_s) (\textrm{ for } j=0,1,...,m_s). $$

I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits: $$ \lim_{n\rightarrow \infty} a_{nj} (\textrm{ for } j=0,...,k-1). $$

Maybe proof or references.

Thanks.

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1 Answer 1

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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 $a=(a_0,a_1,\dots,a_{k-1})\in\Bbb R^k$. 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 $a\in\operatorname{ker} f$, 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 $a\in\Bbb R^k$ iff a sequence $\{f(a^n)\}$ converges to a point $f(a)$.

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    $\begingroup$ 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. $\endgroup$ Commented Dec 15, 2018 at 13:51

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