# Show some theorem concerning of the uniform convergence on compacts of a sequence of polynomials of order $<k$

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.

## 1 Answer

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 $$f(a)$$.

• 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. Commented Dec 15, 2018 at 13:51