TITLE: Prove that a uniformly convergent convergent sequence of $N^\text{th}$ degree polynomials must converge to some $N^\text{th}$ degree polynomial
QUESTION [9 upvotes]: So here's the question I'm trying to answer:

Suppose $p_n(x) = \sum_{k=1}^N a_k^{(n)} x^k$ is a sequence of polynomials such that $p_n \to f$ uniformly over $[0,1]$ for some function $f:[0,1] \to \mathbb{R}$.  Prove that $f$ must itself be an $N^\text{th}$ degree polynomial.

I've already shown that if each $a_k^{(n)} \to a_k$, then $p_n(x) \to p(x) = \sum_{k=1}^N a_k x^k$  uniformly (earlier part of the problem).  I'm thinking that there's some way to show that if $p_n \to f$, then $a_k^{(n)}$ converges for each $k$.  This certainly works for $k = 0$, since we can guarantee that the sequence $a_k^{(n)} = p_n(0)$ is Cauchy.  I've gotten stuck in trying to extend this to other coefficients; I'm thinking there's some trick involving subtracting the $a_0^{(n)}$ off and dividing by $x$, maybe some fancy induction along those lines.
Other potentially helpful thoughts: we can guarantee that $f$ is continuous since it is the uniform limit of continuous functions. Remember also that we have a compact domain, so that all of these functions are bounded and achieve their max/min.
Any comments, hints, or answers are very much appreciated.

REPLY [5 votes]: A proof that for my criterion is elegant is using facts of functional analysis:
Let $P_N$ the set of all polynomials of degrre $N$, this is a vector space and isometrically isomorphic to $\mathbb{R}^{N+1}$ via
$$\begin{array}{rcl}\Phi: P_{N} &\rightarrow & \mathbb{R}^{N+1}\\ a_0+a_1 x +\cdots+a_N x^N & \mapsto & (a_0,\ldots,a_N).\end{array}$$
Given any norm $\left\|\:\:\right\|$  in $\mathbb{R}^{N+1}$ we define the norm in $P_{N}$ for any $p(x)=a_0+a_1 x+\cdots+a_N x^N$ as
$$\left\|p\right\|:=\left\|(a_0,\ldots,a_N)\right\|.$$
Since $\mathbb{R}^{N+1}$ is complete, then $P_N$ is complete with this norm, therefore $(P_N,\left\|\:\:\right\|)$ is a Banach space, but  we know that all norms in a finite-dimensional space are equivalent, therefore, $P_N$ is Banach space with any norm, in particular, with the norm
$$\left\|p \right\|_{\sup}:=\sup_{x\in[0,1]}\left|a_0+a_1 x+\cdots+a_N x^N \right|.$$
From the real analysis courses we know that if $\left\{p_n\right\}$ converges uniformly then it does so in the norm $\left\|\:\right\|_{\sup}$, therefore, $\left\{p_n\right\}$ is a Cauchy sequence in the Banach space $(P_{N}, \left\|\:\right\|_{\sup})$, which allows us to conclude that $f\in P_N$.