If a sequence of polynomials converges then the sequence of the coefficients converges. Let $C[0,1]$ be the set of continous real-valued functions on [0,1]. $C[0,1]$ is a metric space with $d_\infty(f,g) := \sup_{x \in [0,1]}|f(x)-g(x)|$. Let $\mathcal{P}_n$ denote the set of polynomials with degree less or equal to n. I am trying to show $\mathcal{P}_n$ is closed in $C[0,1]$ with the infinity metric. I figured out the proof but at some point I use the following fact which I couldn't proove:
Let $Q_k(x) := a_{0}^{(k)} + a_{1}^{(k)}x +...+a_{n}^{(k)}x^n$ where $a_{i}^{(k)}$ is a real sequence indexed with k. If $Q_k(x)$ converges to some function $f$ in C[0,1] with the infinity metric then there exists $a_0, a_1,...,a_n \in \mathbb{R}$ such that $a_{i}^{(k)}$ converges to $a_i$ for $i \in \{0,1,...,n\}$ as k goes to infinity.
My attempt to prove:
Because $Q_k(x)$ is convergent it is a cauchy sequence. Let $\epsilon > 0$. There exists $K\in \mathbb{N}$ sucht that for all natural numbers $i,j > K$ we have $d_\infty(Q_i,Q_j) = \sup_{x\in[0,1]}|a_{0}^{(i)}-a_{0}^{(j)} + (a_{1}^{(i)}-a_{1}^{(j)})x^1 + ...+(a_{n}^{(i)}-a_{n}^{(j)})x^n| < \epsilon$. We evalute the function at $x=0$ and we get $|a_{0}^{(i)}-a_{0}^{(j)}| < \epsilon$. Hence $a_{0}^{(k)}$ is convergent.
I don't know how to proceed from this point. I tried some inductive arguments but I couldn't quite figured it out. If $n = 1$ I can prove $a_{1}^{(k)}$ converges by evaluating the function at x = 1 and using the previously proven fact $a_{1}^{(k)}$ but that is the best I could do.
 A: $\mathcal{P}_n$ is an $(n+1)$-dimensional linear subspace of $C([0,1])$ (the monomials $x^i$ with $i = 0,\ldots,n$ form a basis). The $\sup$-norm $\lVert - \rVert_\infty$ on $C([0,1])$ restricts to a norm on $\mathcal{P}_n$. Another norm on $\mathcal{P}_n$ can be defined by $\lVert a_0 + a_1x + \ldots + a_nx^n \rVert = \max_{i=0,\ldots,n} \lvert a_i \rvert$. Since all norms on finite-dimensional linear spaces are equivalent, convergence in $\mathcal{P}_n$ with respect to $\lVert - \rVert_\infty$ agrees with convergence with respect to $\lVert - \rVert$. But the latter means that the $n+1$ sequences of the coefficients converge.
A: Below there are two solutions not based on equivalence of the norms in finite dimensional spaces.
Solution $\mathbf 1$
Let $\{P_m\}_{m=1}^\infty$ be a sequence of polynomials such that $\deg P_m\le n.$
Assume  that for $n+1$ points $0\le x_0<x_1<\ldots <x_n\le 1$ the sequences   $\{P_m(x_k)\}_{m=1}^\infty $ are  convergent.
We claim that the sequence $P_m(x)$ is convergent  for any $x$ and $\displaystyle\lim_m P_m(x)=f(x)$ is a polynomial such that $\deg f\le n.$
The proof will go by induction with respect to $n.$ For $n=0$ the conclusion is obvious as $P_m$ is  constant for every $m.$ Assume the conclusion holds for $n-1.$ Let $\deg P_m\le n$ and $\{P_m(x_k)\}_{m=1}^\infty$ be convergent for $k=0,1,\ldots ,n.$ Consider the polynomials
$$Q_m(x)={P_m(x)-P_m(x_n)\over x-x_n}$$ Then $\deg Q_m\le n-1$ and the sequences $\{Q_m(x_k)\}_{m=1}^\infty$ are convergent for $k=0,1,\ldots , n-1.$  By induction hypothesis we get that $Q_m(x)$ is  convergent for every $x$ and $\displaystyle \lim_mQ_m(x)=g(x)$ is a polynomial such that $\deg g\le n-1.$
We have
$$P_m(x)=(x-x_n)Q_m(x)+P_m(x_n)$$ Hence $$\lim_mP_m(x)=(x-x_n)g(x)+\lim_m P_m(x_n)=:f(x)$$  and $\deg f\le n.$
Solution $\mathbf 2$ Consider $n+1$ points
$x_k={k\over n},$ $0\le k\le n.$
The polynomial
$$Q_k(x)=\prod_{j=0,j\neq k}^n{x-x_j\over x_k-x_j}$$ satisfies $$\deg Q_k=n,\quad 
Q_k(x_l)=\begin{cases} 0 & k\neq l\\
1 & k=l
\end{cases}
$$ For any polynomial $P(x)$ such that $\deg P\le n$  we have
$$P(x)=\sum_{k=0}^n P(x_k)Q_k(x)$$
Indeed, both sides are polynomials of degree less than or equal $n$ and give the same values at $n+1$ different points $x_k.$ Therefore they coincide for any $x.$
Assume a sequence of polynomials $P_m,$ $\deg P_m\le n,$ is convergent pointwise to a function $f.$ We have
$$P_m(x)= \sum_{k=0}^n P_m(x_k)Q_k(x)\quad (*)$$
Taking the limit $m\to\infty$ of both sides gives
$$f(x)=\sum_{k=0}^nf(x_k)Q_k(x)$$
The right hand side represents a polynomial of degree less than or equal $n.$
A: Any linear map on a finite dimensional normed linear space is continuous.  $\sum\limits_{k=1}^{n} a_k x^{k} \mapsto (a_1,a_2,...,.a_n)$ is a well-defined linear map and hence it is continuous. Note that this works for any norm on $C[0,1]$, not just the sup norm!
