Arzelà- Ascoli Theorem and polynomials I am studying Real Analysis and I need to proove this exercise:
Let $(Pn)$ a sequences of polynomials, with degree $\leq m$. If $(Pn)$ is uniformly bounded on a compact set $K$ then there is a uniformly convergent subsequence of $(Pn)$.
My Attempt:
It's clearly that I need to use the Arzelà- Ascoli Theorem, but I need to show that this sequences of polynomials are equicontinuous and I don't know how I will show that.
My objective is show that $p_n'(x)$ are uniformly bounded so it's done because by Mean Value Theorem we have:
$|P_n(x) - P_n(y)| = |P_n'(c)||x-y|$, since $|x-y| < \delta $
and if $|p_n '(x)| < L$ for all $n \in \mathbb{N}$ then for all $\epsilon>0$ take $\delta = \frac{\epsilon}{L}$.
How to show that the derivarites is uniformly bounded?
 A: I think this problem is harder than it seems at first, since I see no direct way to apply the Arzela-Ascoli theorem off-the-bat.  Here is one possible solution:
Write $P_n(x) = \sum_{k=0}^m a_k^{(n)}x^k$.  We want to show that the coefficients of the $P_n$ are uniformly bounded.  If $K$ is finite the proposition is trivial.  So we assume $K$ is infinite and pick any $m + 1$ distinct points from $K$, say $\{x_1, \ldots, x_{m + 1}\}$.  I claim that for each $k$ where $0 \leq k \leq m$ there exists $c_1, \ldots, c_{m + 1}$ such that $\sum_{i = 1}^{m + 1} c_i P_n(x_i) = a_k^{(n)}$ for all $n$.  Fix $k$ so that $0 \leq k \leq m$.  Note 
$$\sum_{i = 1}^{m + 1} c_i P_n(x_i) = \sum_{i = 1}^{m + 1} c_i \sum_{j = 0}^{m} a_j^{(n)}x_i^j = \sum_{j = 0}^m a_j^{(n)} \sum_{i = 1}^{m + 1} c_i x_i^j.$$
So we can achieve the desired result if we can choose the $c_i$'s such that
$$\sum_{i = 1}^{m + 1} c_i x_i^j = \begin{cases}
1, j = k & \\
0, j \neq k
\end{cases}$$
Well this is true since the Vandermonde matrix
$$
\begin{bmatrix}
1 & x_1 & x_1^2  & \ldots & x_1^m \\
1 & x_2 & x_2^2 & \ldots & x_2^m \\
\vdots & \vdots &  & & \vdots \\
1 & x_{m + 1} & x_{m+1}^2 & \ldots & x_{m+1}^m
\end{bmatrix}
$$
is invertible.  So indeed we can write $$\sum_{i =1}^{m + 1}c_i P_n(x_i) = a_k^{(n)} \quad \text{for all } n$$
for some choice of $c_i$'s.  By this relation we can conclude the sequence $(a_k^{(n)})_{n = 1}^\infty$ is bounded since the $P_n$ are uniformly bounded on $K$.  This argument runs through for any $k$ such that $0 \leq k \leq m$.  Thus the coefficients of the $P_n$ are bounded uniformly, say by the constant $M > 0$.  Then we have
$$|P_n(x) - P_n(y)| = \left|\sum_{k = 0}^m a_k^{(n)} (x^k - y^k)\right| \leq \sum_{k = 0}^m |a_k^{(n)}| |x^k - y ^k| \leq M \sum_{k =0}^{m} |x^k - y^k|.$$
The sum on the right can be controlled, thus you have equicontinuity.  Thus by the Arzela-Ascoli theorem you get the desired result. 
