Problem
Show the Set of Polynomial $(p(x))$ of Degree $\le n$, Uniformly Bounded in a compact $C$ is Equicontinuous in $C$.
I know what to do but I don't understand it very well.
I've seen answers that goes as follows:
We can assume without loss of generality (I don't get why), that our compact $C$ is $[-1,1]$. Now we can use Markov Brothers' Inequality, which gives us:
$$\sup_{x\in [-1, 1]}|p'(x)| \le n^2 \sup_{x\in [-1, 1]}|p(x)|$$
for all polynomial p of degree $\le n$, therefore (I also don't understand the right-hand side of this step)
$$p'(x)\le n^2$$
so, as this holds for all polynomial in my set and for all $x\in [-1,1]$, we have that the set is equicontinuous.
Now, I really don't get the conclusion. How can this show that $\forall \epsilon >0 \enspace \exists \delta >0$ such that $x,x_0 \in \enspace C, |x-x_0|<\delta \Rightarrow |p(x)-p(x_0)|<\epsilon \enspace $ for every polynomial in my set of polynomials?
Sorry if I'm asking too much but I really don't follow the reasoning.
Best Answer
$1)$Every compact set on the real line is contained on some closed interval $[a,b]$
So there is a bijective continuous and differentiable function $s:[a,b] \to [-1,1]$
For instance $s(x)=\frac{x-a}{b-a}+\frac{x-b}{b-a}$
So you can work with the polynomial $p(s(x))$ the same way you worked with $p(x)$
$2)$Since the set $S$ is uniformly bounded ,then exists $M>0$ such that $\sup_{[-1,1]}|p(x)| \leq M,\forall p \in S$
Thus $|p'(x)| \leq n^2M,\forall x \in [-1,1],\forall p \in S$
So it does not matter here if the bound is $n^2$ (which may not be correct) The whole point of the exercise is the finding of a uniform bounded of the derivatives of the polynomials.
$3)$By M.V.T you have for every polynomial $p$ of degree less than $n$ that $|p(x)-p(y)| \leq Mn^2|x-y|$
Since $C=Mn^2$ does not depend on the choice of the polynomial,you can prove now equicontinuity.