A min-max approximation Let $n\ge 1$ be an integer, $\mathcal P_n$ be the vector space of all polynomial functions over $[a,b]$, of degree at most $n$. 
My question is : Is it true that
$$\inf_{x_0,x_1,...,x_n\in[a,b], x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)|=\inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|.$$ 
I can easily see that $$\inf_{x_0,x_1,...,x_n\in[a,b],x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)| \ge \inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|,$$ 
but I don't know about the other reverse inequality.
 A: Take any interval $[a,b]$.  We want to show that
$$
\inf_{a\leq x_{0}<\ldots <x_{n} \leq b} \sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{P \in \mathcal{P}_{n}}\sup_{x \in [a,b]} |x^{n+1}-P(x)|
$$
Notice that $|x^{n+1}-P(x)| = \prod_{j=0}^{n}|x-z_{j}|$ for some $z_{j} \in \mathbb{C}$. Moreover by varying the polynomial $P$ we can make $z_{j}$ to be an abitrary complex numbers (if $\mathcal{P}_{n}$ are real polynomials then whenever $z_{j}$ is complex for some $j$ there will be its conjugate counterpart as well so we will have terms of the form  $|x-z_{j}|^{2}$)
Next since $|x-z_{j}|\geq |x-\Re z_{j}|$ it is always better to choose $z_{j}$ to be real numbers. Thus we only need to verify that 
$$
\inf_{a\leq x_{0}<\ldots <x_{n} \leq b} \sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{x_{0},x_{1}, \ldots, x_{n}}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}|.
$$
Becuase of the symmetry in the right hand side without loss of generality we can assume that $x_{0}\leq x_{1}\leq \ldots \leq  x_{n}$. It is clear that the right hand side is attainable for some $x,x_{0},\ldots, x_{n}$.  Next, consider the function $f(x)= \prod_{j=0}^{n}|x-x_{j}|$ on $[a,b]$, and fix points $x_{0}\leq \ldots \leq x_{n}$. Let $x_{\ell}$ be the smallest point such that $x_{\ell}>b$. By moving $x_{\ell}$ towards the point $b$ the value of the function $f(x)$ decreasis no matter where $x$ is located in $[a,b]$. Similarlry if $x_{q}$ is the largest point such that $x_{q}<a$ then by moving $x_{q}$ towards $a$ the value of $f(x)$ decreasis. This means that 
$$
\inf_{x_{0},x_{1}, \ldots, x_{n}}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}| = \inf_{a\leq x_{0}\leq x_{1}\leq  \ldots\leq  x_{n}\leq b}\sup_{x \in [a,b]} \prod_{j=0}^{n}|x-x_{j}|
$$
and this proves the claim because it really does not matter whether we take infimum in the righ hand side over $x_{0}<x_{1}...$ or $x_{0}\leq x_{1}...$ by continuity. For example if you have a contiuous function $g(x_{1}, x_{2})\geq 0$  then one can easily see that $\inf_{a\leq x_{0}<x_{1}\leq b}g(x_{1},x_{2}) =\inf_{a\leq x_{0}\leq x_{1}\leq b}g(x_{1},x_{2})$ and you can iterate this equality for the rest of the variables. 
A: When $[a,b]=[-1,1]$, the $\inf$ on the right-hand side is attained (uniquely) by the monic Chebyshev polynomial $T_{n+1}$. It is well known that its roots belong to $[-1,1]$ and are simple. 
For a general interval $[a,b]$, using the linear map 
$$f(x)=\left(\frac{2}{b-a}\right)x-\left(\frac{b+a}{b-a}\right)$$
that sends $[a,b]$ onto $[-1,1]$, it is clear that the $\inf$ on the right-hand side is attained by
$$
\left(\frac{b-a}{2}\right)^{n+1} T_{n+1}(f(x)),$$
which, of course, has all its (simple) roots in $[a,b]$.
Actually, a more general result holds true. Replace $[a,b]$ with any compact subset $K$ of the complex plane $\mathbb{C}$. Set, for a polynomial $P$,
$$\|P\|_{K}=\sup_{z\in K}|P(x)|,$$
and let $T_{n}$ be a polynomial that achieves the minimum of $\|P\|_{K}$ among all monic polynomials of degree $n$.
Claim: All the zeros of $T_{n}$ belong to the convex hull of $K$.
Indeed, assume that $z_{0}$ is a root of $T_{n}$ that does not belong to the convex hull of $K$. Then, $K$ lies in a cone with vertex at $z_{0}$, of opening $<\pi$. Choose a $z_{1}$ on the bisector $L$ of that cone, sufficiently close to $z_{0}$ so that $K$ lies in the half-plane, delimited by the perpendicular to $L$ at $z_{1}$ (and not containing $z_{0}$). Since
$$|z-z_{1}|<|z-z_{0}|,\quad z\in K,$$
one gets $\|\tilde T_{n}\|_{K}<\|T_{n}\|_{K}$ where 
$\tilde T_{n}(z)=T_{n}(z)(z-z_{1})/(z-z_{0})$, which contradicts the assumption on $T_{n}$. 
