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 min-max approximation
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 $T\_{n}$. 


