Pointwise Remez inequality

The classical Remez inequality provides an exact estimate for a polynomial on a given interval if it is known that the polynomial is bounded by one on a subset of this interval of the given Lebesgue measure. To be precise, let $\Pi_n$ denote the set of polynomials of degree at most $n$. For a subset $E$ of the interval $[-1,1]$ let
$$ \Pi_n(E):=\{P_n\in \Pi_n:\ |P_n(x)|\le 1,\ x\in E\}. $$
The Lebesgue measure of $E$ is denoted by $|E|$. For $\delta\in(0,1)$ we define
$$ M_{n,\delta}=\sup_{E:|E|=2-2\delta}\sup_{P_n\in\Pi_n(E)}\sup_{x\in[-1,1]}|P_n(x)|. $$
According to Remez $M_{n,\delta}=T_n\left(\frac{1+\delta}{1-\delta}\right)$, where $T_n(z)$ is the classical Chebyshev polynomial, $T_n(z)=\frac 1 2(\zeta^n+\zeta^{-n})$, $z=\frac 1 2 (\zeta+\zeta^{-1})$.

Since the mid-90s, based on the previous results of T. Erdélyi, E. B. Saff and himself, at several international conferences Vladimir Andrievskii raised the following problem. Find
$$ L_{n,\delta}(x_0)=\sup_{E:|E|=2-2\delta}\sup_{P_n\in\Pi_n(E)}|P_n(x_0)|, \quad x_0\in [-1,1]. $$
Note that $\sup_{x_0\in[-1,1]}L_{n,\delta}(x_0)=M_{n,\delta}$.

In the talk we present a solution of Andievskii's problem on the pointwise Remez inequality.