# Chebyshev polynom problem.

Consider $$P_n = \{Q_n(x) : \deg Q_n = n; \|{Q_n}\| = \max_{[a,b]}|Q_n(x)| = M >0\}$$.

Now consider $$\bar{T}(x) = M T_n(\frac{2x - (b+a)}{b-a}) -$$ Chebyshev polynomial (normed on space $$P_n)$$.

We want to prove that $$\forall \xi \notin (a,b)$$ $$\nexists Q_{n}(x) \in P_n : |Q_n(\xi)| > |\bar{T_n}(\xi)|$$.

My attempt :

First of all assume that there is a point $$\xi \notin (a,b)$$(suppose $$\xi > b$$) and $$Q_n(x) : |Q_n(\xi)| > |\bar{T_n}(\xi)|$$.

If there is exists such point, hence we have there is exists $$[\xi - h,\xi + h]$$, where $$\|Q_n\| > \|\bar{T}\|$$, also there is exists $$\xi_0$$ ,where $$|Q_n(\xi_0)| = |\bar{T}_n(\xi_0)|$$ (let's suppose it's point, where $$Q_n(\xi_0) = \bar{T}_n(\xi_0)$$, if no - we make symmetry of $$Q_n(x)$$).

Hence we have $$a_n[\bar{T}_n(x)] < a_n[Q_n(x)]$$ (the major coefficient).

Now I guess there should be contradiction, but I don't know how to prove that major coefficient of $$\bar{T}(x)$$ should be the largest among $$Q_n(x) \in P_n$$?