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 $Q_n(x) \in P_n$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.