# Find the maximum of $|a_{p}|$, if $a_0+a_1x+\dots+a_nx^n:[-1,1]\mapsto [-1,1]$

Let $n$ be a given positive integer, and let $f(x)=\displaystyle\sum_{k=0}^{n}a_{k}x^k$, where $a_{i}\in \mathbb{R}$, $0 \le i \le n$. If $$|f(x)|\le 1,\qquad \text{for } ~|x|\le 1,$$ what is the maximum of the $|a_{p}|$ for a fixed $p$？

I conjecture that the answer is $|[x^p]T_{n}(x)|$, where the $T_{n}(x)$ are the Chebyshev polynomials. Can we find the closed form for it? Thanks

• Of course not always: Chebyshev polynomials are either odd or even, so many coefficients are simply equal to 0. The formula for the coefficients is written on your link, in the section "Explicit expressions". – Fedor Petrov Dec 23 '16 at 12:34

Let $T_{n}(x)=\sum_{\nu=0}^{n}t_{n,\nu}x^{\nu}$ denote the Chebyshev polynomial (of the first kind) of degree $n$ and let $x_{n,\nu}=\cos\nu\pi/n$ for $0\leq\nu\leq n$.

The answer follows from the following two results :

1) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that $$|f(x_{n,\nu})|\leq1,\qquad 0\leq\nu\leq n.$$ Then $$|a_{n-2\mu}|\leq|t_{n,n-2\mu}|,\qquad0\leq\mu\leq n/2,$$ which becomes an equality for $f=T_{n}$.

2) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that $$|f(x_{n-1,\nu})|\leq1,\qquad 0\leq\nu\leq n-1.$$ Then $$|a_{n-2\mu-1}|\leq|t_{n-1,n-2\mu-1}|,\qquad0\leq\mu\leq (n-1)/2,$$ which becomes an equality for $f=T_{n-1}$.

These are Theorems 16.3.1 and 16.3.2 in :

Rahman, Q. I.; Schmeisser, G. Analytic theory of polynomials. London Mathematical Society Monographs. New Series, 26. The Clarendon Press, Oxford University Press, Oxford, 2002.

The proof of the first assertion consists in considering a one-parameter family of polynomials constructed from $f$ and $T_{n}$ and depending on a parameter $\theta\in(-1,1)$, and then using Descartes' rule of signs. The second assertion is a simple consequence of the first one.