# 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". Dec 23, 2016 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.