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](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Trigonometric_definition). Can we find the closed form for it? Thanks



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


