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Let $f(x)=\sum_{k=0}^{n}a_kx^k\in\mathbb{R}[x]$ with $a_n\neq0$ and $I\subseteq\mathbb{R}$ a nonempty interval, prove that $$\sup_{x\in I}|f(x)|\geq\frac{a_n(\sup I-\inf I)^n}{2^{2n-1}},$$ where the lower bound is optimal.

My attempt:

The result is easily seen to be equivalent to the following proposition:

Let $f(x)=\sum_{k=0}^{n}a_kx^k\in\mathbb{R}[x]$ with $a_n\neq0$ such that $|f(x)|\leq1$ for all $-1\leq x\leq1$, prove that $a_n\leq2^{n-1},$ where the upper bound is optimal.

The first step is to prove the inequality. I started by trying out a few $n$'s. For $n=1$, this is trivial. For $n=2$, observe that $$|a_2|=\Bigg|\frac{f(1)-2f(0)+f(-1)}{2}\Bigg|\leq\frac{|f(1)|+2|f(0)|+|f(-1)|}{2}\leq2.$$ For $n=3$, again, observe that $$|a_3|=\Bigg|\frac{2f(1)-4f(1/2)+4f(-1/2)-2f(-1)}{3}\Bigg|\leq4.$$ Thus, for the general case, I guess we can construct an identity of the form $$|a_n|=\Bigg|\sum_{k=0}^{n}c_kf(\cos\frac{k\pi}{n})\Bigg|,$$ where $\sum_{k=0}^{n}|c_k|=2^{n-1}$. Expanding the expression leads to a system of $n$ linear equations in $n$ variables with augmented matrix $$\left(\begin{array}{@{}cccc:c@{}} \cos^0\dfrac{0\pi}{n}&\cos^0\dfrac{1\pi}{n}&\cdots&\cos^0\dfrac{n\pi}{n}&0\\ \cos^1\dfrac{0\pi}{n}&\cos^1\dfrac{1\pi}{n}&\cdots&\cos^1\dfrac{n\pi}{n}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ \cos^{n-1}\dfrac{0\pi}{n}&\cos^{n-1}\dfrac{1\pi}{n}&\cdots&\cos^{n-1}\dfrac{n\pi}{n}&0\\ \cos^n\dfrac{0\pi}{n}&\cos^n\dfrac{1\pi}{n}&\cdots&\cos^n\dfrac{n\pi}{n}&1\\ \end{array}\right)$$ We need to show that this system has a solution that satisfies $\sum_{k=0}^{n}|c_k|=2^{n-1}$. I am unable to proceed with my poor knowledge of linear algebra, so, any hints will be appreciated.

Edit: In this question, the same result is proved using Chebyshev polynomials. But is it provable following my attempt?

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  • $\begingroup$ Wouldn't you be done with just $\sum |c_k| \leq 2^{n-1}$? That seems easier, as all you need is a small normed solution. $\endgroup$ May 14 '17 at 16:05
  • $\begingroup$ @ArtimisFowl Well, the bound is optimal, so I think we cannot have $\sum|c_k|<2^{n-1}$. Also I'm unclear how to obtain any solution. Could you please elaborate on this? $\endgroup$
    – Colescu
    May 15 '17 at 3:04
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    $\begingroup$ A stronger result is proved in math.stackexchange.com/questions/281315/… $\endgroup$ May 15 '17 at 7:35

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