Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:

Question

Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:

1. $|a_{n}|+|a_{n-1} | \leqslant 2^{n-1}$.
2. $|a_{n}|+|a_{n-1}|+\cdots+|a_{1}|+|a_{0}|\leqslant\frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}$.

Answer 1

Part (a)

By replacing $f(x)$ with $\pm f(\pm x)$ if necessary, we may assume that $a_n, a_{n-1}$ are both non-negative. Let $T_n(x)$ be the Chebyshev polynomial of the first kind, of degree $n$, i.e. it satisfies $$T_n(\cos \theta) = \cos (n\theta)$$ Notice that $T_n(x)$ is a polynomial of degree $n$ with leading coefficient $2^{n-1}$, and that $T_n(x) = \pm 1$ for $x = \cos \frac{0\pi}{n}, \cos \frac{\pi}{n}, \cdots, \cos \frac{n\pi}{n}$.

We first show that $a_n \leq 2^{n-1}$. Suppose the contrary, i.e. $a_n > 2^{n-1}$. Then consider $$g(x) = \frac{a_n}{2^{n-1}}T_n(x) - f(x)$$ would be positive at $\cos \frac{0\pi}{n}$, negative at $\cos\frac{\pi}{n}$, positive at $\cos\frac{2\pi}{n}$ and so forth. (Here we used $|f(x)| \leq 1$ whenever $|x| \leq 1$) Intermediate value theorem says that $g(x)$ would have at least $n$ distinct roots, but $g(x)$ has degree less than $n$. Thus $g(x) \equiv 0$, i.e. $f(x) = \frac{a_n}{2^{n-1}}T_n(x)$. But then $$f(1) = \frac{a_n}{2^{n-1}} T_n(\cos 0) = \frac{a_n}{2^{n-1}} > 1$$ Contradicting the assumption of $f$.

We then show that $a_n + a_{n-1} \leq 2^{n-1}$. Suppose the contrary, i.e. $$a_{n-1} > 2^{n-1} - a_n \hspace{5mm} (*)$$ Consider $$g(x) = (1+\epsilon) T_n(x) - f(x)$$ where $\epsilon > 0$ is sufficiently small and will be picked later.

The leading terms of $g(x)$ are $$g(x) = ((1+\epsilon)2^{n-1} - a_n)x^n - a_{n-1} x^{n-1} + \cdots$$ This implies that the sum of roots of $g(x)$ are $\frac{a_{n-1}}{(1+\epsilon)2^{n-1} - a_n}$. By picking a small enough $\epsilon$ and (*), the sum of roots of $g(x)$ is at least 1.

The condition on $f(x)$ (i.e. $|f(x)| \leq 1$ whenever $|x| \leq 1$) would again imply that $g(x)$ is positive at $\cos \frac{0\pi}{n}$, negative at $\cos\frac{\pi}{n}$, positive at $\cos\frac{2\pi}{n}$ and so forth. Intermediate value theorem says that $g(x)$ would have at least $n$ roots, one lying between $\cos \frac{k\pi}{n}$ and $\cos \frac{(k-1)\pi}{n}$ for $k = 1, \cdots, n$. But $g(x)$ has degree $n$, so these are exactly the roots of $g(x)$. But then the sum of roots is strictly smaller than

$$\cos \frac{0 \pi}{n} + \cos \frac{\pi}{n} \cdots + \cos \frac{(n-1) \pi}{n} = 1$$

Contradiction. This shows part (a).