How prove $ \; |f(1)|\le 2004\;$ if $\sqrt {x(1 - x)}\; \Big|f(x)\Big|\le 334$ for $f(x) = Ax^2+ Bx + C $

Question

Let $ \; A,B, C\in {\mathbb R} ,\;$ and $ \; f(x) = Ax^2+ Bx + C$ and

$ \sqrt {x(1 - x)} \left|f(x)\right|\le 334,\;\forall x\in [0,1]\;$.

How prove $ \; \left|f(1)\right|\le 2004\;$ ?

Answer 1

We will make the linear transformation: $x = \frac{1-t}{2}$ for $t \in [-1,1]$,and $g(t) = \frac{f\left(\frac{1-t}{2}\right)}{668}$ to rewrite the problem as: $$\sqrt{1-t^2}|g(t)| \le 1 \textrm{ for }t \in [-1,1] \tag{1}$$

and make the more general statement that, for any polynomial $g$ with degree $n-1$, satisfying $(1)$, then: $$|g(t)| \le n \textrm{ for } t \in [-1,1] \tag{*}$$

We use the Lagrange Interpolation on the Chebyshev points, $t_j = \cos\frac{(2j-1)\pi}{2n}$ for $j=1,2,\cdots,n$ the roots of the $n^{th}$ Chebyshev Polynomial: $T_n(t)$, to obtain the identity:

$$g(t) = \frac{1}{n}\sum_{i=1}^n (-1)^{i-1}\sqrt{1-t_i^2}g(t_i)\frac{T_n(t)}{t-t_i} \tag{2}$$

Take $t \in [-1,1]$ and if $t \in [t_n,t_1]$, that is $[-t_1,t_1]$, then from $(1)$:

$$|g(t)| \le \frac{1}{\sqrt{1-t^2}} \le \frac{1}{\sqrt{1-t_1^2}} = \frac{1}{\sin \frac{\pi}{2n}} \le n$$

by the Jordan Inequality.

If, $t > t_1$ (similarly for $t < -t_1$), the triangle inequality on $(2)$ gives

$$|g(t)| \le \frac{1}{n}\sum_{i=1}^n \frac{T_n(t)}{t-t_i} = \frac{T'_n(t)}{n}$$

Since, $T_n(\cos \theta) = \cos n\theta \implies T_n'(\cos \theta) = n\frac{\sin n\theta}{\sin \theta}$ but, $|\sin n\theta| \le n|\sin \theta|$

Therefore, $|T_n'(\cos \theta)| \le n^2 \implies |g(t)| \le n$ for $t > t_1$ (or $t < -t_1$) as well.

Equality holds in $(*)$ when $g(t) = \pm U_{n-1}(t)$, ($U_{n}$ being the Chebyshev Polynomial of $2^{nd}$ kind ) at $t=\pm 1$.

Since, $f$ is a polynomial of degree $2$, we have: $$|f(x)| \le 3 \times 688 = 2004 \textrm{ for } x \in [0,1]$$ and equality holds when $f(x) = 688(4(1-2x)^2 -1)$, at $x = 1$.