Kvant Math Problem 859

Verified: no
Verdicts: SKIP + SKIP
Solve time: 55s
Source on kvant.digital

Problem

Find the smallest positive number $a$ such that for every quadratic polynomial $f(x)$ satisfying the inequality $|f(x)|\le1$ for $0\le x\le1$, the inequality $|f'(1)|\le a$ holds.

V. P. Pikulin

Exploration

Let

$$f(x)=ux^2+vx+w.$$

We seek the largest possible value of $|f'(1)|=|2u+v|$ under the condition

$$|f(x)|\le 1 \qquad (0\le x\le 1).$$

The smallest admissible constant $a$ is exactly this maximum.

A quadratic is determined by its values at three points. Since the interval is $[0,1]$, a natural choice is

$$f(0),\quad f!\left(\frac12\right),\quad f(1).$$

Write

$$A=f(0),\qquad B=f!\left(\frac12\right),\qquad C=f(1).$$

Then $|A|,|B|,|C|\le1$.

Using Lagrange interpolation at the nodes $0,\frac12,1$, compute

$$f(x)=A(2x-1)(x-1)-4B,x(x-1)+C,x(2x-1).$$

Differentiating and setting $x=1$ gives

$$f$$