4. Determine real numbers $a$ and $b$ such that for any real numbers $u$ and $v$ the following holds:

$$
\max _{0 \leq x \leq 1}\left|x^{2}-u x-v\right| \geq \max _{0 \leq x \leq 1}\left|x^{2}-a x-b\right| .
$$
Solution. It is easily proven that $\max _{0 \leq x \leq 1}\left|x^{2}-x+\frac{1}{8}\right|=\frac{1}{8}$. We will prove that for any real numbers $u$ and $v$, $\max _{0 \leq x \leq 1}\left|x^{2}-u x-v\right| \geq \frac{1}{8}$. Suppose for some numbers $u$ and $v$ the maximum of the function $f(x)=\left|x^{2}-u x-v\right|$ on the interval [0,1] is less than $\frac{1}{8}$, i.e., that $f(x)<\frac{1}{8}$ for every $x \in[0,1]$. Then

$$
|v|=f(0)<\frac{1}{8},\left|\frac{1}{4}-\frac{u}{2}-v\right|=f\left(\frac{1}{2}\right)<\frac{1}{8},|1-u-v|=f(1)<\frac{1}{8}
$$

from which it follows

$$
\left.\frac{1}{2}=\left|(1-u-v)-2\left(\frac{1}{4}-\frac{u}{2}-v\right)-v\right| \leq |1-u-v|+2\left| \frac{1}{4}-\frac{u}{2}-v\right|+| v \right\rvert\,<\frac{1}{8}+2 \cdot \frac{1}{8}+\frac{1}{8}=\frac{1}{2}
$$

which is a contradiction. Therefore, the required numbers are $a=1, b=-\frac{1}{8}$.

## III and IV year