IMC 2005 · Day 2 · P11

To get an upper bound for $r$, set $f(x, y) = x - \dfrac{x^2}{2} + \dfrac{y^2}{2}$. This function satisfies the conditions, since $\operatorname{grad} f(x, y) = (1 - x, y)$, $\operatorname{grad} f(0, 0) = (1, 0)$ and $|\operatorname{grad} f(x_1, y_1) - \operatorname{grad} f(x_2, y_2)| = |(x_2 - x_1, y_1 - y_2)| = |(x_1, y_1) - (x_2, y_2)|$.

In the disk $D_r = \{ (x, y) : x^2 + y^2 \le r^2 \}$ $$f(x, y) = \frac{x^2 + y^2}{2} - \left( x - \frac{1}{2} \right)^2 + \frac{1}{4} \le \frac{r^2}{2} + \frac{1}{4}.$$ If $r > \frac{1}{2}$ then the absolute maximum is $\frac{r^2}{2} + \frac{1}{4}$, attained at the points $\left( \frac{1}{2}, \pm\sqrt{r^2 - \frac{1}{4}} \right)$. Therefore, it is necessary that $r \le \frac{1}{2}$ because if $r > \frac{1}{2}$ then the maximum is attained twice.

Suppose now that $r \le 1/2$ and that $f$ attains its maximum on $D_r$ at $u, v$, $u \ne v$. Since $|\operatorname{grad} f(z) - \operatorname{grad} f(0)| \le r$, $|\operatorname{grad} f(z)| \ge 1 - r > 0$ for all $z \in D_r$. Hence $f$ may attain its maximum only at the boundary of $D_r$, so we must have $|u| = |v| = r$ and $\operatorname{grad} f(u) = au$ and $\operatorname{grad} f(v) = bv$, where $a, b \ge 0$. Since $au = \operatorname{grad} f(u)$ and $bv = \operatorname{grad} f(v)$ belong to the disk $D$ with centre $\operatorname{grad} f(0)$ and radius $r$, they do not belong to the interior of $D_r$. Hence $|\operatorname{grad} f(u) - \operatorname{grad} f(v)| = |au - bv| \ge |u - v|$ and this inequality is strict since $D \cap D_r$ contains no more than one point. But this contradicts the assumption that $|\operatorname{grad} f(u) - \operatorname{grad} f(v)| \le |u - v|$. So all $r \le \frac{1}{2}$ satisfies the condition.