Putnam 2011, Problem A-5

Let $F: \mathbb{R}^2 \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be twice continuously differentiable functions with the following properties:
- $F(u,u) = 0$ for every $u \in \mathbb{R}$;
- for every $x \in \mathbb{R}$, $g(x) > 0$ and $x^2 g(x) \leq 1$;
- for every $(u,v) \in \mathbb{R}^2$, the vector $\nabla F(u,v)$ is either $\mathbf{0}$ or parallel to the vector $(g(u), -g(v))$.

Prove that there exists a constant $C$ such that for every $n \geq 2$ and any $x_1, \ldots, x_{n+1} \in \mathbb{R}$, we have
$$\min_{i \neq j} |F(x_i, x_j)| \leq \frac{C}{n}.$$

Solution:
(by Abhinav Kumar) Define \(G : \mathbb{R} \to \mathbb{R}\) by \(G(x) = \int_0^x g(t) \, dt\). By assumption, \(G\) is a strictly increasing, thrice continuously differentiable function. It is also bounded: for \(x > 1\), we have  
\[
0 < G(x) - G(1) = \int_1^x g(t) \, dt \leq \int_1^x dt/t^2 = 1,
\]  
and similarly, for \(x < -1\), we have \(0 > G(x) - G(-1) \geq -1\). It follows that the image of \(G\) is some open interval \((A,B)\) and that \(G^{-1} : (A,B) \to \mathbb{R}\) is also thrice continuously differentiable.  
Define \(H : (A,B) \times (A,B) \to \mathbb{R}\) by \(H(x,y) = F(G^{-1}(x), G^{-1}(y))\); it is twice continuously differentiable since \(F\) and \(G^{-1}\) are. By our assumptions about \(F\),  
\[
\frac{\partial H}{\partial x} + \frac{\partial H}{\partial y} = \frac{\partial F}{\partial x}(G^{-1}(x), G^{-1}(y)) \cdot \frac{1}{g(G^{-1}(x))} + \frac{\partial F}{\partial y}(G^{-1}(x), G^{-1}(y)) \cdot \frac{1}{g(G^{-1}(y))} = 0.
\]  
Therefore \(H\) is constant along any line parallel to the vector \((1,1)\), or equivalently, \(H(x,y)\) depends only on \(x - y\). We may thus write \(H(x,y) = h(x - y)\) for some function \(h\) on \((-(B - A), B - A)\), and we then have \(F(x,y) = h(G(x) - G(y))\). Since \(F(u,u) = 0\), we have \(h(0) = 0\). Also, \(h\) is twice continuously differentiable (since it can be written as \(h(x) = H((A + B + x)/2, (A + B - x)/2)\)), so \(|h'|\) is bounded on the closed interval \([-(B - A)/2, (B - A)/2]\), say by \(M\).  
Given \(x_1, \dots, x_{n+1} \in \mathbb{R}\) for some \(n \geq 2\), the numbers \(G(x_1), \dots, G(x_{n+1})\) all belong to \((A,B)\), so we can choose indices \(i\) and \(j\) so that \(|G(x_i) - G(x_j)| \leq (B - A)/n \leq (B - A)/2\). By the mean value theorem,  
\[
|F(x_i, x_j)| = |h(G(x_i) - G(x_j))| \leq M \frac{B - A}{n},
\]  
so the claim holds with \(C = M(B - A)\).