$F: \mathbb{R}^2 \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are twice continuously differentiable functions such that $x^2 g(x) \le 1$ and $F(x, x) = 0 < g(x)$ for all $x.$ Furthermore, $\nabla F(x,y)$ is either $\mathbf{0}$ or parallel to $(g(x), -g(y))$ for all $x,y.$ Prove there exists $C$ such that for all $x_1, \dots, x_{n+1} \in \mathbb{R},$ we have $\min\limits_{i \ne j} |F(x_i, x_j)| \le \frac{C}{n}.$
My progress: The parallelity condition is equivalent to there always existing $c(x,y)$ such that $F_x = c(x,y)g(x), F_y = -c(x,y)g(y).$ Even if we assume or prove $c(x,y)$ is continous, I can't seem to get much use out of $F = \int c(x,y)g(x) \, dx = -\int c(x,y)g(y) \, dy.$ Green's Theorem gives us $\int\int_D (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}) \, dx \, dy = 0$ for any region $D$ bound by a closed curve where $L = c(x,y)g(x), M = c(x,y)g(y),$ but what's next? From $F_{xy} = F_{yx},$ we have $g(x) c_y = -g(y) c_x,$ which makes the integral turn into $0=0.$ Thus, Green's Theorem doesn't give us anything new. And all of this is assuming we can even differentiate $c.$
The idea of the problem seems to be that $F$ does not grow very quickly away from the line $y=x,$ on which it is zero. However, I don't have any intuition about how adding more and more points will guarantee a smaller function value for some coordinate pair formed by a pair of different points. The case $n=2$ is $|F(x_1, x_2)| \le C/2,$ so we need to show $F$ is bounded. If we show $F$ is bounded and we show a reverse triangle inequality in the form $|F(a,c)| \ge |F(a,b)| + |F(b,c)|,$ that would solve the problem. I don't see a reason for the reverse triangle inequality to be true. It seems we need to somewhow connect the parallelity condition and the fact that $x^2 g(x) \le 1$ together in order to show $F$ is bounded. Loosely speaking, $g$ decays faster than $1/x^2,$ so its gradient might be very small too. But the word "parallel" completely wipes out any guarantee on size, making it impossible to figure out large or small $\nabla F$ is. In light of all of this, I'm not sure how to combine the $2$ constraints on $g.$
More observations: Since $g \ge 0,$ the gradient of $F$ is always in the 2nd or 4th quadrant. If "parallel" means $c(x,y) \ge 0,$ then the gradient of $F$ is always in the 4th quadrant. In either case, since the gradient points in the direction of greatest change, we can expect $F$ to increase or decrease the most as we move away from the line $y=x.$ But we already knew this since $F = 0$ on that line. If $F$ is unbounded, there exists $v_1, v_2, \dots \in \mathbb{R}^2$ such that $|F(v_k)| > k.$ There's no analogue of the mean value theorem allowing us to estimate $\nabla F$ between the $v_i,$ and even if there were, the estimate immediately goes out the window because as mentioned earlier, a vector parallel to $(g(x), -g(y))$ can be as large or small as we wish.
In fact, is there any way to use the fact that $g$ decays quickly? Right now, it seems to be a red herring. The only thing I can imagine is fixing $y$ and letting $x \to \infty,$ or fixing $x$ and letting $y \to \infty.$ But then we only get information about $\nabla F$ on one line, and it's not much info.
In summary, I have many observations about each of the conditions, but I haven't been able to find anything significant because I can't connect the conditions together. How should I proceed? How can you prepare yourself to solve such problems?
The function $c$ is indeed $C^1$: $g$ is assumed to be $C^2$ nonvanishing and $F$ being $C^2$ means $F_x$ is $C^1$, so $c\colon(x,y)\mapsto F_x(x,y)/g(x)$ is $C^1$.
But yes, the function $F$ need not even be bounded. It is easy to see that, if $G$ is an antiderivative of $g$, then $G$ is bounded, and $H(x,y)=G(x)-G(y)$ satisfies the condition given for $F$. This is probably what the question intends to do, but fails miserably because of the "parallel to": for any $\varphi\in C^2(\mathbb{R})$ with $\varphi(0)=0$, we have $F=\varphi\circ H$ satisfies the condition in the question. But $H(\mathbb{R}^2)$ is an open interval because $g>0$ everywhere, so you can obviously make $F$ unbounded by, e.g., $$ F(x,y)=\tanh^{-1}\frac{H(x,y)}{\int_{-\infty}^\infty g(t)\,\mathrm{d}t}. $$