For functions $f,g: I \to \mathbb{R}$ increasing on interval $I$, prove that for the solution of the system of equations $$ \begin{cases} \begin{align} f(x_{1}) &= g(x_{2}) \\ f(x_{2}) &= g(x_{3}) \\ &\;\;\vdots \notag \\ f(x_{n}) &= g(x_{1}) \end{align} \end{cases} $$ $x_1 = x_2 = · · · = x_n$ is true.
I believe there might be a lemma for this, unfortunately, I can't find anything. Is there an elementary way to prove this?
On
Well, assume WLOG (the other cases are similar) that $x_n \ge x_{n-1} \ge \dots \ge x_2 \ge x_1$, because $f$ is increasing, we get $$\ \ \ \ \ \ \ \ \ \ \ f(x_n) \ge f(x_{n-1}) \ge \dots \ge f(x_2) \ge f(x_1)$$ $$\implies g(x_1) \ge g(x_{n}) \ge \dots \ge g(x_3) \ge g(x_2)$$ But since $g$ is also increasing, we have $$g(x_n) \ge \dots \ge g(x_3) \ge g(x_2) \ge g(x_1)$$ This naturally means $$g(x_n) = \dots = g(x_3) = g(x_2) = g(x_1)$$ and thus (increasing means injective) $$x_n=x_{n-1}=\dots=x_2=x_1$$
Suppose that $x_2\ge x_1$. Then $f(x_2)\ge f(x_1)$. Hence $g(x_3)\ge g(x_2)$, whence $x_3\ge x_2$. Hence $x_4\ge x_3$, and so on: $x_n\ge x_{n-1}$ and $x_1\ge x_n$, but that implies $x_1=x_2-...=x_n$.
Now suppose $x_2\le x_1$. Then similarly $x_3\le x_2$,..., $x_n\le x_{n-1}$ and $x_1\le x_n$, so again $x_1=x_2=...=x_n$.