Got a super super interesting contest problem. $f(x,y,z)$ is a continuous function of three variables; it is required that $x\leq y\leq z$ (this means that $f$ is only defined on the 3-tuples such that $x\leq y\leq z$).
$f$ is increasing in the sense that $x'>x, y'>y,$ and $z'>z$ implies $f(x',y',z')>f(x,y,z)$.
Is it true that:
$$f(x,y,y)=f(x,x,y) \iff f(x,y,z)=g(x,z)?$$
A friend made this problem for practice purposes and I guess it could be a little bit loosely defined but the idea is there!
My try:
The statement is false and here is a counterexample. Let $A$ be a set; define the "secondary minimal function": $\min_2(A)=\{a\in A||a\leq A\setminus \min A; \}$ if nonempty; and if empty then $\min_2 A=0$.
Let $f(x,y,z)=\min\{x,y,z\}+\min_2\{x,y,z\}$.
for example: $f(0,0,1)=0+1=f(0,1,1)$ but $f(1,2,3)=1+2\neq g(1,3)$
My solution is too complicated. Is there a simple solution/counterexample?
By continuity of $f$ on $S := \left\{(x,y,z) \in \mathbb{R}^3 \mid x \leq y \leq z\right\}$ and the monotonicity condition, we have $f(x,y,z) \leq f(x',y',z')$ for all $\big((x,y,z), (x',y',z')\big) \in S^2$ such that $x \leq x'$, $y \leq y'$ and $z \leq z'$.
Let $x \leq z$ be fixed, let $y \in [x,z]$. Then we have: $$f(x,z,z) = f(x,x,z) \leq f(x,y,z) \leq f(x,z,z)$$ As such: $$\forall y \in [x,z],\quad f(x,y,z) = f(x,z,z)$$ Therefore, $f$ is indeed only a function of $x$ and $z$.