Recently, I came across the following problem (Problem 5, Interuniversity Iberoamerican Math. Competition (CIIM)):
Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be the operator defined by $$T(x,y,z) = (\sin(y) + \sin(z) - \sin(x),\sin(z)+\sin(x)-\sin(y),\sin(x)+\sin(y)-\sin(z)).$$ Determine all points $(x,y,z) \in [0,1]^3$ such that all iterations of this operator $T^{(n)}(x,y,z) \in [0,1]^3.$
I have been a little puzzled about this problem for over a week now, but no full solution has arisen.
My conjecture: the conjecture is that $x=y=z$. This is sufficient, as a computation shows.
1 - My first attempt was based on trying to 'measure' the set $$ A = \{(x,y,z)\in [0,1]^3; T^{(n)}(x,y,z) \in [0,1]^3, \,\forall n \in \mathbb{N}\}.$$ This is an invariant set of $[0,1]^3$ under $T$, and therefore maybe we could draw some conclusion about its volume from there. Even if we can, this argument could only show, potentially, that the set has zero measure, and we would not now exactly what it looks like.
2 - I was trying to explicitly see what the second iteration is. Id est, if $\sin(z) = Z, \sin(y) = Y,\sin(x) = X,$ and $Z \le Y \le X,$ a consequence of $T^2(x,y,z) \in [0,1]^3$ we get
$$ 2\sin(Z)\cos(X-Y) - \sin(X+Y-Z) \ge 0.$$
On the other hand, it is simple to show that the expression above is bounded by $\sin(Z).$ This alone, however, did not allow me to draw the conclusion. One can try to fiddle around with further formulae, but I could not get anything that allows one to conclude from them.
Does anyone have another idea on how to tackle this? Maybe my conjecture is wrong, in which case I would be very interested in knowing what is the right one.
I guess I actually do have an answer (after having talked to a friend about it for a while):
Notice that the difference between the maximal coordinate and minimal coordinates of $T(x,y,z)$ is $$ 2 (\sin x - \sin z),$$ where $\sin x \ge \sin y \ge \sin z.$ But we can easily estimate $$ 2 (\sin x - \sin z) = 2 \int_x^z \cos t \,dt \ge 2\cos 1 (x-z) > (x-z),$$ whenever $x >z.$
As $T^{(n)}(x,y,z) \in [0,1]^3$ for all $n \in \mathbb{N},$ this difference has to be at most $1,$ and, in particular, bounded. But, if $x>z,$ then this sequence of differences (which we shall call $s_n(x,y,z)$) satisfies $$ \frac{s_{n+1}(x,y,z)}{s_n(x,y,z)} > 2 \cos 1 > 1.$$ Therefore, it diverges, a contradiction.
This implies $x=z,$ and, with $x \ge y \ge z,$ that the three of them have to be equal, confirming the conjecture.