Fixed point problem with a monotone vector as a fixed point? Suppose $F : [0,1]^n \to [0,1]^n$ is continuously differentiable and $0 < \frac{\partial F_1}{\partial x_i} \leq \dots \leq \frac{\partial F_n}{\partial x_i} < \beta < 1$ for all $i =1,\dots,n$. Conjecture: there exists unique $x^* = F(x^*)$, and moreover, $x_1^* \leq \dots \leq x_n^*$.
Proof of the first part is quite straightforward: one can easily verify that $F$ is a contraction mapping and then apply the contraction mapping theorem. I would need some help with the second claim.
 A: Here is the question by @TomH (exact quote from the above):
Suppose $F : [0,1]^n \to [0,1]^n$ is continuously differentiable and $0 < \frac{\partial F_1}{\partial x_i} \leq \dots \leq \frac{\partial F_n}{\partial x_i} < \beta < 1$ for all $i =1,\dots,n$. Conjecture: there exists unique $x^* = F(x^*)$, and moreover, $x_1^* \leq \dots \leq x_n^*$.`
Proof of the first part is quite straightforward: one can easily verify that $F$ is a contraction mapping and then apply the contraction mapping theorem. I would need some help with the second claim.

Let me provide a counter-example $\ F : [0;1]^2\rightarrow [0;1]^2\ $ in dim 2 (it can be written in $n$ variables too):
Let an auxiliary function $\ g:[0;1]^2\rightarrow \mathbb R\ $ be given as follows:
$$g(x\ y)\,\ :=\,\ \frac14\cdot(x-\frac12)\ +\ \frac12\cdot(y-\frac14)$$
Then $\ -\frac14\le g(x\ y) \le \frac12\ $ for every $\ (x\ y)\in [0;1]^2;\ $ and $\ g(\frac12\ \frac14)\ =\ 0.\ $ Consider $\ F:[0;1]^2\rightarrow [0;1]^2\ $ defined by:
$$F(x\ y)\ :=\ \left(\frac12 + g\left(x\ y\right),\ \frac14 + g\left(x\ y\right)\right)$$
Indeed, the values of the first coordinate of $F$ belong to the interval $\ \left[\frac14;1\right],\ $ and of the second coordinate to $\ \left[0;\frac34\right].\ $ Thus the range of $F$ is in $\ [0;1]^2\ $. Also, the partial derivatives, with respect to $x$, of the two coordinate functions are the same; and the same is true about $y$. Of course both derivatives (constants) belong to a proper closed subinterval of $\ [0;1].\ $ Finally
$$F(\frac12\ \frac14)\ =\ (\frac12\ \frac14)$$
where $\ \frac12 > \frac14.\ $ That's it.
A: So I see that you are assuming that the $n^2$ first order partial derivatives are all strictly between $0$ and $1$. The fixed point can be found by repeatedly applying $F$ starting (almost?) anywhere. It would be sufficient then that the property $x_1 \le x_2 \le \cdots \le x_n$ is preserved by application of $F$. On way to have that happen is to have conditions such as $\frac{\partial F_i}{x_j} \le \frac{\partial F_{i+1}}{x_{j}}$ although that is more than is needed. Indeed, if any subset of the set of points with ordered coordinates( such as the line $x_1=x_2=x_3=\cdots=x_n$ or the curve ($t^n,t^{n-1},\cdots,t^1)$) is mapped to itself, that suffices. So all the $F_i$ equal would be enough.
In a way that is trivial since the minimal condition is that there is a $1$ point set mapped into itself and that point is ordered. So what is your function or at least what flavor of properties are you considering? 
