Let $f_i(x_1, x_2, ..., x_n)$ for $i=1,...,n$, be real-valued differentiable functions with the following properties:

1) $f_i(x_1, x_2, ..., x_n)=0$ if $x_i=0$.

2) $f_i(x_1, x_2, ..., x_n)=1$ if $x_i=1$.

3) $\frac{\partial f_i(x_1, x_2, ..., x_n)}{\partial x_i}>0$, for $0<x_i<1$

4) $\frac{\partial f_j(x_1, x_2, ..., x_n)}{\partial x_i}<0$, for $j\neq i$ and $0<x_i<1$

Fix $p_1, ..., p_n$, with $0<p_i<1$ for $i=1,...n$. I want to prove that there is a unique solution $x_1, ..., x_n$, with $0<x_i<1$ for $i=1,...n$, such that $f_i(x_1, x_2, ..., x_n)=p_i$ for all $i=1,...n$.

Are the listed properties enough? If not, what extra properties do I require of the functions $f_i$?


closed as off-topic by Willie Wong, Stefan Kohl, Peter Humphries, Ryan Budney, Joonas Ilmavirta Nov 10 '15 at 19:06

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    $\begingroup$ (a) No, they are not enough. Set $n = 2$ you can easily find an example where $\nabla f_1$ and $\nabla f_2$ are parallel on an open subset of $(0,1)^2$. (Start with $f_1 = 1/2 + x_1 - x_2$ and $f_2 = 1/2 + x_2 - x_1$ and modify it outside $(1/3,2/3)^2$. This can be generalized to higher dimensions.) (b) This question has nothing to do with functional analysis. (c) You should ask this at math.stackexchange.com . $\endgroup$ – Willie Wong Nov 6 '15 at 16:02
  • $\begingroup$ (d) You are almost asking for $f$ to be a $C^1$ diffeomorphism of $[0,1]^n$ to itself. So you need that its derivative $Df$ is everywhere surjective. $\endgroup$ – Willie Wong Nov 6 '15 at 16:06
  • $\begingroup$ Thanks for the feedback. How about the existence of a solution? Can I prove that? Probably using something like that for all $x_2, ... x_n$, I can find a $x_1$ such that $f_1(x)=p_1$, but I'm not sure how to formalize this. $\endgroup$ – MthQ Nov 10 '15 at 8:49
  • $\begingroup$ There probably is a degree theory argument for existence. $\endgroup$ – Willie Wong Nov 10 '15 at 14:31

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