Proof of existence and uniqueness of solution to f(c)=0

Question

I have a function $f:R^n_+\rightarrow R^n$ for which I want to show the following:

$$\exists c\in R^n_+ \quad \forall i,j\,\,f_i(c)=f_j(c)$$ Where $f_i (c)$ are the different coordinates of $f$.

$f$ has the following properties:

1. $\frac{\partial f_i}{\partial c_j}>0 \Leftrightarrow i=j$, furthermore the partial derivatives are never $0$.
2. $\lim_{c_i\rightarrow \infty} f_i(c) = \infty$
3. $\lim_{c_j\rightarrow \infty} f_i(c) = A_i$, $A_i$ here does not depend on $j$ (and of course $j\neq i$).
4. $\forall t\in R_+ \quad f(c) = f(t\cdot c)$

For $n=2$ this is very easy, basically just the intermediate value theorem. For higher dimension it gets more complicated. The idea is the following: If there does not exist such a $c$, then the image of $f$ is contained in $R^n-\{x\in R^n|x_1 = x_2\cdots=x_n\}$, which is topicologically different from $R_+^n$ (the preimage). All we have to show is something like we have a circle around this line, that can't then be contracted.

A bit more formally: We define $\tilde{e}_i = [1,\dots,1,1/\epsilon,1,\dots,1]$, where $1/\epsilon$ is at the position $i$. With these points we have $f(\tilde{e}_i)\approx [A_1, A_2, \cdots, B_i,\dots,A_n]$, with $B_i$ being a huge number. We can then define path $p_{ij}:[0,1]\rightarrow R_+^n$, $p_{ij}(t) = t\cdot \tilde{e}_i +(1-t)\cdot \tilde{e}_i$. Then we connect all the path $f(p_{ij})$. These path will then from a closed path "around" the removed line $\{x\in R^n|x_1 = x_2\cdots=x_n\}$. This path could not be contracted if this line were not in the image of $f$. Therefore we have at least one such point.

Questions:

1. Does simply connected suffice also for higher dimensions or do I need algebraic topology?
2. Is there a way to proof that this point is unique?
3. Is there a more beautiful way to proof this? In its current version it's quite a mess.

Answer 1

Put $\mathbb{R}^n_0=\{x\in\mathbb{R}^n:\sum_ix_0=0\}$, and let $\pi\colon\mathbb{R}^n\to\mathbb{R}^n_0$ be the orthogonal projection. You have a map $f\colon(0,\infty)^n\to\mathbb{R}^n$, and we can define $g\colon\mathbb{R}^n_0\to\mathbb{R}^n_0$ by $$ g(x) = \pi(f(e^{x_1},\dotsc,e^{x_n})). $$ You want to show that $0$ is in the image of $g$, and the only way you can realistically hope to do that is by proving that $g$ is surjective.

If you want to use methods of algebraic topology, the main thing that you need to check is that $g$ is proper, or in other words that the preimage of any compact set is compact. Your differential conditions are probably designed to do that; I am not sure whether they succeed. Anyway, it is probably better to go back to whatever context you were looking at, and see whether you can prove properness directly.

If $g$ is proper, then it has a unique continuous extension $g_\infty\colon\mathbb{R}^n_0\cup\{\infty\}\to\mathbb{R}^n_0\cup\{\infty\}$, and $\mathbb{R}^n_0\cup\{\infty\}$ is homeomorphic to the sphere $S^{n-1}$, so the $(n-1)$'th homology group is $\mathbb{Z}$. This means that $g_\infty$ acts on that homology group as multiplication by some integer $d$, called the degree. If $d\neq 0$ then $g_\infty$ (and thus $g$) will be surjective.

If your differential conditions on $f$ ensure that the Jacobian of $g$ is nowhere zero (and you also have properness) then $g$ will be a covering map and you will have $d=\pm 1$.

Suppose instead that you can find a point $b$ such that

• $g_\infty^{-1}\{b\}$ is some finite set $\{a_1,\dotsc,a_r\}$
• The Jacobian of $g_\infty$ at $a_i$ is not zero for any $i$.

Let $d_+$ be the number of $a_i$ where the Jacobian is positive, and let $d_-$ be the number of $a_i$ where the Jacobian is negative. It can then be shown that $d=d_+-d_-$, so if $d_+\neq d_-$ then $g_\infty$ will be surjective.