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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.
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  • $\begingroup$ Conditions 2 and 4 are incompatible; let $\{\mathbf e_k\}$ be the standard basis of $\mathbb R^n$. By 2 and 4, $$f_1(\mathbf e_1)=f_1(2^n\mathbf e_1)=\lim_nf_1(2^n\mathbf e_1)=\infty$$ $\endgroup$ – Meisam Soleimani Malekan Jun 1 '18 at 13:29
  • $\begingroup$ @MeisamSoleimaniMalekan The standard vectors are not allowed, as they contain non positive entries ($c\in\mathbb{R}^n_+$). Basically assumption 4 is why only positive vecotrs are allowed. $\endgroup$ – Jürg Merlin Spaak Jun 1 '18 at 13:34
  • $\begingroup$ 1. Why algebraic topology? Is your $f$ algebraic? 2. Do you have an example of such an $f$? 3. Quoting from en.wikipedia.org/wiki/Simply_connected_space : "The image of a simply connected set under a continuous [or even entire analytic -- I.P.] function need not be simply connected. Take for example the complex plane under the exponential map: the image is C - {0}, which is not simply connected". $\endgroup$ – Iosif Pinelis Jun 1 '18 at 14:11
  • $\begingroup$ @MeisamSoleimaniMalekan The following functions proof, that 2 and 4 are not incompatible: $f_i (c) = \frac{c_i^n}{\Pi_j c_j}$ $\endgroup$ – Jürg Merlin Spaak Jun 1 '18 at 18:13
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    $\begingroup$ @PietroMajer They are assumed to be fixed. I guess uniformity of these limits would be inconsystent with 4. ( just intuition, without any deep thoughts). $\endgroup$ – Jürg Merlin Spaak Jun 2 '18 at 18:34
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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.

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  • $\begingroup$ Thanks, seems promising. I have however some questions: 1. There's a typo in the definition of $\mathbb{R}_0^n$, the sum should go over $x_i$. 2. I assume the usual combinations of proper functions are still proper, is this correct? 3. I only have to show, that $g$ is proper and that it's Jacobian vanishes nowhere, then the theorem is proven, correct? 4. Do you have any recommendation where I could read this up and possibly cite? $\endgroup$ – Jürg Merlin Spaak Jun 1 '18 at 19:56

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