Existence of a fixed point to a vector-valued mapping

Question

I have the following system: $$\begin{cases} F_1(x_1,x_2) \colon= (c_1 - a_{11}x_1 - a_{21}x_2)^{a_{11}}(c_2 - a_{12}x_1 - a_{22}x_2)^{a_{12}} = x_1 \\ F_2(x_1,x_2) \colon= (c_1 - a_{11}x_1 - a_{21}x_2)^{a_{21}}(c_2 - a_{12}x_1 - a_{22}x_2)^{a_{22}} = x_2 \end{cases}$$ where $c_i, a_{ij}\geq 0$ and $c_1 + c_2 = a_{11}+a_{12} = a_{21} + a_{22} = 1.$ I strongly suspect the above system has a solution $(x_1,x_2)$ with $x_i>0,$ but not quite sure how to prove it.

The Jacobian after taking $\ln$ from both sides is positive definite which can be seen easily. But I am not sure how that would help if we wanted to use Implicit Function Theorem, for instance. I thought about using Banach's fixed point theorem but the map $F = (F_1, F_2)$ does not seem to be a contraction since the derivative will be unbounded due to the exponents $a_{ij}\leq 1.$

I think a vector version of the Intermediate Value Theorem is probably most promising but all I can find is Poincare-Miranda Theorem, which I am not sure is immediately applicable. For what its worth we can assume the domain I am interested is: $$D = \{(x_1,x_2)\vert\, a_{1i}x_1 + a_{2i}x_2 < c_i \}\subseteq\mathbb{R}^2_{>0}$$

Finally, its one-dimensional version is an immediate consequence of the Intermediate Value Theorem

Answer 1

Yes, the function $F$ is not a contraction, so we can't use Banach fixed point theorem. There are of course many generalisations of it but I would rather go for topological fixed point theorems.

First observe that the domain of $F$ is equal to $$D_F = \{(x_1,x_2)\in \Bbb R^2:c_1 - a_{11}x_1 - a_{21}x_2\geq 0,\ c_2 - a_{12}x_1 - a_{22}x_2\geq 0\}.$$

Let $$M:=\{(x_1,x_2)\in\Bbb R^2:x_1,x_2\geq 0\}\text{ and }K:=M\cap D_F.$$ These are closed convex sets and $K$ is bounded (hence compact). Indeed, if $(x_1,x_2)\in K$ then $$ (a_{11}+a_{12})x_1 + (a_{21}+a_{22})x_2\leq c_1+c_2,\text{ so } x_1+x_2\leq 1. $$ This together with $x_1,x_2\geq 0$ gives boundedness of $K$.

Case $c_1\cdot c_2=0$.

Assume that $c_1=0$ and that there exists a positive fixed point $(x_1,x_2)\in D_F$ of $F$, i.e. $x_1,x_2>0$ and $F(x_1,x_2)=(x_1,x_2)$. If $a_{11}>0$ or $a_{21}>0$ then $$0\leq c_1 - a_{11}x_1 - a_{21}x_2< 0,$$ a contradiction. Therefore $a_{11}=0$ and $a_{21}=0$. Therefore $x_1=F_1(x_1,x_2)=0$, a contradiction. This shows that in this case there are no positive fixed points (the point $(0,0)$ is a nonnegative fixed point, though). Similarly we can treat the case $c_2=0$.

Case $c_1,c_2>0$.

Now assume $c_1,c_2>0$.

The idea of what will be done

Having this we can consider many fixed point theorems of functions $F\colon K\to \Bbb R^2$. Unfortunately $K$ isn't invariant, so we can't use a Brouwer fixed point theorem. However there are many theorems that deal with such cases but some additional assumptions are needed (which is obvious, since for example $f\colon [0,1]\to \Bbb R$, $f(x)=x+2$ doesn't possess fixed points). These additional assumptions are mainly about the behaviour of $F$ on the boundary of $K$, for example:

• compression / expansion
• tangency.

The idea is that we can consider the retraction $r\colon \Bbb R^2\to K$ (there are such retractions for convex closed sets, for example a metric projection) and the function $r\circ F\colon K\to K$, being $K$ invariant, has a fixed point (from Brouwer f.p.t.). Therefore $r(F(x_1,x_2))=(x_1,x_2)$ for some $(x_1,x_2)$. This additional assumption can give us that this can't happen on $\partial K$ and therefore $r(F(x_1,x_2))=F(x_1,x_2)$.

In our case we have a situation of tangency, that is $F(x,y)$ is tangent to $K$, that is $F(x,y)$ is always in the so called Bouligand/Clarke tangent cone of $K$. I'm going to avoid these notions and to solve this problem directly, without using the more sophisticated theorems than Brouwer's.

Back to proof

Of course

• $F(M)\subset M$.
• $F_1(x,y)=0\iff F_2(x,y)=0$

Let $r\colon M\to K$ be a metric projection, that is to any $(x,y)\in M$ it assigns the closest point in $K$. Therefore $r|_K = \mathrm{id}|_K$ and if $r(p)\neq p\iff p\notin K$ then $r(p)\in\partial_M K$ (boundary with respect to $M$).

Then $r\circ F\colon K\to K$ has a fixed point $(x_1,x_2)$, that is $$r(F(x_1,x_2))=(x_1,x_2).$$

• If $r(F(x_1,x_2))=F(x_1,x_2)$ then we get a fixed point of $F$ on $K$.
• If $r(F(x_1,x_2))\neq F(x_1,x_2)$ then $(x_1,x_2)=r(F(x_1,x_2))\in \partial_M K$. Therefore $c_1 - a_{11}x_1 - a_{21}x_2=0$ or $c_2 - a_{12}x_1 - a_{22}x_2=0$ and then $F(x_1,x_2)=(0,0)$. This is however impossible, since then $(x_1,x_2)=r(F(x_1,x_2))=r(0,0)=(0,0)$ and then $F(x_1,x_2)\neq (0,0)$.
• Now we know that $F$ has a fixed point $(x_1,x_2)$ in $K$. It sufficies to show that $x_1,x_2>0$. If for example $x_1=0$ then $F_1(x_1,x_2)=0$ then $F_2(x_1,x_2)=0$, then $x_2=0$ then $F_1(x_1,x_2)=F_1(0,0)\neq 0$, a contradiction.

Final remarks Here is the picture of the situation for $a_{11}=0.07$, $a_{22}=0.31$, $c_1=0.37$.

The green polgon is the set $K$. The curved shape is the set $F(K)$. We see that $F(K)\not\subset K$. The cross (red/black) is the value $F(p)$ of the dot $p$ (red/black). More in Desmos.

Answer 2

Just an idea! Why don't you use Brouwer's fixed point theorem?

The function $\tilde{F}=(F_{1},F_{2})$ seems perfectly continuous. So

using any compact convex set $K$, there is always a fixed point! $\tilde{F}(z)=z$