Suppose that $f: \mathbb{D} \subseteq \mathbb{R}^n \to \mathbb{R}^n$, where $\mathbb{D}$ is a half-space or a convex cone. We wish to show the existence of a fixed point $x = f(x)$.

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Suppose we know that $f$ is continuous, single-valued and always maps into this half space (or convex cone), given any point or subset of it. Then the only problem is the lack of compactness for Brouwer's fixed point theorem.

Is there any way to deal with these cases? 

(Motivation: rather large classes of functions satisfy this property, such as $\exp(-x): [0, \infty) \to \mathbb{R}$).

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  • $\begingroup$ This is a really good question. This is in general not an easy thing to prove. Check out the book by Auslender et al. -- they have a great review of existence theorems, including a section on functions who do not map from a compact set to itself. Also, you might look up notions of "recession direction" of a function. You can usually show existence for functions who remain bounded in all recession directions. $\endgroup$ – Zim Jun 19 '20 at 18:56
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    $\begingroup$ If $\mathbb D$ is closed, and $f$ is a contraction, then you can use the Banach fixed point theorem. There might be other more sophisticated fixed point theorems you can appeal to, but I imagine you would still need additional hypotheses on $f$ or $\mathbb D$. $\endgroup$ – Theoretical Economist Jun 19 '20 at 18:56
  • $\begingroup$ Closed half-spaces, or allowing for open ones as well? $\endgroup$ – user762914 Jun 19 '20 at 18:57
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    $\begingroup$ en.wikipedia.org/wiki/Schauder_fixed-point_theorem $\endgroup$ – user251257 Jun 19 '20 at 18:58
  • $\begingroup$ On the other hand, if you impose a partial order on $\mathbb R^n$ where you say that $x \le y$ if and only if $x_i \le y_i$ for all $i=1,\ldots,n$, and you know that $f$ is order-preserving, then you can use Tarski's fixed point theorem. Though you will also need that $\mathbb D$ is closed. $\endgroup$ – Theoretical Economist Jun 19 '20 at 19:00

Suppose we're in $\mathbb R^2$ and $\mathbb D$ is the right half-plane. Then the map $(x,y)\to (x+1,y)$ takes $\mathbb D$ t0 $\mathbb D$ and has no fixed points.

In any convex cone in $\mathbb R^2$ with vertex at the origin, we can do the analogous thing: Define the map $z\to z+v,$ where $v$ points in same direction as the angle bisector of the two angles defining the cone.

Examples in higher dimensions are also not hard to come by.


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