# Guesses on fixed point existence

Let $$\mathcal{X} \subset \mathbb{R}^n$$ be a finite set and the mapping $$\Phi : \mathcal{X} \rightarrow \mathcal{X}$$ be defined as follows:

$$\Phi(x) := \{y \in \mathcal{X} \mid J(y,x) \leq J(z,x), \, \forall z \in \mathcal{X} \}$$

with $$J : \mathcal{X} \times \mathcal{X} \rightarrow \mathbb{R}$$ two-argument function. I'm not sure about the existence of a fixed point for $$\Phi$$, but I do not even have a suitable counterexample. Any suggestions/intuitions? Thank you in advance.

• From the definition, it appears that $\Phi$ maps $X$ to the power set of $X$ so I don't see how it could have a fixed point. Jan 16, 2019 at 15:22
• @saulspatz to be precise, $\Phi$ maps each point $x \in \mathcal{X}$ into some $\mathcal{Y} \subseteq \mathcal{X}$ Jan 16, 2019 at 15:31
• That's exactly what I said. So how could $\Phi$ have a fixed point? Jan 16, 2019 at 15:34
• @saulspatz (Just for information) In this context, $x$ is a fixed point if $x\in \Phi(x)$. Jan 16, 2019 at 15:35
• @Song Thanks. I wasn't aware of that usage. Jan 16, 2019 at 15:36

Without further assumptions I think you cannot guarantee the existence of a fixed point. Consider, for example, $$X= \{1, \ldots, m\}$$ for some $$m>1$$, and let $$J(i, j) := |i - (j+1)|$$, if $$j < m$$, $$J(i, m) := |i - 1|$$. Then $$\Phi(j) = \{j+1\}$$, if $$j < m$$, and $$\Phi(m) = \{1\}$$, and there is no $$j \in \{1, \ldots, m\}$$ such that $$j\in \Phi(j)$$.
If you prefer to use a finite subset of $$\mathbb{R}^n$$ with $$n>1$$, let $$X := \{x_1, \ldots, x_m\} \subset\mathbb{R}^n$$ and define the function $$J$$ as above (with obvious modifications, i.e. $$J(x_i, x_j) := |i - (j+1)|$$ if $$j, etc.).
• Thanks for the scalar example. I was wondering what "weak" regularity properties on $J$ may help to ensure the existence of a fixed point. Jan 16, 2019 at 16:39
• If $X$ is a finite set, you can give a look to this paper: link.springer.com/article/10.1134%2FS000143460707022X Jan 16, 2019 at 16:51
• I've already had a look to that paper, thank you in any case! To be honest, the scalar example does not make sense in my context, since I'm dealing with game theory. Since $J$ is a two argument function in the strategy space, the scalar case means that basically there is only one player playing the game. Hence I should assume $n > 1$, but I guess that if it does not work for $n = 1$, there is no hope for $n > 1$ :-( Jan 16, 2019 at 16:59