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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.

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  • $\begingroup$ 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. $\endgroup$ – saulspatz Jan 16 at 15:22
  • $\begingroup$ @saulspatz to be precise, $\Phi$ maps each point $x \in \mathcal{X}$ into some $\mathcal{Y} \subseteq \mathcal{X}$ $\endgroup$ – GuidoLaremi Jan 16 at 15:31
  • $\begingroup$ That's exactly what I said. So how could $\Phi$ have a fixed point? $\endgroup$ – saulspatz Jan 16 at 15:34
  • $\begingroup$ @saulspatz (Just for information) In this context, $x$ is a fixed point if $x\in \Phi(x)$. $\endgroup$ – Song Jan 16 at 15:35
  • $\begingroup$ @Song Thanks. I wasn't aware of that usage. $\endgroup$ – saulspatz Jan 16 at 15:36
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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 $j<m$, etc.).

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  • $\begingroup$ Thanks for the scalar example. I was wondering what "weak" regularity properties on $J$ may help to ensure the existence of a fixed point. $\endgroup$ – GuidoLaremi Jan 16 at 16:39
  • $\begingroup$ If $X$ is a finite set, you can give a look to this paper: link.springer.com/article/10.1134%2FS000143460707022X $\endgroup$ – Rigel Jan 16 at 16:51
  • $\begingroup$ 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 $n > 1$ :-( $\endgroup$ – GuidoLaremi Jan 16 at 16:59
  • $\begingroup$ I think you can modify the example in any dimension, see the edited answer. $\endgroup$ – Rigel Jan 16 at 17:14

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