Consider a circle $C$ with radius of $r$, we place $m$ balls(treated as point) randomly on it, and each ball $i$ has the mass $m\_i$. We define a function $\varphi:C\rightarrow C$ which maps $x\in C$ to the centroid of those balls with distance less than $l<2\pi r$ to $x$(work on arcs).


My question is that, for any point $x\in C$ whether the sequence $\{\varphi(x),\varphi\circ\varphi(x),\cdots,\varphi^n(x)\}$ terminated at some point $x\_0$?(no cycles).


If the circle is not discrete but continuous with a density distribution function, what would be the answer?



Fixed point of a function on the circle
Call a point clockwise (resp. anticlockwise) if $\varphi(x)$ lies in clockwise (resp. anticlockwise) direction of $x$. Note that if $x$ is clockwise then all points on $[x,\varphi(x)]$ are clockwise as well. Now if there would be a cycle all points of the circle would have to be passed over and hence all points of the circle are clockwise (or all points anticlockwise). Consider now a probability density function $f$. Then $x$ clockwise is equivalent to e.g. $g(x)=\int\_{x-l}^{x+l} (y-x)f(y)dy>0$. Note that
$$0<\int\_C g(x)dx=\int\_C \int\_{x-l}^{x+l} (y-x)f(y)dydx=\int\_{C} f(y) \int\_{y-l}^{y+l} (y-x)dxdy=0$$
which is a contradiction. Hence there are no cycles. Note that this proof also works for the discrete case.


