# Fixed point of a function on the circle

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?

(I'm not a English native speaker, so sorry for my poor English)

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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.
Very nice answer! Thank you @user35593! I didn't notice the points $\in [x,\phi(x)]$ are clockwise also. –  Paul Apr 24 '14 at 6:17