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Is there an example of a discontinuous function, $F$, defined on some complete subset $X\subset R^n$ such that under some metric $d$, $\sum\limits_{n=1}^\infty \|F^n(x)-F^n(y)\|<\infty$ and 

Either for multiple $x_i\in X$ where $i\in I$ and $I$ is any arbitrary indexing set, we have $F(x_i)=x_i$.

Or For all $x\in X, F(x)\neq x$?

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usually the notation $||.||$ is used for a norm, do you mean norm or metric? What do you assume about $x$ and $y$ in the inequality you stated? Is this supposed to hold for any $x, y$? – user20266 Dec 11 '11 at 17:10
The parts "multiple $x_i$" and "any arbitrary indexing set" seem to contradict each other -- is $I$ really completely arbitrary, or does "multiple $x_i$" imply that $I$ has at least two elements? – joriki Dec 11 '11 at 17:40
Also the sentence seems to be missing a verb -- do you mean "Is there an example of a discontinuous function ..."? – joriki Dec 11 '11 at 17:58
@Thomas: I meant a metric. Yes, for any x,y. Thanks. – Will Dec 11 '11 at 19:18
@joriki: Sorry for the ambiguities. I has to bhave at least 2 elements for the "either" option. Yes, I should probably use of instead. Edited. :) – Will Dec 11 '11 at 19:20
up vote 0 down vote accepted

Yes. Take $X=[0,1]$ and

$$F(x)=\begin{cases}\frac12&x=0\;,\\\frac x2&x\ne0\;.\end{cases}$$

$F$ is discontinuous at $0$. The sum is a geometric series and thus convergent. And $\forall_{ x\in X}F(x)\ne x$.

Note that the discontinuity at $0$ isn't required to make this work; we could introduce arbitrary discontinuities within the interval, as long as the iteration eventually moves beyond them towards $0$.

Clearly we can't have the other case, $F(x_i)=x_i$ for multiple $x_i\in X$, since in that case the sum would diverge for $x=x_1$, $y=x_2$, being the sum over a non-zero constant.

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