Let $K$ be a non-empty compact convex subset of a Banach space $E$, and let $f : K \longmapsto K$ be a continuous function. Fix $u\_0 \in K$, and define by recurrence $u\_{n+1} = \frac{1}{n+1} \sum\_{j=0}^n f(u\_j)$ ; or equivalently $u\_{n+1} - u\_n = \frac{1}{n+1} (f(u\_n) -u\_n)$.



> 
> Is it always true that $(u\_n)$ converges to some fixed point of $f$ ?
> 
> 
> 


* When $E$ is one-dimensional, this is an easy exercise, but all proofs I know use in a crucial way the ordering of the real numbers.
* I don't expect the two-dimensional case to be easier than the general case.



A recurrent sequence related to the Brouwer fixed-point theorem
A counterexample in dimension $2$: take $K$ the unit closed disk of $\mathbb{C}$, and $f:K\to K$ the map
$$f(z)=e^{i\pi/4}\frac{z}{|z|}\min(2|z|,1)$$ 
for $z\neq0$, and $f(0)=0$, which is the only fixed point of $f$. Then, starting by $u\_0\in K$ with $|u \_ 0|\ge1/2$ produces a sequence with $|u\_n|\ge 1/2$. (Reason: if $|u\_n|\ge1/2$ then $u\_ {n+1}$ lies in the segment of endpoints $u \_ n$ and $f(u\_n)=e^{i\pi/4}\frac{u \_ n}{|u \_ n|}$, a segment that is disjoint from the disk $B(0,1/2)$, by elementary geometry considerations).


