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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.
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up vote 4 down vote accepted

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).

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If we start with $u_0 = 1$ then $u_1 = i$ and $u_2 = (i-1)/2$ so that $u_2$ is in the interior of $K$. –  js21 Jun 26 '13 at 19:01
    
More precisely, $|u_n|$ is $\sim \frac{1}{\sqrt{n}}$ in your example. –  js21 Jun 26 '13 at 19:11
    
Sorry. I think it is fixed now. –  Pietro Majer Jun 26 '13 at 20:40
    
Thanks ! This also yield counterexamples in any dimension. –  js21 Jun 27 '13 at 8:14

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