Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous and $x_1 \in \mathbb{R}^n$.

Consider the sequence $x_{k+1} := f(x_k)$, for $k \geq 1$, such that $\lim_{k \rightarrow \infty} x_k = \bar{x} = f(\bar{x})$.

Suppose that there exists a unique fixed point $\bar{x} = f(\bar{x})$.

Now consider the sequence $y_{k+1} := f\left( \frac{1}{k} \sum_{i=1}^{k} y_i \right)$, $k \geq 1$, with $y_1 = x_1$.

Say if $\lim_{k \rightarrow \infty} y_k = \bar{x}$.

Comment. This question is a variant of this one.

  • $\begingroup$ $f$ is continuous? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 28 '14 at 16:40
  • $\begingroup$ Yes, sorry I forgot. $\endgroup$ – user693 Jan 28 '14 at 16:43

For a counterexample, take $n=1$. Let $f(1)=f(-1)=-1$ and $f(0)=0$, put $x_1=1$. Then $x_k=-1$ for all $k\ge2$, $y_2=-1$, and $y_k=0$ for all $k\ge3$.

  • $\begingroup$ Thanks. What instead if there exists a unique fixed point? $\endgroup$ – user693 Jan 28 '14 at 16:51
  • 1
    $\begingroup$ Then it becomes a bit trickier. Of course, if $(y_k)$ converges, then the limit must be the fixed point, so one would have to look for examples in which $(y_k)$ diverges. I am pretty sure you need $n\ge2$ for a counterexample in that case. $\endgroup$ – Harald Hanche-Olsen Jan 28 '14 at 18:31

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