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Consider a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with the following property. There exists $c \in (0,1)$ such that for all $x,y \in \mathbb{R}^n$ it holds that $\left\| f(x) - f(y) \right\| \leq c \left\| x-y \right\|$, where $\left\| \cdot \right\|$ is the Euclidean norm.

From Banach's fixed-point theorem, we know that, for any $x_0 \in \mathbb{R}^n$, the sequence $$ x_{ k+1 } := f \left( x_k \right) $$ converges to the unique fixed-point $\bar{x}$ of $f$, i.e. $\lim_{ k \rightarrow \infty } f( x_k ) = \bar{x} = f( \bar{x} )$.

Now consider $N$ sequences with $N$ initial conditions $y^1_0$, ..., $y^N_0$: $$ y_{k+1}^i = f\left( \frac{1}{ N } \sum_{j =1 }^{N} y_k^j \right) $$ for $i = 1, 2, ..., N$.

Question. Does $\lim_{k \rightarrow \infty} y_k^i \rightarrow \bar{x}$ for all $i$?

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