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Let $X=\{x_i : i\in I\}\subseteq\mathbf{R}^n$, where $I=\{1,\ldots,m\}$. Then for some initialization $\mu^{(t)}$, and $\pi^{(t)}=\{x\in X : \|x-\mu^{(t)}\|\leq r\}$, $r>0$, we want to prove that a mapping $\mu^{(t+1)}=M(\mu^{(t)})$ defined as $$ \mu^{(t+1)}=\textrm{argmin}_{\mu}\sum_{x\in \pi^{(t)}}\|x-\mu\| $$has a fixed point, i.e. there exist $\mu^*$ such that $\mu^*=M(\mu^*)$. We observe Euclidean norm.

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  • $\begingroup$ Why is the minimiser unique? $\endgroup$ – copper.hat Mar 20 at 23:25
  • $\begingroup$ It is unique because we observe Euclidean norm, i.e. for other norms the minimiser can be not-unique (for example $\ell_1$ norm). $\endgroup$ – Vedran Novoselac Mar 20 at 23:46
  • $\begingroup$ If $\pi^{(t)}$ contains two distinct points then any $\mu$ in $[x_1,x_2]$ is a minimiser, regardless of the norm used. $\endgroup$ – copper.hat Mar 20 at 23:56

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