# How to prove the existence of a fixed point of this mapping?

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.

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