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Towards a well-defined median. (English) Zbl 1140.41005
One considers the space $$\mathbb R^{n}$$ $$(n\geq2)$$ endoved with $$l_p$$-norm, $$p\in[1,\infty]$$. Then the diagonal $$\Delta_n= \{x= (x_1,x_2,\dots,x_n)\in\mathbb R^n: x_1=x_2=\dots= x_n\}$$ is a proximinal subset of $$\mathbb R^n,$$ and for $$p\in(1,+\infty]$$ is even Chebyshev with respect to the $$l_p$$-norm. This means that the function $$f_p:\mathbb R^n\rightarrow\mathbb R,$$ $$f_p(x)=\|(x,x,\dots,x)- (a_1,a_2,\dots, a_n)\|_p^s= \sum_{j=1}^n | x-a_j| ^{p}$$ attains its minimum at a unique point $$(\mu_p,\mu_p,\dots, \mu_p)\in \Delta_n$$ if $$p\in(1,\infty]$$, and the minimum point may be nonunique for $$f_1$$ (i.e., for $$p=1$$). In fact for $$a_1\leq a_2\leq\dots \leq a_n$$, $$f_1$$ attains its minimum at the point $$(a_{(n+1)/2},\dots, a_{(n+1)/2})\in \Delta_n$$ if $$n$$ is odd and at any point in the interval $$[a_{n/2},a_{1+n/2}]$$ if $$n$$ is even. The authors proves that $$\lim_{p\rightarrow q}\mu_p= \mu_q,$$ for all $$q\in(1,\infty].$$ If $$p\in(1,\infty]$$ and $$\mu_p= \mu_p(a),$$ $$a=(a_1,a_2,\dots,a_n)$$ is the minimum point for $$f_p,$$ then there exists $$\lim\mu_p$$ as $$p$$ decrease to 1, and this limit, denoted by $$\mu^*$$, is called by the authors the median of $$a=(a_1,\dots, a_n),$$ or of the data set $$\{a_1,a_2,\dots,a_n\}$$.

##### MSC:
 41A50 Best approximation, Chebyshev systems 26E60 Means
##### Keywords:
Best approximation; median
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