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We have $E\subseteq \mathbb{R}^k$ convex Suppose for some $x\in\mathbb{R}^k$, there are two distinct $p_1$ and $p_2$ s.t. $|x-p_1|=|x-p_2|=\min_{y\in E}|x-y|$.

How do i show that the function $f(t)=|x-(tp_1+(1-t)p_2)|$ where $t\in [0,1]$ is a convex ?

Jean Marie
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limitless
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  • I don't understand: here is a unique point that minimizes the distance to a convex set (https://math.stackexchange.com/q/1065880). Moreover, the last sentence could be asked independently of the first one. – Jean Marie May 15 '17 at 20:57
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    In some of the answer to this post they used this function and said it is convex,but i cant understand why. – limitless May 15 '17 at 21:00

1 Answers1

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$f(\alpha t_1 +(1-\alpha) t_2)$

$=|x-[\{\alpha t_1 +(1-\alpha) t_2\}p_1 + \{1-\alpha t_1 -(1-\alpha) t_2\}p_2]|$

$=|(\alpha +(1-\alpha ))x-[\{\alpha t_1 +(1-\alpha) t_2\}p_1 + \{1-\alpha t_1 -(1-\alpha) t_2\}p_2|]$

$=|\{\alpha (x- t_1 p_1 - (1- t_1)p_2\} +\{(1-\alpha) \{ (x- t_2 p_1 - (1- t_2)p_2\}|$

$\geq \alpha|\{ (x- t_1 p_1 - (1- t_1)p_2\}| +(1-\alpha) |\{ (x- t_2 p_1 - (1- t_2)p_2\}|$

$=\alpha f(t_1)+(1-\alpha) f(t_2)$

The second last step uses the triangle inequality.

Juanito
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