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Let $\mathcal{S}$ denote a compact, convex set, $\mathcal{B}$ be it's boundary and $\mathcal{D}(.,.)$ a convex symmetric distance defined on it. i.e.

$$\mathcal{D}(x,y)=\mathcal{D}(y,x)$$ and $$\mathcal{D}(x, \lambda y_1+(1-\lambda)y_2)\leq\lambda\mathcal{D}(x,y_1)+(1-\lambda)\mathcal{D}(x,y_2)$$

for all $\lambda\in[0,1]$ and $x,y,y_1,y_2\in\mathcal{S}$.

Is the following function concave? $$f(x):=\min_{b\in\mathcal{B}}\mathcal{D}(x,b)$$

i.e. can you prove $$f(\lambda x_1+(1-\lambda)x_2)\geq \lambda f(x_1)+(1-\lambda)f(x_2)$$ ?

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No, take $D(x,y) = |x-y|$ and $b=[0,1]$, then $f(x) = \min\{|x|, |x-1|\}$ is neither convex nor concave.

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  • $\begingroup$ Nope, then $f(x)=min\{x, 1-x\}$ which is concave. however if you take $D(x,y)=(x-y)^2$ on the same set, then $f(x)=min\{x^2, (1-x)^2\}$ which is neither convex nor concave. $\endgroup$
    – K. Sadri
    Nov 20, 2018 at 10:08
  • $\begingroup$ @K.Sadri Sorry, I used $b\in \mathcal{S}$ instead of $b\in \mathcal{B}$. Your solution ($f(x)= \min\{x,1-x\}$) allow for negative numbers. I fixed my initial example. $\endgroup$
    – LinAlg
    Nov 20, 2018 at 13:58
  • $\begingroup$ @K.Sadri consider accepting my answer to mark this question as answered $\endgroup$
    – LinAlg
    Dec 7, 2018 at 17:45

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