# A natural concave function

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 $$f(\lambda x_1+(1-\lambda)x_2)\geq \lambda f(x_1)+(1-\lambda)f(x_2)$$ ?

No, take $$D(x,y) = |x-y|$$ and $$b=[0,1]$$, then $$f(x) = \min\{|x|, |x-1|\}$$ is neither convex nor concave.
• 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. Nov 20, 2018 at 10:08
• @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. Nov 20, 2018 at 13:58