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Consider the convex set $$C=\left\{ \mathbf{x}\in \mathbb{R}^N :0\le x_1\le x_2\le\dots\le x_i\le x_{i+1}\le \ldots\le x_N\le \frac 1{N-1}\text{ and } \sum_{k=1}^{N}x_k=1\right\}$$ I need to minimize a strictly concave function $\boldsymbol{\varphi}$ on this set, we know the function will attain its minimum at some extreme point of the set. By an extreme point, I mean a point such that at least one of the above inequalities hold as equality (please correct me if this definition is wrong).

For $N=2$ the extreme points of $C$ are $(0,1)$ and $(1/2,1/2)$. For $N=3$, they are of the form $(x,x,1-2x)$ where $\frac{1}{4}\le x\le \frac{1}{3}$. For $N\ge 4$, the set of extreme points of $C$ seems to become intractable. But $\boldsymbol{\varphi}$ is a symmetric function. Its maximum on $C$ is at $(\frac 1 N, \ldots, \frac 1 N)$.

My guess is the minimum of $\boldsymbol{\varphi}$ on $C$ occurs at $\left(\frac {N-2}{(N-1)^2}, \frac {N-2}{(N-1)^2},\ldots,\frac {N-2}{(N-1)^2}, \frac{1}{N-1}\right). $ Is this correct? If yes, any suggestions, tips, hints on how to prove it?

Sergio Parreiras
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    You are basically right about the minimum being attained at an extreme point. The minimizer will be a point on the boundary of $C$ that cannot be written as a convex combination of the other points in $C$. I don't think without knowing $\varphi$ you can make any better statement. Actually, I think you can cook up different $\varphi$s that have minimum at different extreme points. – S.B. Apr 10 '14 at 20:39
  • @S.B. So my definition of extreme points for $N=3$ is wrong right? It should be only $(\frac{1}{4},\frac{1}{4},\frac{1}{2})$ and $(\frac 1 3,\frac 1 3 ,\frac 1 3)$ instead no? – Sergio Parreiras Apr 10 '14 at 20:45
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    For $N=3$ you draw the set in 3D and vertices of the set you get would be the extreme points. I think (0,0.5,0.5) is one of them, but (1/3,1/3,1/3) is not. – S.B. Apr 10 '14 at 20:58
  • @S.B. Thanks again but $C$ is such that $0\le x_1\le x_2\le x_3 \le \frac{1}{2}$ and $x_1+x_2+x_3=1$ so if we minimize (on $C$) the projection on the last coordinate $\pi_3(\mathbf{x})=x_3$, which is linear, on $C$ we get $(1/3,1/3,1/3)$ as the solution so it is also an extreme point. – Sergio Parreiras Apr 11 '14 at 13:55
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    I was just guessing about $(1/3,1/3,1/3)$, but after doing the calculations I see that it's an extreme point. In addition to the points you mentioned $(0,1/2,1/2)$ is also one of the extreme points. Since the maximum occurs at $(1/3,1/3,1/3)$, your minimum point is either $(0,1/2,1/2)$ or $(1/4,1/4,1/2)$. – S.B. Apr 11 '14 at 14:16

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I think I got the answer (thanks S.B. for the comments) but not sure so I'm placing it as a community wiki feel free to edit:

The minimum occurs at $\left(\underbrace{\frac{1-\frac 1k}{N-1},\ldots,\frac{1-\frac 1k}{N-1}}_{k\text{ times}},\underbrace{\frac 1{N-1},\ldots,\frac 1{N-1}}_{N-k\text{ times}}\right)$ for some $k$.

First show the minimum must have either $x_1=0$ or $x_N=\frac 1{N-1}$. Assume $z\in C$ such that $0<z_1\le\ldots\le z_N<\frac 1{N-1}$. Pick $\varepsilon>0$ and let $\hat z$,$\tilde z$ coincide with $z$ for all entries but the first and last, $\hat z_1=z_1-\varepsilon$ and $\hat z_N=z_N+\varepsilon$ and $\tilde z_1=z_1+\varepsilon$ and $\hat z_N=z_N-\varepsilon$. Notice that we can choose $\varepsilon >0$ such that $\hat z\in C$ but we cannot guarantee that $\tilde z$ is in $C$. Let's assume $\boldsymbol\phi$ is strictly concave. We have $\phi(z)>\frac12\phi(\hat z)+\frac 12\phi(\tilde z)$ since $z=\frac12\hat z+\frac12 \tilde z$. Even when $\tilde z\not\in C$ we can find a permutation of its entries (call it $\tilde z^\pi$) that is in $C$. Now since $\boldsymbol \phi$ is symmetric $\phi(\tilde z)=\phi(\tilde z^\pi)$. So clearly $z$ can not minimize $\phi$ in $C$.

If $x_1=0$ we must have $x_k=\frac{1}{N-1}$ for all $k\ge2$.

If $x_1>0$ and $x_N=\frac 1{N-1}$ we claim either $x_2=x_1$ or $x_{N-1}=\frac{1}{N-1}$ and again the proof of the claim is as above.

Applying the same reasoning we tackle all the intermediate entries.

Sergio Parreiras
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