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?
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.