In the linear case (see here) and under some conditions I've been able to establish that all constraints (except the non negativity constraints) will be binding. I'm seeking to extend this result to non linear constraints.

Say $\mathbf{X}$ is a $n\times 2$ matrix with no negative entries and such that every row and every column have at least a non zero element.

Let $\mathbf{x}^i$ be the $i$-th row of $\mathbf{X}$ and assume further that $\mathbf{x}^i\not\leq\mathbf{x}^j$ and $\mathbf{x}^j\not\leq\mathbf{x}^i\;\forall i,j$. Also, let $\beta>1$.

For $j={1,2}$, let $\mathbf{q}^j$ be the solution to the problem

$$ \begin{array}{ll} \text{minimize} &\quad \sum_{i=1}^nx_{ij}q_i^\beta\\ \text{subject to} & \quad \sum_{i=1}^nq_i=1. \end{array}$$

Let $\mathbf{z}^j=\mathbf{X}^\top\mathbf{q}^j$, and $\langle\mathbf{Z}\rangle$ the cone spanned by $\mathbf{z}^j, j={1,2}$. I believe that if $\mathbf{k}\in\langle\mathbf{Z}\rangle$, the solution to

$$ \begin{array}{ll} \text{maximize} &\quad \mathbf{q}^\top\cdot \mathbf{1}\\ \text{subject to} & \quad \mathbf{X}^{\top}\mathbf{q}^{\beta}\leq\mathbf{k}\\ &\quad\mathbf{q}\geq \mathbf{0}, \end{array}$$

denoted $\mathbf{\bar q}$, is such that $\mathbf{X}^{\top}\mathbf{\bar q}^{\beta}=\mathbf{k}.$



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