most recent 30 from http://mathoverflow.net 2013-05-26T01:05:56Z http://mathoverflow.net/feeds/question/86457 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86457/non-linear-programming Gershon Wolansky 2012-01-23T16:11:26Z 2012-01-23T16:39:32Z

<p>Let $c_{i,j}:R \rightarrow R$ a family of functions, $i,j\in{1, \ldots n}$. Let $$\Lambda:=\left[ \lambda_{i,j}\geq 0, 1\leq i,j\leq n \ ; \ \sum_{i=1}^n \lambda_{i,j}=m_j \ ; \ \ \sum_{j=1}^n\lambda_{i,j}=m^*_i \right]$$ where $\vec{m}=(m_1\ldots m_n)\in R_+^n$, $\vec{m}^*= (m_1^*,\ldots m_n^*)\in R_+^n$ are given with $\sum_1^n m_i=\sum_1^n m^*_i.$</p> <p>$$\Gamma:= \left( \vec{x},\vec{y}\in R^n ; x_i-y_j\leq c_{i,j}(x_i) \ , 1\leq i,j\leq n\right)$$ My question is:</p> <p>Under which conditions on $c_{i,j}$ we get $$\max_{\vec{x},\vec{y}\in \Gamma; \lambda\in \Lambda} \sum_{i=1}^n\sum_{j=1}^n\left[ \lambda_{i,j}(x_i-y_j-c_{i,j}(x_i, x_j)\right] = 0$$ The case of $x$-independent $c_{i,j}$ follows from duality in linear programing. Is the sameresult holds in case $c_{i,j}$ are convex functions on $R$ for each $i,j$?</p>