A question about the lagrangian $L(x,\lambda, \nu)$ in the dual function in Convex Optimization - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:36:49Z http://mathoverflow.net/feeds/question/20375 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20375/a-question-about-the-lagrangian-lx-lambda-nu-in-the-dual-function-in-conve A question about the lagrangian $L(x,\lambda, \nu)$ in the dual function in Convex Optimization ofer 2010-04-05T09:27:45Z 2010-04-05T10:10:26Z <p>Hi. My question is probably very simple to some of you that have experience in Convex Optimization. The dual function is defined as the infimum of the lagrangian $L(x,\lambda, \nu)$ over all $x\$ in the domain. The lagrangian is: $f_0(x)+\sum \lambda_i f_i(x)+\sum \nu_i h_i(x)$</p> <p>My question is, if $x\$ is in the domain, it satisfies the equality constraints $h_i(x)$ and in that case, $h_i(x)=0$. So why do we even have to mention the equality constraints if they zero-out anyway?</p> <p>Thanks a lot, I hope I wrote my question clearly.</p> http://mathoverflow.net/questions/20375/a-question-about-the-lagrangian-lx-lambda-nu-in-the-dual-function-in-conve/20377#20377 Answer by Gjergji Zaimi for A question about the lagrangian $L(x,\lambda, \nu)$ in the dual function in Convex Optimization Gjergji Zaimi 2010-04-05T10:10:26Z 2010-04-05T10:10:26Z <p>The domain in question is the intersection of all the domains of the functions $f_i,h_i$. Not all the points in the domain satisfy the conditions (such points constitute what's called the feasible set). Also keep in mind that the Lagrangian dual is often a relaxation of the original convex optimization and only gives you a lower bound, unless you have strong duality.</p>