Recent content by lementin

Thank you very much, mfb. This really seems to be the case that with some choices of constants within the inner products the expression can be unbounded from above. I meant a partial order on the vector space, with not all vectors being comparable. x\ge y iff \forall i,\,\,\,x_i\ge y_i.

I actually found a walkaround and solved my problem by a completely different method. Concerning the original problem: I did a little research comparing the duals of the original and the derived problems, and they seem to be equivalent (in particular, minumums do coincide). Thus, the answer...

Hi, chiro. Thank you for your reply. The original domain is a simplex and the transformed problem is obtained by incorporating the linear condition and, thus reducing the dimensionality to n-1. That being said the topologies of two domains are different. For curvature and orientation, I'm not...

Hello all, Assume we have an optimization problem and that strong duality holds. Will it also hold for another optimization problem obtained from the initial by a bijective coordinates transform? Thank you in advance.

Thank you for your reply, mfb. In fact I don't compare vectors with real numbers. In the notation \lambda\ge 0,\,\,\xi\ge 0,\,\,\,\,\,0\,\, is actually the 0 vector (that is all its components are 0). I would also like to extend my question here. How to find the closed form solution of...

Thank you very much, mathman. So the level set corresponding to the maximum value is \xi=\lambda, \,\,\lambda\ge 0, \xi\ge 0

Hello, My question is as follows. Is it possible to obtain a closed form solution to \displaystyle \max_{\xi\ge 0, \lambda\ge 0}\,\, -\frac{1}{2}\||\xi\||^2 +(\xi,\,\lambda) -\frac{1}{2}\||\lambda\||^2 Here \xi and \lambda are vectors. Thank you.