Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If there are two sets of linear constraints in different variables, Ax <= b with x_l <= x <= x_u and Cy <= d with y_l <= y <= y_u, and a set of equality constraints of a specific non-linear but convex form relating the two sets of variables e^y - x = 0, is the feasible space convex, and what is the proof behind the answer? The difficulty seems to me that the equality constraint would be equivalently formulated as e^y - x <= 0 and e^y - x >= 0, which means that one of these inequalities would be convex and the other would be concave, which may somehow create concavities in the feasible space.

share|improve this question
    
Of course it's not convex. Since if $e^y=x$ and $e^{y'}=x'$ then in general $e^{(y+y')/2}\neq (x+x')/2$. A more reasonable question would be if the its projection on $x$ or on $y$ is convex. –  Yoav Kallus Jun 6 '13 at 22:39
    
Thanks for the reply. I meant to indicate that the epigraph of the function e^y - x is a convex set, which is indicated by the Hessian of the function being positive semi-definite, which it is. I believe my question would have been better stated as you phrased it, regarding the convexity of its projection in either space. Any additional comments would be appreciated. –  DanZ Jun 7 '13 at 6:41
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.