Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a general convex program, a feasible point is an optimal solution if and only if it lies in a hyperplane whose a normal vector is the gradient to the objective function at this point. Please suggest what will be the form of this result in case of an invex function involving a non-linear function eta.

For an convex function $f$, $x$ is an optimal solution if and only if $\langle \nabla f(x) , y-x \rangle \geq 0$ which explains the the fact that gradient of the objective function is the normal to the hyperplane at x.

For an invex function $f$, $x$ is an optimal solution if and only if $\langle \nabla f(x) , n(y,x) \rangle \geq 0$, where $n(y,x)$ is a nonlinear function.

Looks like in case of an invex function, the gradient of the objective at an optimal point must make an acute angle with all the non linear curves $n$ ?

share|cite|improve this question
Isn't it the fact you're looking for? Sentence about Ben-Israel and Mond. – Ilya Sep 26 '11 at 16:12
@Gortaur: I am looking for a geometrical interpretation of optimality conditions for differential invex functions, just like the ones provided for differentiable convex functions above. – Tav Sep 26 '11 at 16:27
up vote 2 down vote accepted

The Wikipedia article says that if the objective and constraints are invex wrt the same $g(x,u)$, the Karush-Kuhn-Tucker conditions are sufficient for a global minimum. Geometrically the Karush-Kuhn-Tucker conditions say that the gradient of the objective is in the cone generated by the outward normals of the active constraints.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.