In standard convex programming, the objective function and each of the constraint inequalities are convex. in such case, if the KKT condition hold for a point, and Slater condition is also hold for the solution space, that point is global optimum. 
However what if one of the constraint isn't convex but the solution space is convex? 


for example suppose that $g\_1(\mathbf{x}),g\_2(\mathbf{x})$ be two nonconvex functions and the set $S=\{\mathbf{x} | g\_1(\mathbf{x})\le0 ,g\_2(\mathbf{x})\le0\}$ is convex. Is the same result from KKT condition held? or is it hold with a different constraint qualification? or is not generally applicable in these case?


Actually I need a proof which KKT condition (with/without new constraint qualification) is applicable otherwise a counterexample.



Generalization of standard convex problem
Let me sketch a proof that the KKT conditions imply global optimality in the case that the objective $f$ and $S$ is convex. No constraint qualification is needed.


Let us assume that the KKT conditions hold at $x$. For simplicity, let both inequality constraints be active at $x$, i.e., $g\_1(x) = g\_2(x) = 0$.


First, you always have
$$T\_S(x) \subset \{d : g\_1'(x) \, d \le 0, g\_2'(x) \, d \le 0\}.$$
Here, $T\_S(x)$ is the tangent cone of $S$ at $x$.


The KKT conditions imply (use Farkas' Lemma)
$$-\nabla f (x) \in \{d : g\_1'(x) \, d \le 0, g\_2'(x) \, d \le 0\}^\circ.$$
Here, $(\cdot)^\circ$ refers to the polar cone.


Hence,
$$-\nabla f(x) \in T\_S(x)^\circ.$$
In view of convexity of $f$ and $S$ this last condition implies (actually it is equivalent to) global optimality of $x$.


