KKT conditions implication for active constraints

Question

I understand if an inequality constraint within a constrained optimization is not active, i.e. $c_i(x) > 0$, then that means the corresponding lagrange multiplier $\lambda_i$ must be zero. (Likewise if $\lambda_i \neq 0$ then the constraint must be active). However is the converse also true? That is, if the lagrange multiplier of the inequality constraint is zero, does this have to mean constraint is inactive? Could an active constraint have zero lagrange multiplier?

Answer 1

Let us consider the lagrangian associated to the problem

$$ \max_x f(x)\ \ \text{s. t.}\ \ \ g(x) \le 0 $$

as

$$ L(x,\lambda,s) = f(x)+\lambda(g(x)+s^2) $$

where $s$ is a convenient slack variable, converting the inequality into an equivalent equation. The stationary conditions are

$$ \nabla L = \cases{\nabla f(x) +\lambda\nabla g(x) = 0\\ g(x)+s^2=0\\ s\lambda = 0} $$

now if $\lambda = s = 0$ (here $s=0$ indicates thar the restriction is active) we have as a consequence $\nabla f(x) = 0$ indicating that $f(x)$ should be constant, or there exists $x^*$ such that $\nabla f(x^*) = 0\cap g(x^*) = 0$ meaning that if $f(x)$ is not a constant, then the stationary point should be at the restriction boundary.