How to handle the case when KKT condition is always false for this toy case?

Question

For the following simple problem: $$\begin{aligned} \min_x & (x-2)^2\\ s.t. \ \ & x^2=0\\ &x^2 \le 0 \end{aligned}$$

Since there is only one feasible point $x=0$, the answer is obvious. But when writting the KKT condition as: $$L=(x-2)^2+vx^2+wx^2$$ $\partial_x L=2(x-2)+2vx+2wx=0$ if $x=0$, we have $-4=0$, it obviously wrong. I think the solution should fulfill all KKT point. Is there anything wrong?

Answer 1

Recall the KKT conditions.

Suppose that $x^*$ is regular, that is, the gradients of the LHS functions of the constraints which are satisfied as an equality are linearly independent, and suppose that $x^*$ is a local optimum. Then, $x^*$ satisfies the KKT conditions.

In your case, $x^* = 0$ is irregular. Both constraints are satisfies as an equality, and for both the derivative at $x = 0$ is zero. The set $\{0, 0\}$ is clearly not linearly independent.

Both KKT points and irregular points may be optimal, and their union is your set of candidates.