I have a question about the constraint qualification for KKT. As I've seen the theorem stated if $G(x^*)=(g_1(x^*),\dots,g_n(x^*))$ are the binding constraints at a local max $x^*$ then the jacobian $DG(x^*)$ must have rank $n$. Obviously this is impossible if $n>\text{length}(x)$. In the examples I can think of for this case, this simply means that we cannot find a unique multiplier $\lambda$.
E.g. $\max x_2$ subject to $x_1^2 + x_2^2\leq 1$, $x_2\leq x_1 + 1$, and $x_2\leq 1-x_1$ with $x^*=(0,1)$.
So my question is, if per the theorem, KKT requires rank $n$ so that these overdetermined cases fail the constraint qualification, what might be going wrong? The theorem does not require a unique multiplier. Could we relax the constraint qualification to $DG(x^*)$ has full rank?
Regarding the last question. Here is my proof sketch for why I think this is possible.
Suppose the original problem is to maximize $f$ on the constraint set $\mathcal{D}=U\cap \{x\in\mathbb{R}^n:G(x)\geq 0\}$ where $U$ is an open set. Assume all continuously differentiable.
Let $E=\{i:G^i(x^*)=0\}$ and suppose $|E|>n$ and rank($DG_E(x^*)$)$=n$. Then we can find $F\subsetneq E$ where rank($DG_F(x^*)$)$=n$.
Suppose $x^*$ is a local max of $f$ on the constraint set $\mathcal{D}'=U\cap \{x\in\mathbb{R}^n:G_F(x)\geq 0\}$ where $U$ is an open set. Then $x^*$ is also a local max on the constraint sets defined by $G_E$ and $G$ where the former includes superfluous effective constraints and the latter includes ineffective constraints so that the constraint set only gets smaller while still allowing $x^*$.
If we can find a vector $(x^*, \mu^*)$ that solves the KKT conditions from the problem max $f$ on $\mathcal{D}'$, then $(x^*, \lambda^*)$ solves the KKT conditions for the original problem if $\lambda_i=\mu_i$ for $i\in F$ and $\lambda_i = 0$ otherwise. This proves the claim because for $E$ where $|E|>n$ then we must be able to find an $F$ and $(x^*, \mu^*)$ from which we construct the solution to the original problem.
I think I'm just missing optima on $\mathcal{D}$ are also optima on $\mathcal{D}'$.
Edit: Thought of a counterexample--so long as I am not abusing any definitions or the statement of KKT.
$\max x_1$ subject to $g_1(x)=x_1\geq0$, $x_2\geq0$ and $x_2\leq x_1^2$.
The solution is the only point in the feasible region, $(0,0)$, which is therefore trivially a local max. All constraints bind. The Jacobian has rank 2, but the FOCs would require the multiplier on $g_1$ to be negative.
