I was learning the KKT conditions theory and in the notes the author was trying to prove the necessity of KKT conditions with bare hands. But I found a phenomenon which is a bit weird and I seek for some suggestions here.
Suppose $\hat{x}$ solves $$\min_{x\in \mathbb R^m} F(x) \quad\text{subject to } \quad G(x)\le 0\,,$$ where $F,G$ are from $\mathbb R^m$ to $\mathbb R$, continuously differentiable.
Define $\mathcal{L}(x,\mu)=F(x)+\mu G(x)$, where $\mu\in \mathbb R$.
Consider the case that $G(\hat{x})<0$, i.e. $\hat{x} \in \{G(x)<0 \}$, which is open in $\mathbb R^m$.
Then $\nabla_x F(\hat{x})=0$. Recall that the stationarity condition in KKT is, there exists $\hat{\mu}$ such that $\nabla_x F(\hat{x})+\hat{\mu} \nabla_x G(\hat{x})=0$. Therefore we need to have that $\hat{\mu} \nabla_x G(\hat{x})=0$. If we choose $\hat{\mu}=0$, then we are done. But then $\mathcal{L}(x,\hat{\mu})$ reduces to $F(x)$.
It seems like introducing $\mathcal{L(x,\mathcal{\mu})}$ is somehow meaningless. What is the meaning of introducing KKT in this specific case? Thanks for posts and comments.
In general, you don't know before you find them, which, if any, of the minimizers $\ \hat{x}\ $ will satisfy the condition $\ G(\hat{x})<0\ $, and if $\ \hat{x}\ $ is a local minimizer for which $\ G(\hat{x})=0\ $, then it will not necessarily be true that $\ \nabla F(\hat{x})=0\ $. Thus, if you want to be sure of finding all the minimizers, you will normally need to find all solutions of the full KKT conditions, which are \begin{align} \nabla F(\hat{x})+\hat{\mu}\nabla G(\hat{x})&=0\\ \hat{\mu}G(\hat{x})&=0\\ \hat{\mu}&\ge0 \end{align} for this problem. Note that the last condition is also necessary. If $\ \nabla F(\hat{x})+\hat{\mu}\nabla G(\hat{x})=0\ $, $\ \nabla F(\hat{x})\ne0\ $ and $\ \hat{\mu}<0\ $, then $\ \hat{x}\ $ cannot be a minimizer.