MFCQ gives KKT multipliers at a smooth local minimizer

Every $d\in T_C(x_{\ast })$ satisfies $\mathrm{\nabla }f(x_{\ast }{)}^{T}d\ge 0$. For $d=0$ this is immediate. Otherwise take a defining sequence $x_{\ast }+t_k{d}_k\in C$, with $t_k\downarrow 0$ and $d_k\to d$. For large $k$, local minimality gives $f(x_{\ast }+t_k{d}_k)-f(x_{\ast })\ge 0$. Differentiability gives $f(x_{\ast }+t_k{d}_k)-f(x_{\ast })=t_k\mathrm{\nabla }f(x_{\ast }{)}^T{d}_k+o(t_k)$; divide by $t_k$ and pass to the limit.

By [F1] and step 1.1, $\mathrm{\nabla }f(x_{\ast }{)}^{T}d\ge 0$ for every $d\in {\mathcal{L}}_C(x_{\ast })$. Form a matrix $B$ with columns $\mathrm{\nabla }{h}_j(x_{\ast }),-\mathrm{\nabla }{h}_j(x_{\ast })$ for each equality and $-\mathrm{\nabla }{g}_i(x_{\ast })$ for each active inequality, and set $c=\mathrm{\nabla }f(x_{\ast })$. If alternative 1 of [F2] failed, alternative 2 would give $B^{T}d\ge 0$ and $c^{T}d<0$. The paired equality columns force $Dh(x_{\ast })d=0$, while the inequality columns force $D{g}_i(x_{\ast })d\le 0$ for every active $i$, so $d\in {\mathcal{L}}_C(x_{\ast })$, contradicting step 1.1. Thus there are nonnegative coefficients representing $c$ as a combination of the columns of $B$.

Write the equality-column coefficients as ${\alpha }_j,{\beta }_j\ge 0$ and the active inequality coefficients as ${\lambda }_i\ge 0$. The representation from step 2.1 is $\mathrm{\nabla }f(x_{\ast })=Dh(x_{\ast }{)}^T(\alpha -\beta )-D{g}_{I(x_{\ast })}(x_{\ast }{)}^T{\lambda }_I$. Set $\nu =-(\alpha -\beta )$ and set ${\lambda }_i=0$ for inactive $i$. Then stationarity holds. Feasibility holds because $x_{\ast }\in C$; dual feasibility holds by construction; and complementarity holds because active constraints vanish while inactive multipliers are zero. These are exactly the KKT conditions.