Let
- $d,k,l\in\mathbb N$;
- $x^\ast\in\mathbb R^d$;
- $f:\mathbb R^d\to\mathbb R^k$ and $g:\mathbb R^d\to\mathbb R^l$ both be differentiable at $x^\ast$;
- $A:={\rm D}f(x^\ast)$;
- $C:={\rm D}g(x^\ast)$;
- $S:=\{g\le0\}$;
- $\mathcal J:=\left\{j\in\{1,\ldots,l\}:g_j(x^\ast)=0\right\}.$
Assume $x^\ast\in S$ and that $g$ obeys the Kuhn-Tucker constraint qualification at $x^\ast$$^1$. Say that $x^\ast$ is Pareto critical if$^1$ $$\not\exists x\in S:f(x)<f(x^\ast)\tag3.$$
We can show the following:
Lemma 1: If $x^\ast$ is Pareto critical, then $\not\exists v\in\mathbb R^d$ with \begin{align}\forall i\in\{1,\ldots,k\}&:\langle v,\nabla f_i(x^\ast)\rangle<0\\\forall j\in\mathcal J&:\left\langle v,\nabla g_j(x^\ast)\right\rangle\le 0.\tag4\end{align}
Note that $(4)$ is equivalent to \begin{align}Av&<0\\\pi_{\mathcal J}Cv&\le0\tag5,\end{align} where $\pi_{\mathcal J}$ is the canonical projection of $\mathbb R^l$ onto $\mathbb R^{\mathcal J}$.
Now I would like to show the following:
Theorem 2: $x^\ast$ is Pareto critical iff $\exists(\lambda,\mu)\in[0,\infty)^k\times[0,\infty)^l$ with \begin{align}\overbrace{\sum_{i=1}^k\lambda_i\nabla f_i(x^\ast)+\sum_{j=1}^l\mu_j\nabla g_j(x^\ast)}^{=\:A^\ast\lambda\:+\:C^\ast\mu}&=0\\\sum_{i=1}^k\lambda_i&=1\\\forall j\in\{1,\ldots,l\}:\mu_jg_j(x^\ast)&=0\tag6\end{align}
Using Lemma 1, it is straight forward to show "$\Rightarrow$", but I really struggle to show the other direction.
Maybe (but I'm not sure and we would need to prove this), we can show that necessary condition in Lemma 1 is sufficient as well and hence we could assume that $x^\ast$ is not Pareto critical and infer that there is a $v\in\mathbb R^d$ with $(5)$.
However, even then I'm not able to conclude: By $(6)$ there is a $\lambda_\ast$ with $\lambda_{i_\ast}>0$. By $(5)$ and $(6)$, $$\langle\underbrace{Av}_{=:\:a},\lambda\rangle\le a_{i_\ast}\lambda_{i_\ast}<0\tag7.$$ Now let $c:=Cv$. If it would hold that $$\langle c,\mu\rangle\le0\tag8,$$ we could conclude, since we would obtain by $(6)$ that $$0=\langle v,\underbrace{A^\ast\lambda+C^\ast\mu}_{=\:0}\rangle=\underbrace{\langle a,v\rangle}_{<\:0}+\underbrace{\langle c,\mu\rangle}_{\le\:0}<0\tag9;$$ which is clearly impossible.
The problem is that we only know that $$\left\langle\pi_{\mathcal J}c,\mu\right\rangle\le0\tag{10}$$. And, again, we would still need to show that $v$ as claimed exists at all. But if it does, maybe we can even show that it does not hold $(10)$, but even $(8)$.
$^1$ , i.e. if $v\in\mathbb R^d$ with $$\forall j\in\mathcal J:\left\langle v,\nabla g_j(x^\ast)\right\rangle\le0\tag1,$$ then \begin{align}\gamma(0)&=x^\ast;\\\gamma'(0)&=\alpha v;\\g\circ\gamma&\le0\tag2\end{align} for some $\gamma:[0,1]\to\mathbb R^d$ differentiable at $0$ and some $\alpha\ge0$.
$^1$ componently-wisely strictly smaller.