Consider the problem
\begin{align} \max_{x\in\mathbb{R}^n} f(x)\\ \text{subject to }\quad h(x) = 0\\ x\in X \end{align} where $X$ is a convex and compact subset of $\mathbb{R}^n$. I also know that the Jacobian $\nabla h(x)$ is full rank for all $x$, if that helps. Furthermore, we can assume $f:\mathbb{R}^n\to\mathbb{R}$ and $h:\mathbb{R}^n\to\mathbb{R}^m$ are smooth functions.
Form the partial Lagrangian of the above as \begin{align} L(x,\lambda) = f(x) - \lambda^\top h(x) \end{align}
Using the sign convention of the above Lagrangian, under what conditions on $f$ and $h$ ensure that the vector of Lagrange multipliers, $\lambda$, is positive (that is, $\lambda_i>0$ for all $i$)? How can I show this mathematically?
My attempt: My thought process so far is that the Lagrange multipliers are set as to exactly ''cancel out the forces'' -- that is $\lambda$ is such that $\nabla f - \lambda^\top\nabla h = 0$. So, if I can show $\nabla f$ and $\nabla h$ are both acting in the same direction, then is $\lambda$ positive? (Update: I don't think this is right, due to the existence of the abstract set constraints $x\in X$ the gradient of the Lagrangian may never be zero)
Update: It is known (if we are minimizing $f$) that a necessary condition for $x^*$ to be a local minimum of the Lagrangian $M(x,\lambda) = f(x) + \lambda^\top h(x)$ over $x\in X$ (when $X$ is a convex set) is \begin{align} (\nabla f(x^*) + \lambda^{*\top}\nabla h(x^*))^\top(x-x^*) \ge 0\quad \forall\, x\in X \end{align} This inequality can be thought of as the feasible variations at $x^*$ (the directions that point into the set $X$). I'm unsure how, or if, this is helpful.
Update #2: A useful direction may be the following. Write the constraints as $h(x) = a$. Let $x(a)$ be the minimizer for some $a$ and $\lambda(a)$ be the corresponding Lagrange multiplier (not sure how to show that this even exists, but that's besides the point for now). The objective value is $f(x(a))$. Now if I differentiate $f(x(a))$ with respect to $a_i$ I get the following \begin{align} \frac{\partial f(x(a))}{\partial a_i} &= \frac{\partial}{\partial a_i}L(x(a),\lambda(a))\\ &= (\nabla f(x(a)) - \lambda(a)^\top\nabla h(x(a))\frac{\partial x(a)}{\partial a_i} - \frac{\partial \lambda(a)}{\partial a_i}(h(x(a)) - a) + \lambda_i(a) \end{align} but $h(x(a))=a$. So \begin{align} \frac{\partial f(x(a))}{\partial a_i} = (\nabla f(x(a)) - \lambda(a)^\top\nabla h(x(a))\frac{\partial x(a)}{\partial a_i} + \lambda_i(a) \end{align} So, $\lambda_i(a) = \frac{\partial f(x(a))}{\partial a_i} - (\nabla f(x(a)) - \lambda(a)^\top\nabla h(x(a))\frac{\partial x(a)}{\partial a_i}$. So the Lagrange multipliers that we are after are evaluated at $a=0$. Where can I go from here? Have I messed up somewhere along the way -- it seems like this equation is not useful, as it depends on $\lambda(a)$ itself?
