Exactly!
How do you define your \mu?
In the theorem I stated, \mu_i is the lagrange multiplier corresponding to the i^\mathrm{th} inequality constraint. You typically don’t deal with it explicitly; rather it is handled by the quadratic optimizer under the hood.
Here is a toy example to illustrate my point.
The goal is to minimize F(x) = x_1^2 + x_2^2 + x_1
subject to h(x) = 1 - x_1 = 0
and g(x) = x_1 - x_2 - \psi \le 0
In this example, \psi can be interpreted as a model parameter.
We can work this out analytically. From the equality constraint, x_1 = 1. We then minimize F(x) my making x_2 as close to 0 as possible subject to x_2 \ge 1 - \psi. There are two cases.
Case 1: \psi > 1, then x_2 = 0, and the inequality constraint does not matter.
Case 2: \psi \le 1, then x_2 = 1 - \psi.
Let’s now differentiate the solution, x, with respect to \psi. \mathrm d x_1 / \mathrm d \psi = 0. For x_2 we have
\begin{eqnarray*}
\mathrm d x_2 / \mathrm d \psi & = & 0, \ \ \psi > 1 \\
& = & -1, \ \ \psi < 1 \\
& = & \mathrm{does \ not \ exist \ at} \ \psi = 1
\end{eqnarray*}
You would get the derivative if as you approached the boundary \psi = 1, the derivative of x_2 went to 0. This is what the theorem tells you, but as far as I can tell, this is a difficult condition to enforce – and things get more intricate when you have multiple constraints.
Now here’s the solution using the Lagrangian multipliers. We start by introducing the Lagrangian
\mathcal L(x, \omega, \mu) = F(x) + \omega h(x) + \mu g(x)
and solve for stationarity
\nabla_{x, \omega, \mu} \mathcal L = 0
So instead of solving for x, you solve for the augmented variable \tilde u = (x, \omega, \mu). You have some additional constraints: (i) primal feasibility, g(\tilde u) \le 0 and complementary slackness \mu g(\tilde u) = 0. Plug in the analytical solutions, and you’ll find that they work.
Now the equation f := \nabla \mathcal L = 0 gives you an implicit function theorem for the derivatives, namely \mathrm d \tilde u / \mathrm d \psi = - [\partial f / \partial \tilde u]^{-1} \partial f / \partial \psi. This is the term you need to control at the boundary. I hope this sheds some light, even though I omitted several steps.