I'm attempting to find a proof for a property from Floudas' Nonlinear and Mixed-Integer Optimization book. Consider a nonlinear optimization problem of the form \begin{align} \min_{{\bf x}}&\quad f({\bf x})\\ \nonumber \text{subject to } \quad&{\bf h}({\bf x}) = {\bf 0}\\ \nonumber \quad&{\bf g}({\bf x}) \le {\bf 0}\\ \nonumber \quad&{\bf x}\in {\bf X} \end{align} where ${\bf X}$ is a nonempty convex set. No assumptions on ${\bf h}$ or ${\bf g}$ are made. Form the partial Lagrangian of the above problem as \begin{align} L({\bf x},\lambda,\mu) = f({\bf x}) + \lambda^T{\bf h}({\bf x}) + \mu^T{\bf g}({\bf x}) \end{align} The dual function is formed as \begin{align} \phi(\lambda,\mu) = \inf_{{\bf x}\in{\bf X}}L({\bf x},\lambda,\mu) \end{align} Define the set ${\bf Y}(\lambda,\mu) = \{{\bf x}^*:{\bf x}^* \text{ minimizes }L({\bf x},\lambda,\mu)\text{ over }{\bf x}\in {\bf X} \}$.
Property (4.2.3 - Differentiability of dual function) Let $f({\bf x})$, ${\bf h}({\bf x})$, ${\bf g}({\bf x})$ be continuous functions, and ${\bf X}$ be a nonempty compact set. If the set ${\bf Y}(\bar\lambda,\bar\mu)$ reduces to a single element at the point $(\bar\lambda,\bar\mu)$, then the dual function $\phi(\lambda,\mu)$ is differentiable at $(\bar\lambda,\bar\mu)$ and its gradient is \begin{align} \nabla\phi(\bar\lambda,\bar\mu) = ({\bf h}({\bf x}^*),{\bf g}({\bf x}^*)) \end{align}
My attempt at a solution:
Let \begin{align} {\bf x}^*(\lambda,\mu) = \arg \min_{{\bf x}\in{\bf X}}L({\bf x},\lambda,\mu) \end{align} Then \begin{align} \phi(\lambda,\mu) = f({\bf x}^*(\lambda,\mu)) + \lambda^T{\bf h}({\bf x}^*(\lambda,\mu)) + \mu^T{\bf g}({\bf x}^*(\lambda,\mu)) \end{align} I'm having difficulty forming the gradient of the above. Am I going about this the right way?
In this lecture there is a proof of this proposition, using arguments from convex analysis. The "meat" of the proof starts in page 28, but you should read all the lecture to understand the notation, definitions and such.