Is the global minimum of a nonconvex function over a constrained set the maxmin of the corresponding Lagrangian?

Question

Let a nonconvex differentiable function $f:X\to\mathbb{R}$ and differentiable constraint $g(x)\leq 0$, with $X$ convex.

Does the global minimum of $f$, $f(x^\star)$, with $g(x^\star)\leq 0$ coincide with the maxmin value of the corresponding Lagrangian function?

I.e., does the following hold true: \begin{equation} f(x^\star) \stackrel{?}{=} \max_{\lambda\geq 0} \min_{x\in X} \{ f(x)+ \lambda g(x) \}. \end{equation}

Answer 1

This is sometimes true, but not true in general. The difference $$f(x^*) - \max_{\lambda \ge 0} \min_{x \in X} L(x, \lambda)$$ is called the Duality Gap. If $f$ and $g$ are convex, and $g$ satisfies some mild conditions, such as Slater's Condition, then you can be sure that the duality gap is $0$.