When is the solution to Lagrangian equal to the solution of the constrained optimization problem?

Question

I am trying to understand when the solution found by Lagrangian is equal to the original optimization problem with equality constraints. This post is related to my question but that question is about inequality constraints and the answer states "under a broad range of conditions" the two are equal. Other questions seem to discuss special cases such as optimization of convex functions.

Specifically, consider the following optimization problem: \begin{align} x^\star = \arg \max_{x \in \mathcal{X}} f(x)\\ \text{s.t. } g(x) = 0 \end{align}

• Variable $x$ is a vector.
• $x^\star$ is a unique solution and there exists $x^\star \in \mathcal{X}$ such that $g(x^\star) = 0$.
• Functions $f(x), g(x)$ are differentiable with respect to $x$ but they are not convex/concave and we don't have $g(x) \geq 0$.
• The set $\mathcal{X}$ might not be a continuous set and may just be a finite set so we generally can't use arguments that introduce gradients of $f(x)$ and $g(x).

My understanding is that the Lagrange method forms the following optimization problem: \begin{align} \tilde x, \tilde \lambda = \arg \max_{x \in \mathcal{X}} \arg \min_\lambda f(x) - \lambda g(x). \end{align}

My questions:

1. When is $\tilde x = x^\star$, given that we assumed $x^\star$ is unique and $\tilde x$ is the actual optimum of the Lagrange form not just $\nabla f(\tilde x) = 0$?
2. I am confused about what $\tilde \lambda$ should be. My understanding is if we define $\tilde \lambda(x) = \arg \min_\lambda -\lambda g(x)$, then the solution is

\begin{align} \tilde{\lambda}(x) = \begin{cases} - \infty \quad & \text{if } g(x) > 0\\ + \infty \quad & \text{if } g(x) < 0\\ \text{any number} \quad & \text{if } g(x) = 0 \end{cases} \end{align}

If $\tilde{x} = x^\star$, since we know $g(x^\star) = 0$, does that mean $\tilde \lambda$ can be anything?

3. If No. 2 is true, what happens if we find $\tilde \lambda$ by $\min_{\lambda \in [-B, B]}$ over a bounded interval?

4. In some examples, I have seen a discussion of on strong duality and whether $\max_x \min_\lambda f(x) - \lambda g(x) = \min_\lambda \max_x f(x) - \lambda g(x)$. Does strong duality have anything to do with $\tilde x$ being equal to $x^\star$?