When is the minimizer of the Lagrangian the solution to the primal problem?

Question

This is a problem that came up when I studied the proof of the Newman-Pearson lemma/theorem, both in Van Trees' book "Detection Estimation, and Modulation theory" and Kay's "Fundamentals of statistical signal processing: detection theory".

The problem is (the functions are basically integrals of probability density functions, but that is not important here (I think)) \begin{aligned} \min_x f(x)\\ \text{s.t. } h(x) = 0 \end{aligned} and I wish to find the solution to this problem, $x^*$. In the books, this is solved by forming the Lagrangian: $$L(x,\lambda) = f(x) + \lambda h(x)$$ and minimizing this wrt $x$, giving the dual function $$g(\lambda) = L(\hat{x}(\lambda), \lambda) = \min_x L(x,\lambda) .$$ We can then choose $\lambda^*$ in such a way that the constraint in the first problem is satisfied, i.e., $h(\hat{x}(\lambda^*) = 0.$ (We assume that there exists such a $\lambda^*$.)

My question: when does this procedure ensure that the obtained solution $x^*$ solves the original problem? In other words, what has to be assumed about $f(x)$, $h(x)$, $L(x,\lambda)$ and $g(\lambda)$ for this to hold?

My thoughts: It seems like this is some special case of basically solving the dual problem, instead of the primal problem, but stopping half way through. It seems reasonable to me that if there is a unique minimizer for the Lagrangian, this should be the one minimizing the original problem, but I base this on (probably) flawed intuition.

A related question, asking about (basically) the converse to my question is here.

Edit I feel I need to clarify something. To me, the way this problem is solved seems to be kind of a "shortcut" to solving the dual problem. Instead of finding $\lambda^*$ by maximizing the dual function, it is found by looking at the primary constraint. Is this always the case? Or does this only hold if we have strong duality? Or is there another sufficient condition?

Answer 1

You basically, If I got you write, ask when Strong Duality holds.

Well, there are countless of edge cases but the simplest one is Slater Condition.

Simply:

1. If the objective function is convex.
2. If the Equality Constraints are affine.
3. If the inequality constraints are convex.
4. If the domain all above are defined on is convex.

The above states a Convex Problem.
Slater Condition simply says:

5. The Domain isn't empty.
6. There is a point in the domain which holds all inequality constraints in strict form or all inequality constraints are affine.

If (5) and (6) hold then the problem is guaranteed to hold Strong Duality -> The dual solution is identical to the primal solution.