Assume we have a problem
$$ \min_x \max_y f(x, y) \ \ \text{ s.t. } \ g(x, y) = 0$$
I wonder if it is possible to incorporate this constraint into an objective in Lagrangian-like fashion:
$$ \min_{x} \max_{y, \lambda} \min_{\eta} f(x, y) + \eta \cdot \lambda \cdot g(x,y) $$
but my attempts at getting such an expression failed so far. For example, if we consider $h(x,y,\lambda) = \min_{\eta} f(x, y) + \eta\lambda g(x,y)$, it already takes the $-\infty$ for all values of $(x, y, \lambda)$ such that $g(x, y) \neq 0$ including $\lambda = \pm \infty $, so there is no reason for $\min_x$ not to choose an $x$ s.t. $g(x,y)\neq0$.
It is no suprrise since I was trying to get some characteristic term $ f(x, y) + \chi(g)(x, y)$ that would be "not favorable" to nor minimization or maximization, so it should not be one of $\pm \infty$.
If constraints are not joint, i.e. $g(x)=0$ and $k(y)=0$ then adding two multiplier terms seems to work $p(x,y,\lambda,\eta) = \max_\eta \min_\lambda f(x, y) + \eta \cdot g(x) + \lambda \cdot k(y)$ is equal to $+\infty$ if $k(y) \neq 0$ and $-\infty$ if $g(x)\neq 0$ and $\infty$ if both do not hold (which is weird, but reasonable). But this is not the case I am interested in.
When solving this kind of problem
$$ \min_x \max_y f(x, y) \ \ \text{ s.t. } \ g(x, y) = 0 $$
we are searching for the $f_g(x,y)$ saddle points and those points are handled normally in the Lagrange Multipliers procedures so
$$ L(x,y,\mu) = f(x,y)+\mu g(x,y) $$
and the stationary points are determined by
$$ \nabla L = 0 $$
regarding only saddle points.