Given functions $f_1,f_2:\mathbb{R}^n\rightarrow \mathbb{R}$ and constraints $h_1,h_2:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$ consider the (constrained) Nash equilibrium problem of finding $x^*=(x_1^*,x_2^*)\in\mathbb{R}^{2n}$ such that
$x_1^* \in argmin \{ f_1(x,x_2^*)\quad st \quad h_1(x)=0\},$
$x_2^* \in argmin \{ f_2(x_1^*,x)\quad st \quad h_2(x)=0\}.$
What is the sufficient second order condition in order for $x^*$ to be an equilibrium point in this case?
If it was an optimization problem with an equality constraint then it would suffice that $\nabla^2L(x^*,\lambda^*)>0$, where $L(x,\lambda)$ is the Lagrangian function associated with the problem.
When it comes to Nash equilibriums I'm aware that the necessary 1st order condition is $\nabla_{x_i} L_i(x^*,\lambda_i)=0\in \mathbb{R}^n$, for each of the two associated Lagrangians of the problems ($i=1,2$), but everywhere I look I don't find the 2nd condition, the texts all mention the first order instead and whenever the search returns something on second order and I open the text it's about optimization instead of equilibrium on the context...
So what would be the 2nd order sufficient condition? Is it really just $\nabla^2_{x_ix_i}L_i(x^*,\lambda_i^*)>0$, completely analogous to an optimization problem? I feel like there's something missing here... For instance if you do a Newton method for the optimality conditions with $f_i$ quadratic and organize in a linear system then the matrix of the system becomes of size $2(n+m)\times2(n+m)$ so I feel like it couldn't be simply a condition on two smaller square matrices and that's that, no?
