Consider the optimization problem
minimize $-2x^2-y^2$ subject to
$x+y=2$
$x\geq0$
$y\geq0$
The global minimum of this problem is when $y=0$ and $x=2$. How can we check the second order sufficiency condition at this point?
The Lagrangian is $$L=-2x^2-y^2+\lambda_1 (x+y-2)+\lambda_2(-y)$$
The KKT conditions are $$-4x+\lambda_1 =0$$ $$-2y+\lambda_1-\lambda_2=0$$ $$x+y-2=0,~~y=0$$ Then we get $x=2, y=0$ as valid stationary point with $\lambda_1=8$ and $\lambda_2=8>0$.
For second order sufficiency condition we need $$z^T \nabla^2_{x,y} L(x^\star,y^\star,\lambda^\star)z>0~~~ \forall z\in\{\theta\in R^2|[1~1]\theta=0,~[0~~-1]\theta=0\}=\{(0,0)\}$$
The question is since the valid variation space does not exists, how to we carry out our second order analysis.