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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.

Eric Brown
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  • The optimal point has to be on the boundary by the maximum principle (minimum of a concave function on a convex set). – LinAlg Mar 06 '19 at 17:46
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    Yes the minimum point is in the boundary. But we assess the second order necessary condition that the Hessian of the Lagrangian over the valid variation space has to be positive definite, the valid variation space is just point zero and we cannot asses the second order sufficiency condition. – Eric Brown Mar 07 '19 at 01:41
  • I added the KKT conditions – Eric Brown Mar 12 '19 at 04:45
  • I do not understand why $y=0$ is a KKT condition. Should you not get $\lambda_2y=0$, $x\geq 0$, $y\geq 0$? Since the KKT conditions are necessary, you can just enumerate all solutions to the KKT conditions and check whichs one is a minimum and which one is a maximum. – LinAlg Mar 12 '19 at 13:55

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