Second order optimal condition for this question I am trying to solve a very simple constrained optimization problem below:
P: $\min{x_1 + x_2}$
subject to
$ x_1 \geq 0$
$ x_2 \geq 0$
$ (2x_1+x_2)^2 = 4$
By solving the KKT condition, I have the KKT point as $(x_1,x_2,\lambda_1,\lambda_2,\mu)=(1,0,0,\frac{1}{2},-\frac{1}{8})$
and I know it is a global minima. However, when I checked the second order condition (sufficient condition), it turned out the Tagent Space (Cone) $\{d\neq0: \nabla h\cdot d=0, \nabla g_1 \cdot d\leq0, \nabla g_2\cdot d=0\}$ is empty, and thus cannot tell whether the Hessian of restricted Lagragian function is positive definite or not. In fact, if I don't use the Cone, the Hessian of the restricted Lagragian $\nabla_{xx}\bar{L}(x, \bar{\lambda},\bar{\mu})$ is negative definite, which contradicts the fact that the solution is a local min.
Could anyone help to answer my question? What should we do if the cone set is empty?
Thanks in advance.
 A: Rewritting the problem:
$$
\mathscr{P}: \min x_1+x_2, \\(2x_1+x_2)^2-4=0, \\-x_1\le0,\\ -x_2\le0
$$
Then the KKT conditions are:
$$
L=(x_1+x_2)+\mu_1(-x_1)+\mu_2(-x_2)+\lambda_1((2x_1+x_2)^2-4)
$$
Stationarity:
$$
\nabla_{x_1}L=1-\mu_1            +\lambda_14(2x_1+x_2)=0\\
\nabla_{x_2}L=1      -\mu_2      +\lambda_12(2x_1+x_2)=0
$$
Feasibility (this is normally redundant and serves only to expose $\nabla L$)
$$
\nabla_{\mu_1}L=-x_1\le0\\
\nabla_{\mu_2}L=-x_2\le0\\
\nabla_{\lambda_1}L=(2x_1+x_2)^2-4=0
$$
Slackness:
$$
\mu_1x_1=0\\
\mu_2x_2=0
$$
The specified point of the full problem is: $[x_1 \ x_2 \ \mu_1 \ \mu_2 \ \lambda_1]=[1 \ 0 \ 0 \ 1/2 \ -1/8]$.
The tangent cone for the solution is evaluated for all the active constraints (Mangasarian-Fromovitz constraint qualification), which are only the active inequalities plus the equalities (which will always be active!):
$$
\nabla g_1 \cdot d = \le 0\\
\nabla h_1 \cdot d = 0\\
$$
The condition in this form is trivially met for the given point.
The full Hessian of the active constraints must be semi-positive $d'Ld>0$ for $\nabla g_1 d=0$ and $\nabla h_1 d=0$, the orthogonal directions of the active constraints.
Hence in the full Hessian: 
$$
\nabla^2 L=
\left[
\begin{array}{ccccc}
 8\lambda_1 &4\lambda_1 &-1 &0 &4(2x_1+x_2)\\
 4\lambda_1 &2\lambda_1 &0 &-1 &2(2x_1+x_2)\\
 -1 &0 &0 &0 &0\\
 0 &-1 &0 &0 &0\\
 4(2x_1+x_2) &2(2x_1+x_2) &0 &0 &0
\end{array}
\right]
=
\left[
\begin{array}{c}
-1 &-2 &-1 &0 &8\\
 -1/2 &-1/4 &0 &-1 &4\\
 -1 &0 &0 &0 &0\\
 0 &-1 &0 &0 &0\\
 8 &4 &0 &0 &0
\end{array}
\right]
$$
we discard the 3rd row and column for considering the direction $[1 \ 2 \ 0 \ 0 \ 0]$ for the $h_1$ constraint active, and the rest inactive, and we have the reduced hessian, which is positive definite:
$$
L(d)=d'[-1 2][-1 -2;-1/2 -1/4][-1;2]d=3d^2
$$
Note that we cannot choose any other direction without breaking feasibility. So the critical point is a minimum.
A: That is a very good question. It is all a matter of definition. Let $\bar{x}$ be the point being analyzed and
$$K= \{d\in \Bbb R^n: \nabla h(\bar{x})\cdot d=0, \nabla g_{I(\bar{x})}(\bar{x}) \cdot d\leq0, \nabla g_2(\bar{x})\cdot d=0\},$$  be the tangent cone at $\bar{x}.$ Here $I(\bar{x})=\{i: g_i(\bar{x})=0\}$ is the set of active constraints. Recall that the second order sufficient condition states: 
$$\textrm{ If } \nabla f(\bar{x})=0 \textrm{ and } d^T\nabla^2f(\bar{x})d>0 \;\forall \;d\in K\setminus\{0\}, \textrm{ then } \bar{x} \textrm{ is a local minimum }.$$ Well, you can easily see now that this condition holds by vacuity in your problem because there is no $d\in K\setminus\{0\}.$ Hence the point is in fact a local minimum.
EDIT: Since the OP asked for a reference, I would like to further elaborate on that point. I haven't seen this exactly, in the literature, but it is easy to see this fact from the proof of the second order sufficient condition, see Bazaraa. This proof goes as follows: they assume that $\bar{x}$ is not a global minimum, and from there they build a sequence $\{d_n\}$ such that $\|d_n\|=1,$ and $d_n\to d \in K.$ 
If $K=\{0\},$ we now obtain a contradiction because $\|d\|=1,$ so that $d\neq 0.$ Hope this helps.
