Notes on optimization problem

Optimality criterion for differentiable $f_0$
Given a convex optimization problem with objective $f_0$.  $x$ is optimal iff it is feasible and
$$\nabla f_0(x)^T (y-x) \ge 0 \quad \text{for all feasible } y$$

if $\nabla f_0(x) \neq 0$, it means that $-\nabla f_0(x)$ defines a supporting hyperplane to the feasible set at $x$.

Unconstrained problem: $x$ is optimal iff
$$x\in \text{dom }f_0, \quad \nabla f_0(x) = 0$$
Equality constrained problem:
$$\text{minimize } f_0(x) \quad \text{ subject to } Ax=b$$
$x$ is optimal iff there exists a $\nu$ such that
$$x\in \text{dom }f_0, \quad Ax=b, \quad \nabla f_0(x)+A^T\nu = 0$$
Note: Lagrange multiplier optimality condition.

Minimization over nonnegative orthant
$$\text{minimize } f_0(x) \quad \text{ subject to } x\succeq 0$$
$x$ is optimal iff
$$x\in\text{dom }f_0, \quad x\succeq 0, \quad  \left\{ \begin{array}{lr}                                                                                                 \nabla f_0(x)_i \ge 0, \quad &x_i = 0  \\                                                                                                 \nabla f_0(x)_i = 0, \quad &x_i \ge 0                                                                                       \end{array} \right.$$
Note: The last condition is a complementarity condition.