Why is one of the KKT conditions the same as one of the constraints?

Question

I'm working through an SVM tutorial (from Andrew Ng Stanford course notes). In the brief coverage of Lagrange duality. The primal optimization problem is stated $$ \min_{w} \theta_{\mathcal{P}}(w) = \min_{w} \max_{\alpha, \beta \::\: \alpha_{i} \ge 0} \mathcal{L}(w,\alpha,\beta) $$

The dual optimization problem is stated

$$ \max_{\alpha, \beta \::\: \alpha_{i} \ge 0} \theta_{\mathcal{D}}(\alpha,\beta) = \max_{\alpha, \beta \::\: \alpha_{i} \ge 0} \min_{w} \mathcal{L}(w,\alpha,\beta) $$

Under the appropriate conditions (convex objective, convex inequalities, affine equalities), it's noted that $w^{*}, \alpha^{*}, \beta^{*}$ can be found such that $w^{*}$ is the primal solution and $\alpha^{*}, \beta^{*}$ are the solution to the dual problem and that they satisfy the KKT conditions. One KKT condition in particular is that

$$ \alpha_{i}^{*} \ge 0. $$

Why is this condition notable considering that a constraint was $\alpha_{i} \ge 0$?

Answer 1

The constraints are part of the KKT conditions, by definition.

The idea of KKT conditions (for convex problems) is that: if you solve this system of equations/inequations, independently of the original optimization problem (i.e. forget everything about the original optimization problem: objective function, constraints, etc...), then the obtained solutions are the solutions to the primal and dual of the original optimization problem.

Thus the constraints of the original problem should be added to the KKT conditions because we want to find solutions that satisfy these constraints.