Convex KKT optimality certificate ($f(x_{\ast })\le f(z)(z\text{ feasible})$)

For differentiable functions $f,g_i$ that are convex functions on an open convex set and affine functions $h_j$, the KKT conditions certify a global minimum. Once Lagrange multipliers satisfy them, no extra Slater condition is needed.^[ref]

The certificate is a short chain of inequalities. Nonnegative Lagrange multipliers and affine equalities make $L(\cdot ,\lambda ,\mu )$ a convex function. Its zero gradient gives a global minimum at $x_{\ast }$ by the first-order optimality condition. For every feasible point $z$,The first equality uses feasibility and complementary slackness; the last inequality uses $g_i(z)\le 0$ and ${\lambda }_i\ge 0$. Equivalently, the Lagrange dual function attains the same value as the primal objective function, proving strong duality. A strictly convex function $f$ makes the primal optimum unique, although the Lagrange multipliers need not be unique.

Affine equality optimization constraints matter: an arbitrary signed multiple of a nonlinear equality can destroy convexity of the optimization Lagrangian. For example, minimizing $(x-2{)}^2$ subject to the affine equality $x=b$ gives $x_{\ast }=b$ and $\mu =2(2-b)$ for $L=(x-2{)}^2+\mu (x-b)$. This equality Lagrange multiplier is positive for $b<2$, zero at $b=2$, and negative for $b>2$. Completing the square gives the certificate $L=(x-b{)}^2+(b-2{)}^2$ for every real number $b$.

Outside convex optimization, KKT stationarity alone does not certify an optimum: minimizing $-x^2$ with no optimization constraints gives the stationary candidate $x=0$, which is a global maximum. The Lagrangian sufficiency theorem remains applicable whenever a global minimum of the optimization Lagrangian can be proved directly.