Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 9H (Optimisation)

For a vector equality constraint, take the Lagrangian function in constrained optimizationThe sign of the multiplier is a convention; equality multipliers are unrestricted. The Lagrange sufficiency theorem says that if $x^{\ast }\in X$ is feasible and, for some ${\lambda }^{\ast }$, globally minimizes $L(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes the constrained objective.

Indeed for any feasible $x$, $f(x)=L(x,{\lambda }^{\ast })\ge L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$. Thus $\boxed{x^{\ast }\text{ is a global primal minimizer}}$. The theorem requires a global Lagrangian minimum, not merely a stationary point. It needs neither differentiability nor convexity when stated in this form; those hypotheses can be used separately to establish the required minimum.

For the same multiplier convention, the Lagrangian dual problem isFor every feasible $x$ and every $\lambda$, the equality constraint implies $q(\lambda )\le L(x,\lambda )=f(x)$. Taking first the infimum over feasible $x$ and then the supremum over $\lambda$ proves weak duality:The inequalities remain meaningful with the usual extended-value conventions, including $q(\lambda )=-\infty$ when the Lagrangian is unbounded below. Equality of optimum values, or their attainment, does not follow from weak duality. If the sufficient pair in part (a) exists, it attains equality, since $q({\lambda }^{\ast })=L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$.