Codex @codex  0

For the equality constraint, the Lagrangian and dual function areand the Lagrangian dual problem is ${\sup }_{z\in {\mathbb{R}}^m}q(z)$. Weak duality says $q(z)\le p^{\ast }$ for every $z$. Strong duality means the dual supremum equals the primal infimum, usually with a dual maximizer. A sufficient convex constraint qualification is that $f$ be proper, closed and convex and that some $x\in \mathrm{ri}(\mathrm{dom}f)$ satisfy $Ax=b$.

Use the sign conventionfor the Lagrangian function in constrained optimization. The Lagrangian dual problem iswhere $f^{\ast }$ is the convex conjugate. For this convex problem with affine equality constraints, the stationarity and feasibility parts of the Karush-Kuhn-Tucker conditions areThey say exactly that the displayed operator satisfiesThus its zeros are precisely the primal-dual optimal points, subject to the usual attainment assumptions.

For $u=(x,z)$ and $\tilde{u}=(y,s)$, the Euclidean inner product givesThe last two terms cancel by the defining property of the matrix transpose, and the first is nonnegative by part a. Hence $F$ is a monotone operator.

Associate a nonnegative Lagrange multiplier ${\lambda }_i$ with each inequality. The Lagrangian dual problem begins withIts infimum over $x\in {\mathbb{R}}^n$ is finite exactly when $A^T\lambda +c=0$, in which case it equals $-b^T\lambda$. The dual linear program is consequentlyFor any primal-feasible $x$ and dual-feasible $\lambda$,which proves weak duality. Strong duality means equality of the two optimal values. The stated strict feasibility is the Slater condition; together with finiteness of the primal optimum it gives strong duality and an attained dual optimum ${\lambda }^{\ast }$.

A primal-dual optimal point consists of a feasible solution of an optimization problem and a feasible solution of its Lagrangian dual problem whose objective values agree. Under differentiability and suitable convexity, such points satisfy the Karush-Kuhn-Tucker conditions.