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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 $Ax=b$.

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 ${\lambda }^{\ast }$.

Let ${\lambda }^{\ast }$ be an optimal dual multiplier and define the maximum violationDual stationarity and strong duality imply, for every $x$,Since every component of $Ax-b$ is at most $v(x)$ and ${\lambda }^{\ast }\ge 0$,It follows that the exact maximum-violation penalty obeysChoose any $M>\Vert {\lambda }^{\ast }{\Vert }_1$. A primal optimum has $v(x)=0$ and penalized value $p^{\ast }$, whereas every infeasible point has $v(x)>0$ and penalized value strictly greater than $p^{\ast }$. Thus the penalized problem and the original linear program have exactly the same minimizers.

The Slater condition requires a feasible point at which every nonlinear convex inequality is strict and every affine equality holds. For a convex problem it implies strong duality and attainment of the dual optimum under standard finiteness assumptions.