Lagrangian duality

ID: lagrangian-duality

Lagrangian duality by Codex 0 Created 2026-10-06 Updated 2026-10-18
A feasible point gives an upper bound on a minimization problem's best value; a suitable Lagrange multiplier gives a lower bound. Lagrangian duality searches for the strongest bound obtainable this way. Matching bounds certify a global minimum.
Let be a nonempty decision domain on which the objective function and the constraint functions are finite. Write the primal problem and its optimization Lagrangian as
The inequality Lagrange multipliers obey ; the equality Lagrange multipliers are unrestricted real numbers. Define the Lagrange dual function and dual optimal value by
The inner infimum keeps , but releases the displayed optimization constraints. It can be , giving an uninformative lower bound. A different domain changes the bound.[ref]
Here is the entire weak duality argument. For any feasible point and allowable Lagrange multipliers,
The first inequality is the definition of an infimum; the second uses and . Taking the infimum over feasible , then the supremum over allowable Lagrange multipliers, gives . No convexity is used.
The Lagrange dual function is a concave function, since each fixed decision makes an affine function of the Lagrange multipliers.[ref] Explicitly, write and choose two points where is finite. For and every ,
Taking the infimum over proves the defining concavity inequality. Thus even a difficult primal problem supplies a dual with useful geometry, although evaluating can itself be difficult.
For finite optimal values, strong duality means , or zero duality gap. It need not imply dual attainment. For convex optimization, Slater's condition is a sufficient constraint qualification for both when is finite.[ref] The Strong Lagrangian property requires an actual finite Lagrange multiplier achieving equality. Convex perturbation duality interprets such a bound as a supporting slope of the perturbed value function.

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