Karush-Kuhn-Tucker

Bruce C. Dieffenbach ()
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Bruce C. Dieffenbach: Independent author

Chapter 9 in Conjugate Duality in Economic Analysis, 2026, pp 73-79 from Springer

Abstract: Abstract We consider the Karush-Kuhn-Tucker problem $$\sup _{\boldsymbol {x}}\left [ g\left ( \boldsymbol {x}\right ) -\delta _{\mathrm {epi\ } f}\left ( \boldsymbol {x},z\right ) \right ] \!,$$ in which the perturbation base value z=0. Here g concave, and f is convex. The Lagrangian is $$L\left ( \boldsymbol {x},z^{\ast }\right ) =g\left ( \boldsymbol {x}\right ) -z^{\ast }f\left ( \boldsymbol {x}\right ) -\delta _{\mathrm {dom\ }f}\left ( \boldsymbol {x}\right ) + \delta _{z^{\ast }\geq 0}.$$ For $$\left ( \boldsymbol {x},z^{\ast }\right )$$ to be a saddle point of the Lagrangian is necessary and sufficient for x to solve the Karush-Kuhn-Tucker problem and for z* to solve its dual. The “Lagrange multiplier” is the dual solution set, which may contain zero as well as positive values. Compared to the traditional ad hoc Lagrangian analyzed by economists, this expression contains the extra term $$\delta _{\mathrm {dom\ }f}\left ( \boldsymbol {x}\right )$$ , which comes into play for the exceptional case z*=0. The dual implies several results concerning the exceptional case. If x solves the Karush-Kuhn-Tucker primal such that $$f\left ( \boldsymbol {x}\right )

Date: 2026
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DOI: 10.1007/978-3-032-21396-9_9

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