Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 65 / 2 / a

Use $z\in {\mathbb{R}}^m$ for the perturbation variable and $y\in {\mathbb{R}}^m$ for its dual variable. The full Fenchel conjugate is $f^{\ast }(v,y)={\sup }_{x,z}(⟨v,x⟩+⟨y,z⟩-f(x,z))$. One signed-marginal convention isThe primal problem of convex perturbation duality is ${\inf }_x\phi (x)=p(0)$ and its dual problem of convex perturbation duality is ${\sup }_y\psi (y)=q(0)$. The signed dual marginal $q$ is concave; some conventions instead use $-q$ as a convex marginal. The definitions above fix all signs. Weak duality follows from $f(x,0)+f^{\ast }(0,y)\ge 0$. Also $p^{\ast }(y)=f^{\ast }(0,y)$, so $q(0)=p^{\ast \ast }(0)$.

For a sufficient strong duality condition, assume $f$ is jointly proper convex, $p$ is proper with $p(0)$ finite, and $p$ is finite and continuous in a neighborhood of $0$. More generally $0\in \mathrm{ri}(\mathrm{dom}p)$ suffices in finite dimensions. A supporting subgradient $y^{\ast }\in \partial p(0)$ then exists, andThus the dual is attained with no gap. This condition does not by itself assert attainment of the primal infimum; that needs an additional compactness or coercivity argument.

The same subgradient describes sensitivity analysis in convex perturbation duality: $p(z)\ge p(0)+⟨y^{\ast },z⟩$ bounds the optimum's change under perturbation. When $p$ is finite convex near zero, its one-sided directional derivative is $p^{\prime }(0;d)={\max }_{y\in \partial p(0)}⟨y,d⟩$. If $\partial p(0)=\{y^{\ast }\}$, then $p$ is differentiable there andThese conditions allow first-order sensitivity predictions; without differentiability the subgradient set gives directional bounds. Multipliers can measure the value of relaxing constraints, quantify changes in noise tolerance, and guide parameter choice without resolving every perturbed problem. For the constraint convention in part (b), the noise-budget sensitivity is negative the nonnegative constraint multiplier.