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A jointly proper convex function $f(x,z)$ represents an optimization problem at perturbation $z=0$ and a family of nearby problems at other $z$. Its primal value function is $p(z)={\inf }_{x}f(x,z)$. Relaxing an inequality by $z$ gives the example $f(x,z)=J(x)+{\delta }_{(-\infty ,0]}(r(x)-\sigma -z)$, using an indicator functional. The resulting convex perturbation duality expresses multipliers as supporting subgradients of the value function.

If two nonnegative finite-valued convex functions are defined on all of a finite-dimensional space, their infimal convolution is finite, convex and continuous everywhere. Convex perturbation duality then gives strong duality and dual attainment at every argument. It does not require or imply attainment of the primal decomposition infimum.

More precisely, let $f,g:{\mathbb{R}}^n\to [0,\infty )$ be finite everywhere and write $h=f\square g$. Nonnegativity supplies a lower bound, while the two trial splits $(x,0)$ and $(0,x)$ giveMixing nearly optimal splits proves convexity, and continuity of a convex function applies because the effective domain is the entire finite-dimensional vector space. These statements concern the optimized value, rather than existence of an optimal split. For the continuity theorem and dual attainment under the equality constraint $u+v=x$, see Bertsekas, Convex Optimization Theory: A Summary, Propositions 1.3.11 and 5.3.3(a). The equality has a feasible point in the whole space for every $x$, and the displayed bound makes its optimal value finite.

A finite value without any optimal split. On $\mathbb{R}$, take the convex functions $f(u)=e^u$ and $g(v)=0$. Both are nonnegative and finite everywhere, butfor every $x$. The splits $(u,v)=(-k,x+k)$ have costs $e^{-k}\to 0$, yet the exponential function is strictly positive at every finite $u$. No split attains the infimum. The value function is even constant and differentiable; regularity of the value function does not prevent optimizing splits from escaping to infinity. For example, $k=1,3,6$ gives costs approximately $0.367879$, $0.049787$ and $0.002479$.

A sufficient extra condition for attainment at a fixed $x$ is that $u\mapsto f(u)+g(x-u)$ be a coercive function, meaning it tends to $+\infty$ as $\Vert u\Vert \to \infty$. Under the finite-valued assumptions it is a continuous function, so a sufficiently large closed ball contains a global minimum by the extreme value theorem. If this cost is also a strictly convex function, the optimal split is unique. The quadratic example above satisfies both conditions; the exponential function example does not satisfy coercivity.

Nonnegativity is sufficient, not necessary: finite lower bounds $f\ge A$, $g\ge B$ would give $h\ge A+B$ by the same argument. Removing all lower-bound control can lose finiteness even for finite convex functions: $f(u)=u$ and $g(v)=-v$ give $f(u)+g(x-u)=2u-x$, hence $h(x)=-\infty$. Nor does finiteness alone guarantee differentiability: choosing $f=g=|\cdot |$ gives $h(x)=|x|$ by the triangle inequality and the split $(x,0)$.

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 $X$ be a nonempty decision domain on which the objective function $f$ and the constraint functions $g_i,h_j$ are finite. Write the primal problem and its optimization Lagrangian asThe inequality Lagrange multipliers obey ${\lambda }_i\ge 0$; the equality Lagrange multipliers ${\nu }_j$ are unrestricted real numbers. Define the Lagrange dual function and dual optimal value byThe inner infimum keeps $X$, but releases the displayed optimization constraints. It can be $-\infty$, giving an uninformative lower bound. A different domain changes the bound.^[ref]

Here is the entire weak duality argument. For any feasible point $z$ and allowable Lagrange multipliers,The first inequality is the definition of an infimum; the second uses ${\lambda }_i{g}_i(z)\le 0$ and $h_j(z)=0$. Taking the infimum over feasible $z$, then the supremum over allowable Lagrange multipliers, gives $d_{\ast }\le p_{\ast }$. No convexity is used.

The Lagrange dual function is a concave function, since each fixed decision makes $L$ an affine function of the Lagrange multipliers.^[ref] Explicitly, write $y=(\lambda ,\nu )$ and choose two points where $q$ is finite. For $0\le t\le 1$ and every $x\in X$,Taking the infimum over $x$ proves the defining concavity inequality. Thus even a difficult primal problem supplies a dual with useful geometry, although evaluating $q$ can itself be difficult.

For finite optimal values, strong duality means $d_{\ast }=p_{\ast }$, or zero duality gap. It need not imply dual attainment. For convex optimization, Slater's condition is a sufficient constraint qualification for both when $p_{\ast }$ 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.

Use the maximizing-dual convention. For the perturbation function $f(x,u)$, defineThe primal problem is ${\inf }_x\phi (x)=p(0)$; the Lagrangian dual problem is ${\sup }_y\psi (y)=q(0)$. The sign convention makes the dual marginal $q$ concave for jointly convex data; some formulations negate $q$ to express a minimization problem.

The central convex perturbation duality calculation isThe last inequality is weak duality. A sufficient finite-dimensional condition for strong duality with dual attainment is: $f$ is jointly proper and convex, $p$ is proper with $p(0)$ finite, and $p$ is finite on a neighborhood of zero. More generally $0\in \mathrm{ri}(\mathrm{dom}p)$ suffices.

Indeed continuity of a convex function on the interior of its domain gives a supporting subgradient $y_{\ast }\in \partial p(0)$, soTherefore $\psi (y_{\ast })=p(0)$ and the dual maximum is attained. This qualification establishes equality and dual attainment; it does not by itself establish a primal minimizer.

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.

A perturbation function encodes a family of objectives and constraints indexed by a perturbation $u$, with the original problem at $u=0$. Extended-real values encode constraints. Its marginal $p(u)={\inf }_{x}f(x,u)$ measures the optimal value's response to perturbation; jointly convex data give convex perturbation duality.

The original mathematical optimization problem relative to a chosen Lagrangian dual problem. In convex perturbation duality, its objective is $\phi (x)=f(x,0)$ and its value is $p(0)$. Strong duality concerns optimal values; primal attainment additionally requires an actual minimizing point.