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A finite-dimensional convex value function has a support at relative-interior points

Statement

Let $V$ be a finite-dimensional Euclidean space and let $p : V \to (- \infty, + \infty]$ be convex, never $- \infty$ , and finite somewhere. If $0 \in ri (dom p)$ and $p (0) \in R$ , then there is a linear functional $ℓ : V \to R$ such that $p (z) \geq p (0) + ℓ (z) (z \in V) .$

Facts & Assumptions

Given: The finite-dimensional Euclidean space, convex value function, relative-interior point, and finite value in the Statement. The conventions for domains, affine hull, relative interior and epigraph are those of Finite-dimensional convex program, Lagrangian dual, and Slater condition.

[F1]

A finite convex real-valued function on an open convex subset of a finite-dimensional Euclidean space is locally Lipschitz and hence continuous (A convex function on an open convex set is locally Lipschitz).

[F2]

Every point has a nearest point in a nonempty closed convex Euclidean set (Every point has a unique nearest point in a nonempty closed Euclidean convex set), and the projection satisfies $⟨ z - Π_{C} z, y - Π_{C} z ⟩ \leq 0$ for every $y \in C$ (Metric projection onto a closed convex set satisfies the variational inequality).

[F3]

The unit sphere in a finite-dimensional normed space is sequentially compact, including the choice-free finite-dimensional case (Unit spheres in finite-dimensional normed spaces are sequentially compact).

Proof

1.1givenF1

Let $W = aff (dom p)$ . Since $0$ belongs to the domain, $W$ is a linear subspace; relative interior gives a ball about $0$ in $W$ contained in $dom p$ . If $W = {0}$ , take $ℓ = 0$ . Otherwise identify $W$ with a Euclidean coordinate space. The restriction of $p$ to that open ball is finite and convex, so [F1] makes it continuous at $0$ .

2.1step 1.1givenalgebra

In $W \times R$ , let $E = {(z, t) : p (z) \leq t}$ and let $\overline{E}$ be its closure. The set $E$ is convex, $(0, p (0)) \in E$ , and $\overline{E}$ is nonempty, closed and convex. Continuity from step 1.1 shows that $(0, p (0) - ϵ) \notin \overline{E}$ for every $ϵ > 0$ : choose $δ > 0$ so $p (z) > p (0) - ϵ / 3$ whenever $∥ z ∥ < δ$ , while a sufficiently small open ball about $(0, p (0) - ϵ)$ has $∥ z ∥ < δ$ and $t < p (0) - 2 ϵ / 3$ , so it misses $E$ . Thus $(0, p (0))$ is a boundary point of $\overline{E}$ .

3.1F2F3step 2.1construct

For each positive integer $k$ , project $z_{k} = (0, p (0) - 1 / k)$ onto $\overline{E}$ , obtaining $y_{k}$ . Since $(0, p (0)) \in \overline{E}$ , the projection distance is at most $1 / k$ , so $y_{k} \to (0, p (0))$ . The unit vectors $a_{k} = (y_{k} - z_{k}) / ∥ y_{k} - z_{k} ∥_{2}$ have a convergent subsequence by [F3], with unit limit $a = (s, α) \in W \times R$ . For each $y \in \overline{E}$ , [F2] gives $⟨ a_{k}, y - y_{k} ⟩ \geq 0$ ; passing along that subsequence gives $⟨ a, y - (0, p (0)) ⟩ \geq 0$ .

4.1step 1.1step 3.1algebra∎

The epigraph contains every upward vertical ray, so the support inequality forces $α \geq 0$ . If $α = 0$ , points $(z, p (z)) \in E$ for all sufficiently small $z$ and $- z$ in the ball from step 1.1 force both $⟨ s, z ⟩ \geq 0$ and $⟨ s, - z ⟩ \geq 0$ . Hence $s = 0$ , contradicting $∥ a ∥_{2} = 1$ . Therefore $α > 0$ . Setting $ℓ (z) = - ⟨ s, z ⟩ / α$ , the support inequality at $(z, p (z))$ yields $p (z) \geq p (0) + ℓ (z)$ for $z$ in the domain; outside the domain the inequality is automatic because $p (z) = + \infty$ .

Depends on

Used by

Slater gives zero duality gap and an attained dual certificate Theorem

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Boyd and Vandenberghe, Convex Optimization (standard reference, not scraped)
Zhiyuan Li, Convex Optimization, Lecture 5: Lagrange Multipliers and KKT (standard reference, not scraped)