TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Finite-dimensional Fenchel-Moreau biconjugation

Statement

Let $d \geq 1$ and let $f : R^{d} \to \overline{R}$ be closed, proper, and convex in the sense of Proper, convex, and lower-semicontinuous extended-real functions on Euclidean space. With the Fenchel conjugate and biconjugate of Fenchel conjugate, biconjugate, and subdifferential, one has

$f^{* *} (x) = f (x) (x \in R^{d}) .$

Facts & Assumptions

Given: The closed proper convex function and conjugates in the Statement.

[F1]

Proper, convex, and lower-semicontinuous extended-real functions on Euclidean space: proper means $dom f \neq \emptyset$ and $f > - \infty$ ; convexity is the extended chord inequality; lower semicontinuity means closed weak sublevel sets.

[F2]

Extended-real convexity and lower semicontinuity are characterized by the epigraph: under [F1], $epi f$ is nonempty, closed, and convex.

[F3]

Strictly separate a point from a nonempty closed convex Euclidean set: a point outside a nonempty closed convex subset of $R^{d + 1}$ has a strict linear separator.

[F4]

Fenchel conjugate, biconjugate, and subdifferential: $f^{*} (p)$ and $f^{* *} (x)$ are the stated extended suprema over the effective domain and finite conjugate domain.

[L1]

Every complete ordered field is Archimedean: the canonical natural numbers are cofinal in $R$ ; for every $q \in R$ there is a positive integer $k$ with $q < k$ .

Proof

technique · direct

1.1F4givenalgebra

For any $x$ with $f (x) < + \infty$ and any $p$ with $f^{*} (p) < + \infty$ , the defining supremum gives $f^{*} (p) \geq ⟨ p, x ⟩ - f (x)$ , hence $⟨ p, x ⟩ - f^{*} (p) \leq f (x)$ ; taking the supremum over $p$ yields $f^{* *} (x) \leq f (x)$ . If $f (x) = + \infty$ , the same upper bound is automatic.

1.2F2F3givenalgebra

Fix $x \in dom f$ and a real $t < f (x)$ . Apply [F3] to $(x, t)$ and the nonempty closed convex set $epi f$ , and negate the resulting separator to write $⟨ a, y ⟩ + b s \geq c > ⟨ a, x ⟩ + b t$ for all $(y, s) \in epi f$ . Upward closure forces $b \geq 0$ . If $b = 0$ , the inequality at $(x, f (x)) \in epi f$ would give $⟨ a, x ⟩ \geq c > ⟨ a, x ⟩$ , impossible. Thus $b > 0$ .

1.3F2F3F4algebra

Now fix $x \notin dom f$ and $t \in R$ . Apply [F3] to $(x, t)$ and negate the separator to obtain $⟨ a, y ⟩ + b s \geq c > ⟨ a, x ⟩ + b t$ for all $(y, s) \in epi f$ . If $b > 0$ , then $f (y) \geq - ⟨ a / b, y ⟩ + c / b$ for every $y \in dom f$ , so with $p = - a / b$ one has $f^{*} (p) \leq - c / b < + \infty$ and $f^{* *} (x) \geq ⟨ p, x ⟩ - f^{*} (p) \geq (c - ⟨ a, x ⟩) / b > t$ .

2.1F4step 1.2algebra

Substituting $(y, f (y))$ for every $y \in dom f$ gives the affine minorant $ℓ (y) = - ⟨ a / b, y ⟩ + c / b \leq f (y)$ . With $p = - a / b$ , [F4] gives $f^{*} (p) \leq - c / b$ , hence $f^{* *} (x) \geq ⟨ p, x ⟩ - f^{*} (p) \geq ℓ (x) > t$ . Since this holds for every $t < f (x)$ , $f^{* *} (x) \geq f (x)$ on $dom f$ .

2.2F1F2F3F4L1step 1.3choosealgebra

In the $b = 0$ case of step 1.3, the separator gives $⟨ a, y ⟩ \geq c > ⟨ a, x ⟩$ for every $y \in dom f$ . Choose $y_{0} \in dom f$ and a finite $t_{0} < f (y_{0})$ . Apply [F3] to $(y_{0}, t_{0})$ and $epi f$ , reversing the separator orientation to obtain $⟨ a_{0}, y ⟩ + b_{0} s \geq c_{0} > ⟨ a_{0}, y_{0} ⟩ + b_{0} t_{0}$ for every $(y, s) \in epi f$ . Upward closure gives $b_{0} \geq 0$ , and $b_{0} = 0$ contradicts the inequality at $(y_{0}, f (y_{0}))$ , so $b_{0} > 0$ . Set $p_{0} = - a_{0} / b_{0}$ and $r_{0} = c_{0} / b_{0}$ ; substituting $(y, f (y))$ gives the affine minorant $ℓ_{0} (y) = ⟨ p_{0}, y ⟩ + r_{0} \leq f (y)$ on $dom f$ , so [F4] gives $f^{*} (p_{0}) \leq - r_{0}$ . For each positive integer $k$ , define $ℓ_{k} (y) = ℓ_{0} (y) + k (c - ⟨ a, y ⟩) = ⟨ p_{0} - k a, y ⟩ + (r_{0} + k c)$ . Since $c - ⟨ a, y ⟩ \leq 0$ on $dom f$ , this is still a minorant; [F4] gives $f^{*} (p_{0} - k a) \leq - (r_{0} + k c) < + \infty$ and hence $f^{* *} (x) \geq ℓ_{k} (x) = ℓ_{0} (x) + k δ$ , where $δ : = c - ⟨ a, x ⟩ > 0$ by step 1.3. For any prescribed $B \in R$ , apply [L1] to $q = (B - ℓ_{0} (x)) / δ$ to choose a positive integer $k$ with $q < k$ ; multiplying by $δ > 0$ gives $ℓ_{0} (x) + k δ > B$ . Hence $f^{* *} (x)$ exceeds every real bound, so $f^{* *} (x) = + \infty = f (x)$ .

3.1step 1.1step 2.1step 2.2∎

The upper bound from step 1.1 and the lower bounds from steps 2.1 and 2.2 cover respectively finite and infinite values of $f (x)$ , proving equality everywhere.

Finite-dimensional Fenchel-Moreau biconjugation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources