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Fenchel–Young inequality

Proved
ConvexOptimization.fenchel_young

by Shuze Chen · Aug 11, 2026 · Mathlib c5ea003 (Lean v4.30.0)

convexanalysisconvexoptimizationlog-concavity

Fenchel's inequality (also called the Fenchel–Young inequality).

For f:Rn→Rf : \mathbb{R}^n \to \mathbb{R}f:Rn→R let f∗(y)=sup⁡x(⟨x,y⟩−f(x))f^{*}(y) = \sup_x(\langle x,y\rangle - f(x))f∗(y)=supx​(⟨x,y⟩−f(x)) be its Fenchel conjugate, with values in (−∞,+∞](-\infty,+\infty](−∞,+∞]. Then for all x,y∈Rnx, y \in \mathbb{R}^nx,y∈Rn,

⟨x,y⟩  ≤  f(x)+f∗(y).\langle x, y\rangle \;\le\; f(x) + f^{*}(y).⟨x,y⟩≤f(x)+f∗(y).

The inequality holds for every function fff — no convexity, continuity or measurability is needed — since it is nothing but the definition of the supremum applied at the point xxx. Equality holds exactly when yyy is a subgradient of fff at xxx, which is how the inequality is used to characterize the solutions of dual pairs of problems.

Specialized to f(x)=∣x∣p/pf(x) = |x|^{p}/pf(x)=∣x∣p/p on R\mathbb{R}R, whose conjugate is ∣y∣q/q|y|^{q}/q∣y∣q/q with 1/p+1/q=11/p + 1/q = 11/p+1/q=1, it becomes Young's inequality xy≤xp/p+yq/qxy \le x^{p}/p + y^{q}/qxy≤xp/p+yq/q, and hence — after integration — Hölder's inequality.

Formalization Note The inequality is stated in EReal, adding the real value f(x)f(x)f(x) (coerced) to the EReal-valued conjugate, so the case f∗(y)=+∞f^{*}(y) = +\inftyf∗(y)=+∞ needs no separate treatment. Source: B&V §3.3.1, p. 94.

Preamble
import Mathlib
import Definitions.Def_fenchelConjugate

open scoped RealInnerProductSpace ENNReal
open MeasureTheory
Formal statement
theorem ConvexOptimization.fenchel_young {n : ℕ} (f : EuclideanSpace ℝ (Fin n) → ℝ)
    (x y : EuclideanSpace ℝ (Fin n)) :
    ((⟪x, y⟫ : ℝ) : EReal) ≤ (f x : EReal) + fenchelConjugate f y := by
  sorry
Source
Boyd & Vandenberghe 2004, Convex Optimization, Cambridge University Press (seventh printing with corrections, 2009), https://web.stanford.edu/~boyd/cvxbook/, pp. 94, §3.3.2 (Fenchel's inequality; stated unnumbered in the text)
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What the Lean code literally says, in plain math · claude-fable-5

Theorem statement. For every nnn, every total function f:Rn→Rf : \mathbb{R}^n \to \mathbb{R}f:Rn→R — no convexity, continuity, or any other hypothesis — and every pair of points x,y∈Rnx, y \in \mathbb{R}^nx,y∈Rn, the inequality ⟨x,y⟩≤f(x)+f∗(y)\langle x, y\rangle \le f(x) + f^*(y)⟨x,y⟩≤f(x)+f∗(y) holds in the extended reals, where f∗(y)f^*(y)f∗(y) is the file's fenchelConjugate: sup⁡x′∈Rn(⟨x′,y⟩−f(x′))\sup_{x' \in \mathbb{R}^n}\big(\langle x', y\rangle - f(x')\big)supx′∈Rn​(⟨x′,y⟩−f(x′)) taken in [−∞,+∞][-\infty,+\infty][−∞,+∞], a value in R∪{+∞}\mathbb{R} \cup \{+\infty\}R∪{+∞}. The left side is the finite real ⟨x,y⟩\langle x, y\rangle⟨x,y⟩ coerced into the extended reals; the right side is extended-real addition of the finite f(x)f(x)f(x) and f∗(y)f^*(y)f∗(y), which equals +∞+\infty+∞ whenever f∗(y)=+∞f^*(y) = +\inftyf∗(y)=+∞ (making the inequality automatic in that case). The particular choice x′=xx' = xx′=x inside the supremum already gives the term ⟨x,y⟩−f(x)\langle x, y\rangle - f(x)⟨x,y⟩−f(x), so the statement's content is exactly this one-term lower bound on the supremum.

Human review
  • Endorsed by Community (Bot) · Aug 11, 2026

  • Endorsed by Shuze Chen · Aug 11, 2026

    Confirmed by the mission captain (proposal self-audit).

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