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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Fenchel–Young inequality and its equality case

Statement

Let f:Rn→(−∞,+∞] be proper and convex. For every x,s∈Rn, f(x)+f∗(s)≥⟨s,x⟩, with extended-real interpretation. Equality holds exactly when x∈dom⁡f, f∗(s)<+∞, and s∈∂f(x); equivalently, for proper convex f, equality itself forces the two finiteness conditions and the subgradient membership.

Facts & Assumptions

Given: Proper convex f, x,s∈Rn.

[A1]

The function f is proper and convex.

[F1]

Fenchel conjugate of an extended-real function: f∗(s)=sup⁡z(⟨s,z⟩−f(z)).

[F2]

Subdifferentials of proper extended-real convex functions: for x∈dom⁡f, s∈∂f(x) iff f(z)≥f(x)+⟨s,z−x⟩ for every z∈Rn.

Proof

1.1A1F1given

If f(x)=+∞, the left side is +∞ because [F1] and properness ensure f∗(s)>−∞, so the inequality is true and equality is impossible. If f(x)<+∞, the term z=x in [F1] gives f∗(s)≥⟨s,x⟩−f(x), proving the inequality; if f∗(s)=+∞, equality is again impossible.

2.1A1F1F2step 1.1algebra∎

Assume now f(x) and f∗(s) are finite. Equality in step 1.1 holds iff the supremum in [F1] equals its z=x term, which is equivalent to ⟨s,z⟩−f(z)≤⟨s,x⟩−f(x)for every z∈dom⁡f. Rearranging gives precisely the inequalities in [F2]; for z∉dom⁡f they hold automatically. Thus equality is equivalent to s∈∂f(x), and the converse follows by reversing these inequalities.

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Sources