The Legendre transformation #
Let f be a convex function, differentiable on an open set C. Its Legendre conjugate is the
pair (D, g) with D = ∇f(C) the image of the gradient mapping and g y = ⟨x, y⟩ - f x for any
x with ∇f x = y. The value of g does not depend on which such x is chosen, and g is
nothing but the restriction of the Fenchel conjugate f* to D; in particular D ⊆ dom f*.
Main results #
legendreDom— the setD, the image of the gradient mapping, withlegendreDom_subset_dom_conjforD ⊆ dom f*.conj_eq_of_hasGradientAt—f*(∇f x) = ⟨x, ∇f x⟩ - f x: both the formula forgand, at a stroke, its well-definedness (Theorem 26.4 in [^1]).
Implementation notes #
There is deliberately no legendreConj definition: y ↦ ⟨(∇f)⁻¹ y, y⟩ - f ((∇f)⁻¹ y) would need
a choice function and would then have to be proved equal to conj B f on D anyway, and
conj_eq_of_hasGradientAt is that equality without the detour.
References #
[^1]: R. T. Rockafellar, Convex Analysis, Princeton University Press, 1970, §26.
A function is finite wherever it has a gradient.
Where f is differentiable, Fenchel's inequality holds with equality at (x, ∇f x).
The Legendre conjugate at ∇f x is f* at ∇f x, given by the formula
⟨x, ∇f x⟩ - f x.
Well-definedness: ⟨x, y⟩ - f x is the same for every x with ∇f x = y, because it is
f* y. So ∇f need not be one-to-one for g to be single-valued.
The domain D of the Legendre conjugate, the image of the gradient mapping. Every gradient is
attained at an interior point of dom f (HasGradientAt.mem_interior_dom), so no interiority side
condition is needed here.
Equations
- Tdaf.ConvexAnalysis.legendreDom f = {y : StrongDual ℝ E | ∃ (x : E), Tdaf.ConvexAnalysis.HasGradientAt f y x}
Instances For
Every gradient of f is a point where f* is finite: D ⊆ dom f*.