advanced31 min studyLesson 276 in Mathematics

Fenchel conjugacy and biconjugation

Encode the best affine lower bounds by slope and recover a closed proper convex objective from them.

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What you’ll understand

  • Compute conjugates using supremum definitions and domain tests.
  • Derive Fenchel–Young inequality and its subgradient equality condition.
  • State the finite-dimensional Fenchel–Moreau theorem with closedness and properness.

Instead of recording a convex function by its value at every location, record how much intercept must be sacrificed to support each possible slope. This is the conjugate function. The operation exchanges a location variable for a slope variable and turns constraints into directional penalties. It also supplies a precise inequality linking primal and dual quantities. For closed proper convex functions in finite dimensions, conjugating twice reconstructs the original function. Dropping closedness can change a boundary value; dropping convexity can fill in entire nonconvex regions beneath the original graph.

Find the cost of a proposed slope

For a proper extended-real function f on Rd\mathbb R^d, define f* at y as the supremum of ⟨y,x⟩−f(x)\langle y,x\rangle-f(x) over finite-domain x. This is the smallest intercept cost making ⟨y,x⟩−f∗(y)\langle y,x\rangle-f^*(y) a global affine lower bound. Some slopes have infinite cost and provide no finite lower bound. Since a conjugate is a supremum of continuous affine functions of y, it is convex and lower semicontinuous regardless of whether the original f is convex. A conjugate can fail to be proper for a badly behaved original function.

For f(x)=x²/2, completing the square makes the supremum y²/2, attained at x=y. For f(x)=|x|, slopes in [−1,1] have supremum zero, while a slope of magnitude greater than one makes the expression grow without bound on a suitable ray. Hence its conjugate is the indicator of [−1,1]. The conjugate of an indicator is the support function of its set. These examples show why an infinite conjugate value often encodes a geometric slope restriction rather than a numerical difficulty.

f∗(y)=sup⁡x∈dom⁡f{⟨y,x⟩−f(x)},(∣⋅∣)∗(y)=δ[−1,1](y)f^*(y)=\sup_{x\in\operatorname{dom}f}\{\langle y,x\rangle-f(x)\},\qquad (|\cdot|)^*(y)=\delta_{[-1,1]}(y)
For |y|≤1, yx−|x|≤0 with equality at x=0. For |y|>1, the supremum is infinite along the sign of y.↔ Scroll wide equations horizontally.

The gap is a support certificate

The supremum definition immediately gives Fenchel–Young inequality: f(x)+f*(y) is at least the inner product of x and y whenever the sum is meaningful. Equality holds exactly when y is a subgradient of f at x. To prove this equivalence, rewrite the subgradient inequality as an upper bound on ⟨y,z⟩−f(z)\langle y,z\rangle-f(z) and observe that z=x attains it. Conversely, equality of the supremum gives that same upper bound for every z. This converts an apparently abstract conjugate calculation into a checkable optimality relation.

For closed proper convex f, the equality condition is symmetric: y belongs to the subdifferential at x if and only if x belongs to the subdifferential of f* at y. Differentiability is unnecessary, so corners and flat pieces exchange naturally. A corner with an interval of slopes becomes a flat region in the conjugate. If both sides are differentiable and their gradients are mutually invertible on the relevant domains, this relation reduces to inverse gradients, but that stronger smooth picture must not be assumed for every convex function.

f(x)+f∗(y)≥⟨x,y⟩,f(x)+f∗(y)=⟨x,y⟩  ⟺  y∈∂f(x)f(x)+f^*(y)\ge\langle x,y\rangle,\qquad f(x)+f^*(y)=\langle x,y\rangle\iff y\in\partial f(x)
The equality statement concerns finite f(x) and f*(y). It identifies a supporting slope, not merely a stationary point of an arbitrary nonconvex function.↔ Scroll wide equations horizontally.

Why conjugating twice reconstructs a closed function

The biconjugate is the supremum of all affine lower bounds indexed by y. Therefore it never exceeds f. The finite-dimensional Fenchel–Moreau theorem states that a proper convex lower semicontinuous f equals its biconjugate everywhere, including outside its effective domain. The converse reconstruction uses separation of points below the closed convex epigraph; this is the geometric reason closedness and convexity appear. For a proper convex function that is not lower semicontinuous, biconjugation instead produces its lower semicontinuous closure. This theorem is stated here; the earlier separation argument explains its mechanism rather than serving as a complete proof of every epigraph boundary case.

As a diagnostic, take the indicator of the open interval (−1,1). Its conjugate is |y|, since the endpoint supremum is approached even though not attained. Conjugating again gives the indicator of [−1,1]. Thus the operation retains all affine lower information but cannot remember missing endpoints invisible to suprema. In applications, conjugacy often turns an awkward nonsmooth loss into a simple constraint. The benefit depends on tracking domains carefully: completing a square without checking whether the maximizing point is allowed can give the wrong conjugate formula.

A shifted quadratic exchanges slope and position

Compute the conjugate of f(x)=x²/2+2x and verify Fenchel equality at x=3.

  1. For a fixed y, maximize (y−2)x−x2/2(y-2)x-x^2/2. Completing the square gives −(x−(y−2))2/2+(y−2)2/2-(x-(y-2))^2/2+(y-2)^2/2.

  2. The unique maximizing point is x=y−2, so f∗(y)=(y−2)2/2f^*(y)=(y-2)^2/2 for every real y.

  3. At x=3, the supporting slope is f'(3)=5. Then f(3)=21/2 and f*(5)=9/2, whose sum is 15=3·5. The equality verifies the conjugate and the slope relation simultaneously.

The result

The conjugate is finite everywhere and its derivative y−2 recovers x. The linear term in f translates the conjugate’s argument; it does not add the same linear term to f*.

Explore the model

Interactive lab

Interactive lab · function

Match a slope to a quadratic conjugate

Pause and Reset, then set a=0.5,x=1.5. Predict primal value 1.125, slope y=1.5, and conjugate value y²/2=1.125. Verify the two costs sum to xy=2.25. For a fresh comparison Reset and set a=1,x=1.5; now slope y=3 and the conjugate y²/4=2.25. The graph shows the primal quadratic and tangent, not a separate conjugate graph.

Match a slope to a quadratic conjugateThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe touching tangent identifies the conjugate’s maximizing primal point and makes Fenchel’s gap zero.

Check your understanding

What is the biconjugate of the indicator of (−1,1) on the real line?

Try it yourself

Compute the conjugate of f(x)=3|x| and characterize all pairs (x,y) with zero Fenchel gap.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Test the two unbounded rays before differentiating anything. At x=0, use the full supporting-slope interval.

Compare with the solution

If |y|≤3, then yx−3|x|≤0 and x=0 attains zero. If y>3, let x tend to positive infinity; if y<−3, let x tend to negative infinity. Thus f*=δ_[−3,3]. Equality holds at x>0 precisely for y=3, at x<0 precisely for y=−3, and at x=0 for every y in [−3,3]. These are exactly the pairs with y∈∂f(x); no equality pair uses a slope outside the finite conjugate domain.

Take it with you

  • A conjugate records the intercept cost of each global supporting slope.
  • Fenchel equality is equivalent to a subgradient relation.
  • Biconjugation recovers proper closed convex functions and repairs missing convex boundary information.

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