advanced33 min studyLesson 277 in Mathematics

Fenchel–Rockafellar duality

Derive a dual lower bound, verify a relative-interior qualification, and distinguish dual attainment from primal existence.

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What you’ll understand

  • Derive the Fenchel dual of f(x)+g(Ax) with correct adjoint signs.
  • Check a sufficient finite-dimensional relative-interior qualification.
  • Use two Fenchel gaps to certify a primal–dual solution.

A dual problem is useful because every feasible dual point bounds the original optimum from below. Equality of the best bound with the primal infimum is a stronger conclusion, and existence of points attaining those values is stronger again. Fenchel–Rockafellar duality packages these distinctions for objectives made from two convex pieces linked by a linear map. The central calculation is short, but the qualification is part of the theorem rather than optional fine print. We will derive weak duality directly and state precisely when a finite-dimensional strong-duality conclusion is justified.

Split the objective with a linear measurement

Let f on Rn\mathbb R^n and g on Rm\mathbb R^m be proper closed convex functions, and let A be linear from the first space to the second. The primal value is the infimum of f(x)+g(Ax). Applying Fenchel–Young to f with slope −Aᵀy and to g with slope y cancels the two inner products. The resulting bound is −f∗(−A⊤y)−g∗(y)-f^*(-A^\top y)-g^*(y). Maximizing it defines the dual. A dual point is useful only when both conjugate terms are finite; otherwise its bound may simply be negative infinity.

The signs can also be derived by introducing z=Ax and a multiplier term ⟨y,Ax−z⟩\langle y,Ax-z\rangle. Infimizing over x produces −f*(−Aᵀy), and infimizing over z produces −g*(y). Writing this split carefully is safer than memorizing a sign pattern. The dual objective is concave because it is the negative sum of convex conjugates composed with linear maps. Weak duality holds without the later qualification, so every finite dual value remains a valid certificate even when the best attainable certificate leaves a gap.

p∗=inf⁡x{f(x)+g(Ax)},d∗=sup⁡y{−f∗(−A⊤y)−g∗(y)}≤p∗p^*=\inf_x\{f(x)+g(Ax)\},\qquad d^*=\sup_y\{-f^*(-A^\top y)-g^*(y)\}\le p^*
The transpose is the Euclidean adjoint. The two Fenchel inner products cancel, giving a lower bound for every dual y.↔ Scroll wide equations horizontally.

State a sufficient qualification, then its conclusion

Use the following finite-dimensional sufficient condition: there exists x₀ in the relative interior of dom f such that Ax₀ lies in the relative interior of dom g. The relative interior is taken in each domain’s own affine hull. Assume also that the primal infimum is a finite real number. Under these hypotheses, Fenchel–Rockafellar duality gives p*=d* and the dual supremum is attained. This is a theorem statement, not a derivation from weak duality. Its proof uses separation and regularity of the convex perturbation value function.

The conclusion does not assert that the primal infimum is attained. For f(x)=exp(x), g identically zero and A the identity, the domains are all of R and the qualification holds. The primal value zero is approached as x tends to negative infinity but has no minimizing point. The dual is attained at y=0 and has value zero. A separate compactness or coercivity argument is needed for a primal minimizer. Nor is the sufficient relative-interior condition necessary: its failure means this particular theorem cannot be invoked, not that a positive duality gap has been proved.

∃x0∈ri⁡(dom⁡f) with Ax0∈ri⁡(dom⁡g),p∗∈R ⟹ p∗=d∗ and the dual maximum exists\exists x_0\in\operatorname{ri}(\operatorname{dom}f)\ \text{with }Ax_0\in\operatorname{ri}(\operatorname{dom}g),\quad p^*\in\mathbb R\ \Longrightarrow\ p^*=d^*\ \text{and the dual maximum exists}
This is a sufficient finite-dimensional qualification for proper closed convex f and g. It includes a finite-value assumption and does not promise primal attainment.↔ Scroll wide equations horizontally.

Read optimality as two vanishing gaps

For finite primal and dual values at x and y, their difference is the sum of two nonnegative Fenchel gaps. It vanishes exactly when −Aᵀy belongs to ∂f(x) and y belongs to ∂g(Ax). These conditions certify both sides optimal directly, even if one did not first establish a global qualification. They also reveal the meaning of the dual variable: y is a supporting slope of the measurement-space cost, and its pullback balances a supporting slope of the original-space cost. Constraints fit naturally because indicator subgradients are normal cones.

An approximate computational certificate should report feasible finite x and y and the numerical value of their gap. A small gradient residual alone is different: it can be tiny at a point with substantial objective error, especially without strong convexity or an error bound. Checking domains is essential when g is an indicator or a conjugate imposes a norm constraint. An infeasible dual vector does not provide the claimed finite lower bound merely because substituting it into a simplified polynomial happens to return an appealing number.

Certify a soft-thresholded scalar optimum

Solve min_x (x−2)²/2+|x| and certify it using its Fenchel dual with A=1.

  1. For f(x)=(x−2)²/2, completing the square gives f*(u)=u²/2+2u. For g(x)=|x|, g*=δ_[−1,1]. Both primal domains are R, so the relative-interior qualification holds.

  2. The dual is maximizing −y2/2+2y-y^2/2+2y over |y|≤1. Its derivative 2−y is positive throughout this interval, so y=1 maximizes it with value 3/2.

  3. At x=1 the primal cost is 1/2+1=3/2. Its slopes satisfy −y=x−2=−1 and y=1∈∂|x|. The matching feasible values prove optimality on both sides.

The result

The primal minimizer x=1 and dual maximizer y=1 have zero gap. The closed conjugate domain is essential: the unconstrained dual quadratic would choose y=2 and yield an invalid bound.

Explore the model

Interactive lab

Interactive lab · function

Fenchel equality as tangent contact

Pause and Reset, then set a=0.5,x=1. Predict slope y=1 and primal cost 0.5. For this self-conjugate quadratic, f*(y)=0.5 and xy=1, so the gap is zero. For a fresh comparison Reset and set a=0.5,x=2; predict y=2 and equality 2+2=4. The instrument checks one Fenchel relation through a tangent; it neither solves the two-term dual example nor verifies relative-interior qualifications.

Fenchel equality as tangent contactThis 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 forDual optimality joins two support equalities whose linear terms cancel.

Check your understanding

Under the stated qualification and finite primal infimum, what does the theorem guarantee?

Try it yourself

For f(x)=x²/2, g(z)=(z−3)²/2, and Ax=2x, find a primal and dual pair with matching values.

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

Give me a hint

The conjugates are f*(u)=u²/2 and g*(y)=y²/2+3y. Keep the factor two when applying Aᵀ.

Compare with the solution

The primal derivative is x+2(2x−3)=5x−6, giving x=6/5. Its value is 18/25+9/50=9/10. The dual objective is −(−2y)²/2−y²/2−3y=−5y²/2−3y, maximized at y=−3/5 with value 9/10. The support conditions hold: −2y=6/5=f'(x), while y=2x−3=−3/5=g'(2x). Both domains are all real, so feasibility and qualification are immediate.

Take it with you

  • Fenchel duality cancels two support inequalities through an adjoint map.
  • Relative-interior qualification plus finite primal value guarantees equality and dual attainment here.
  • A feasible zero-gap pair certifies both optima; it does not require a uniqueness claim.

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