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A max-affine kink produces one null and one serious bundle step

Example

Run two trials of Finite-memory-free proximal bundle method with serious and null steps on $f (x) = ∣ x ∣ = \max {x, - x}$ over $C = R$ , with $t = 2$ , $m_{s} = 1 / 2$ , and initial center $x_{0} = 1$ . At $1$ choose subgradient $1$ ; at the first trial point $- 1$ choose subgradient $- 1$ ; at $0$ choose subgradient $0$ . The first trial is null and the second is serious, with exact trial points $- 1$ and $0$ respectively.

Facts & Assumptions

Given: The serious/null algorithm and subgradient definition Finite-memory-free proximal bundle method with serious and null steps, Subgradients and the subdifferential of a convex function.

[A1]

For $λ \in [0, 1]$ , $f ((1 - λ) x + λ y) = \max {(1 - λ) x + λ y, - (1 - λ) x - λ y} \leq (1 - λ) \max {x, - x} + λ \max {y, - y}$ , because each affine branch is bounded by the right side; hence $f$ is convex.

[F1]

For every $y$ , $∣ y ∣ \geq y$ , $∣ y ∣ \geq - y$ , and $∣ y ∣ \geq 0$ , so the chosen slopes $1, - 1, 0$ support $f$ at $1, - 1, 0$ respectively (Subgradients and the subdifferential of a convex function).

[F2]

The finite maximum-affine model plus its positive quadratic has a unique minimizer (The proximal bundle model subproblem has a unique minimizer).

Verification

1.1givenA1F1F2algebra

The initial cut is $m_{0} (y) = y$ . The model objective is $y + (y - 1)^{2} / 4 = (y + 1)^{2} / 4$ , so its unique minimizer is $z_{0} = - 1$ . Then $M_{0} (z_{0}) = - 1$ , $P_{0} = f (1) - M_{0} (- 1) = 2$ , and actual decrease is $f (1) - f (- 1) = 0 < m_{s} P_{0} = 1$ . Thus this is a null step and the cut at $- 1$ is $m_{1} (y) = - y$ .

2.1givenF1F2step 1.1algebra∎

The new model is $M_{1} (y) = \max {y, - y} = ∣ y ∣$ . For $y \geq 0$ , $∣ y ∣ + (y - 1)^{2} / 4 = (y + 1)^{2} / 4 \geq 1 / 4$ , with equality only at $y = 0$ . For $y \leq 0$ , its difference from $1 / 4$ is $y (y - 6) / 4 \geq 0$ because both factors are nonpositive, with equality only at $y = 0$ . Thus its unique minimizer is $z_{1} = 0$ , and $P_{1} = f (1) - M_{1} (0) = 1$ , while actual decrease is $1 \geq m_{s} P_{1} = 1 / 2$ . The second step is serious and sets the center to $0$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Mateo Díaz and Benjamin Grimmer, Optimal Convergence Rates for the Proximal Bundle Method, §§2–3 (standard reference, not scraped)
Dominikus Noll, Olivier Prot, and Aude Rondepierre, A proximity control algorithm to minimize nonsmooth and nonconvex functions, §§2–4 (standard reference, not scraped)