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A redundant active margin with zero KKT multiplier

Example

Consider the two positive examples x1=(1,0) and x2=(1,1) in R2 with bias-free score w↦⟨w,x⟩. Their hard-margin constraints are w1≥1 and w1+w2≥1. The unique minimum-norm feasible vector is w∗=(1,0); both constraints are active, but the second active constraint has a KKT multiplier equal to zero.

Facts & Assumptions

Given: The two-point positive sample, its bias-free linear score, and the hard-margin and Clarke-KKT definitions.

[F1]

Bias-free linear margins and the hard-margin SVM: the signed margin is mi(w)=⟨w,xi⟩ for a positive label, and the hard-margin problem minimizes 12∥w∥22 subject to every margin being at least one.

[F2]

A strictly separable finite sample has a unique hard-margin vector: a finite strictly separable sample has a unique feasible minimum-norm hard-margin vector.

[F3]

KKT conditions and MFCQ for margin constraints: KKT stationarity represents w∗ as a nonnegative combination of active margin subgradients, with zero multipliers allowed for active constraints.

[F4]

Clarke subdifferential and gradient-flow trajectory: for a differentiable linear margin, its Clarke subdifferential is the singleton containing its ordinary gradient.

Proof

technique · verify strict separability, solve the two scalar inequalities, and exhibit a valid multiplier vector with a zero coordinate
1.1F1F2givenalgebra

The vector (1,0) gives both margins equal to one, so the sample is strictly separable. Feasibility forces w1≥1, hence ∥w∥22=w12+w22≥1, with equality only for w=(1,0). Thus w∗=(1,0) is the unique hard-margin minimizer, consistent with [F2].

2.1F1F3F4step 1.1algebra∎

Both margins at w∗ equal one. Their ordinary gradients, and hence their Clarke subgradients by [F4], are (1,0) and (1,1). The multipliers λ1=1 and λ2=0 are nonnegative and satisfy λ1(1−1)=λ2(1−1)=0 and λ1(1,0)+λ2(1,1)=w∗. These are the KKT conditions in [F3], while the second constraint is active with zero multiplier.

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