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The soft-margin SVM primal attains its minimum, including the unpenalized-bias boundary

Statement

Let n≥1, let C>0, and let (xi,yi)i=1n be a finite labeled sample with yi∈{−1,+1} in a real Hilbert feature space H, with feature map ϕ. The soft-margin SVM primal has a finite attained minimum. At least one minimizing tuple has its weight vector in M:=span⁡{ϕ(xi):1≤i≤n}. This conclusion includes samples with only one label and does not assert uniqueness of the bias or slack variables.

Facts & Assumptions

Given: The sample, a real Hilbert feature space, its feature map, and C>0.

[A1]

The soft-margin primal minimizes 12∥w∥2+C∑iξi over w,b,ξ subject to yi(⟨w,ϕ(xi)⟩+b)≥1−ξi and ξi≥0 (Hard- and soft-margin support-vector machines with an unpenalized bias).

[F1]

Projection onto a finite-dimensional subspace preserves inner products with that subspace and does not increase norm (Projection onto a finite-dimensional subspace of an arbitrary real inner-product space).

[F2]

∣⟨u,v⟩∣≤∥u∥∥v∥ in a real inner-product space (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

[F3]

Every finite-dimensional real inner-product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[F4]

A nonempty closed bounded subset of Rd, d≥1, is compact, and every continuous real-valued function on it attains a minimum (For a nonempty subset of Rn with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

Proof

1.1givenA1constructalgebra

The tuple w=0, b=0, and ξi=1 is feasible and has objective Cn. Every feasible objective is nonnegative, so the primal infimum lies in [0,Cn].

1.2givenF1A1

Let M=span⁡{ϕ(xi):1≤i≤n}. For any feasible tuple, project w onto M using [F1]. The residual is orthogonal to every ϕ(xi), so all sample scores and constraints are unchanged, while the norm does not increase. Thus minimizing over M×R×Rn has the same infimum as minimizing over the full feature space.

2.1step 1.1A1construct

If all labels are +1, then w=0, b=1, and ξ=0 is feasible with objective 0, which is globally minimal by step 1.1. If all labels are −1, then w=0, b=−1, and ξ=0 gives the same conclusion.

3.1step 1.1step 1.2A1F2F3F4chooseconstructalgebra∎

Suppose both labels occur. Set R=max⁡i∥ϕ(xi)∥ and let S be the subset of M×R×Rn consisting of feasible tuples with objective at most Cn. By step 1.1, S is nonempty. It is closed because the constraints are continuous affine inequalities and the objective is continuous; continuity of each score follows from [F2]. Every tuple in S satisfies ∥w∥≤2Cn and 0≤ξi≤n. Choose indices i+ and i− with labels +1 and −1. Their constraints and [F2] give b≥1−n−2CnR and b≤n−1+2CnR, so b is bounded. Equip the product space with the inner product ⟨(w,b,ξ),(w′,b′,ξ′)⟩=⟨w,w′⟩+bb′+ξTξ′. It has finite dimension d=dim⁡M+1+n≥1, so [F3] gives an orthonormal basis identifying it with Rd. Hence S is a nonempty closed bounded subset of that Euclidean space. By [F4], the continuous objective attains a minimum on S. Any feasible tuple outside S has objective greater than Cn, whereas S contains a feasible tuple of objective Cn; therefore this minimum is also the global primal minimum. By step 1.2 it is a full-space minimizer with w∈M.

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