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A singular shifted Hessian produces two hard-case minimizers

Example

Take $n = 2$ , $Δ = 1$ , $B = diag (- 2, 1)$ , and $g = (0, 1)$ . The gradient is orthogonal to the most negative eigenspace. The singular multiplier leaves a free component in that eigenspace, and the boundary condition produces two distinct global minimizers.

Facts & Assumptions

Given: The model definition Quadratic trust-region models and the actual-to-predicted reduction ratio and the necessary-and-sufficient global optimality result Global trust-region minimizers satisfy the shifted-Hessian conditions, including the hard case.

[F1]

The model quadratic part is $q (p) = g^{T} p + \frac{1}{2} p^{T} B p$ and its feasible set is $∥ p ∥_{2} \leq 1$ (Quadratic trust-region models and the actual-to-predicted reduction ratio).

[F2]

Feasible points satisfying the shifted-Hessian conditions are global minimizers (Global trust-region minimizers satisfy the shifted-Hessian conditions, including the hard case).

Proof

1.1givenF1algebra

The least eigenvalue is $μ = - 2$ with eigenspace $E_{μ} = span {(1, 0)}$ , and $g ⊥ E_{μ}$ . The equation $(B - μ I) p_{0} = - g$ with $p_{0} ⊥ E_{μ}$ gives $p_{0} = (0, - 1 / 3)$ , so $∥ p_{0} ∥_{2} = 1 / 3 < 1$ . The missing boundary component has squared norm $1 - 1 / 9 = 8 / 9$ .

2.1givenF1F2step 1.1algebra∎

Let $p_{\pm} = (\pm 2 \sqrt{2} / 3, - 1 / 3)$ . Each has norm $1$ , and $(B + 2 I) p_{\pm} = (0, - 1) = - g$ with $B + 2 I = diag (0, 3) ⪰ 0$ . Thus both satisfy [F2] with $λ = 2$ and are global minimizers. They are distinct, and each has $q (p_{\pm}) = - \frac{1}{3} + \frac{1}{2} (- 2 \cdot \frac{8}{9} + \frac{1}{9}) = - \frac{7}{6} .$ This verifies the singular hard case and its two boundary choices.

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