LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The Cauchy step gives a quantitative model decrease

Statement

In the quadratic model of Quadratic trust-region models and the actual-to-predicted reduction ratio, let $g \neq 0$ , let $B$ be symmetric, and let $Δ > 0$ . Write $M : = ∥ B ∥_{2}$ for the induced Euclidean matrix norm, with the convention $\min {Δ, ∥ g ∥_{2} / M} = Δ$ when $M = 0$ . The Cauchy step $p_{C}$ of The Cauchy step for a quadratic trust-region model minimizes the model on the negative-gradient segment ${- t g : 0 \leq t \leq Δ / ∥ g ∥_{2}}$ and satisfies $m (0) - m (p_{C}) \geq \frac{1}{2} ∥ g ∥_{2} \min {Δ, ∥ g ∥_{2} / M} .$

Facts & Assumptions

Given: A trust-region quadratic model with $g \neq 0$ , symmetric $B$ , radius $Δ > 0$ , and its Cauchy step.

[F1]

The Cauchy step for a quadratic trust-region model: if $c : = g^{T} B g \leq 0$ , then $p_{C} = - (Δ / ∥ g ∥_{2}) g$ ; if $c > 0$ , then $p_{C} = - τ g$ with $τ = \min {∥ g ∥_{2}^{2} / c, Δ / ∥ g ∥_{2}}$ .

[F2]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: $∥ B g ∥_{2} \leq ∥ B ∥_{2} ∥ g ∥_{2}$ .

[F3]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ .

[F4]

Quadratic trust-region models and the actual-to-predicted reduction ratio: $m (p) = f (x) + ⟨ g, p ⟩ + \frac{1}{2} p^{T} B p$ .

Proof

Given: The hypotheses in the Statement.

1.1givenF1F4algebra

Put $a : = ∥ g ∥_{2} > 0$ , $c : = g^{T} B g$ , and $ψ (t) : = m (- t g) - m (0) = - t a^{2} + \frac{1}{2} t^{2} c$ on $0 \leq t \leq Δ / a$ . If $c \leq 0$ , then for $t_{2} > t_{1} \geq 0$ , $ψ (t_{2}) - ψ (t_{1}) = - (t_{2} - t_{1}) a^{2} + \frac{1}{2} c (t_{2}^{2} - t_{1}^{2}) < 0$ ; hence its minimum is at $t = Δ / a$ , which is the first branch in [F1]. The predicted reduction is $P \geq a Δ \geq \frac{1}{2} a \min {Δ, a / M}$ . This case also covers $M = 0$ , since then $c = 0$ .

1.2givenF1F2F3F4algebra

Suppose $c > 0$ . Completing the square in $ψ$ shows its unconstrained minimizer is $t = a^{2} / c$ , so its minimizer on the segment is $τ = \min {a^{2} / c, Δ / a}$ , the second branch in [F1]. Put $s : = τ a$ ; then $s = \min {a^{3} / c, Δ}$ and $s \leq a^{3} / c$ . Thus $P = a s - \frac{1}{2} (c / a^{2}) s^{2} \geq \frac{1}{2} a s$ . By [F2] and [F3], $c = ⟨ g, B g ⟩ \leq a ∥ B g ∥_{2} \leq M a^{2}$ , whence $a^{3} / c \geq a / M$ and $s \geq \min {Δ, a / M}$ . Therefore $P \geq \frac{1}{2} a \min {Δ, a / M}$ .

2.1step 1.1step 1.2∎

The cases $c \leq 0$ and $c > 0$ exhaust all real $c$ . Both establish that the algorithm returns the segment minimizer and prove the stated decrease bound.

Depends on

Used by

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Sources

Jorge Nocedal and Stephen J. Wright, Numerical Optimization, second edition (standard reference, not scraped)
Nick Gould, Continuous Optimization Algorithms, Part 3.2: Trust-Region Methods (standard reference, not scraped)