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The actual-to-predicted ratio approaches one at small radii away from stationarity

Statement

Let m be the model of Quadratic trust-region models and the actual-to-predicted reduction ratio, with g≠0, symmetric B, and radius Δ>0. Suppose ∥B∥2≤MB for a declared MB≥0, and f is C1 on an open set containing the segment from x to x+p. Suppose the gradient of f is L-Lipschitz on that segment, meaning ∥∇f(u)−∇f(v)∥2≤L∥u−v∥2 there. Let 0<θ≤1, let p be feasible, and suppose m(0)−m(p)≥θ [m(0)−m(pC)] for the Cauchy step pC of The Cauchy step for a quadratic trust-region model. Interpret ∥g∥2/(2MB)=+∞ if MB=0, and assume Δ≤∥g∥2/(2MB). Then the predicted reduction is positive and ∣ρ(p)−1∣≤2(L+MB)Δθ∥g∥2. Consequently, for 0<η2<1, if L+MB>0 and Δ≤θ(1−η2)∥g∥22(L+MB), then ρ(p)≥η2. If L+MB=0, then ρ(p)=1.

Facts & Assumptions

Given: The model, the smoothness and Hessian-bound assumptions in the Statement, and a feasible step satisfying the stated fraction of Cauchy decrease.

[F1]

The Cauchy step for a quadratic trust-region model: if c:=gTBg≤0, then pC=−(Δ/∥g∥2)g; if c>0, then pC=−min⁡{∥g∥22/c,Δ/∥g∥2}g.

[F2]

The Cauchy step gives a quantitative model decrease: m(0)−m(pC)≥12∥g∥2min⁡{Δ,∥g∥2/MB}, with the zero-matrix convention from that lemma.

[F5]

The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a): if a scalar function is continuous on a closed interval and differentiable in its interior, its endpoint difference equals its derivative at an interior point times the interval length.

[A1]

∥B∥2≤MB.

[A2]

The gradient Lipschitz inequality in the Statement holds along the segment x+tp, 0≤t≤1.

[A3]

m(0)−m(p)≥θ[m(0)−m(pC)] with 0<θ≤1.

Proof

Given: The hypotheses in the Statement.

1.1givenA1A3F1F2F3F4algebra

Put a:=∥g∥2>0 and c:=gTBg. By [F3] and [F4], c≤∣c∣≤a∥Bg∥2≤MBa2. If c≤0, [F1] gives ∥pC∥2=Δ. If c>0, then MB>0 and a2/c≥1/MB; since Δ/a≤1/(2MB), [F1] again gives ∥pC∥2=Δ. Put M:=∥B∥2. If M>0, [A1] gives a/M≥a/MB; if M=0, use the zero-matrix convention in [F2]. Hence [F2] and [A3] imply m(0)−m(p)≥θ[m(0)−m(pC)]≥12θaΔ>0.

2.1givenA1A2F4F5step 1.1algebra∎

Define ϕ(t):=f(x+tp) and h(t):=ϕ(t)−ϕ(0)−tϕ′(0). For 0<t≤1, the gradient Lipschitz bound and [F4] give ∣h′(t)∣=∣⟨∇f(x+tp)−∇f(x),p⟩∣≤Lt∥p∥22. By [F5] on h over [0,1], ∣h(1)∣≤L∥p∥22. From the model formula and [A1], ∣ared⁡(p)−pred⁡(p)∣≤∣h(1)∣+12∣pTBp∣≤(L+MB)∥p∥22. Since ∥p∥2≤Δ, step 1.1 yields ∣ρ(p)−1∣≤(L+MB)Δ212θaΔ=2(L+MB)Δθa. If L+MB=0, the numerator vanishes and ρ(p)=1; otherwise the stated threshold gives ∣ρ(p)−1∣≤1−η2, hence ρ(p)≥η2.

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