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A lower singular-value bound and Lipschitz Jacobian control local parameter error

Statement

Let F:Θ→Rm be C1 on an open Θ⊆Rp, let θ0∈Θ, and let r0>0 be such that the closed Euclidean ball B‾(θ0,r0) lies in Θ. Suppose ∥DF(θ0)v∥2≥μ∥v∥2 for every v∈Rp, where μ>0, and suppose DF is L-Lipschitz in operator norm on this ball, where L≥0. Define r=r0 if L=0 and r=min⁡{r0,μ/(2L)} if L>0. If y=F(θ0)+e with ∥e∥2≤δ, and θ^∈B‾(θ0,r) satisfies ∥F(θ^)−y∥2≤ε, then ∥θ^−θ0∥2≤2(ε+δ)μ. The bound is conditional on the stated local ball and on the existence of such a fit; it does not assert that a fit exists.

Facts & Assumptions

Given: F,Θ,θ0,r0,μ,L,y,e,θ^,ε,δ with the hypotheses in the Statement.

[F1]

Finite nonlinear observation designs, local identifiability, and weighted information: the Jacobian is the total derivative of the observation map; a lower singular-value bound μ means ∥DF(θ0)v∥2≥μ∥v∥2 for every v.

[F2]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): for a differentiable path γ, D(F∘γ)(t)=DF(γ(t))Dγ(t).

[F3]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: ∥Av∥2≤∥A∥2∥v∥2 for a linear map A and vector v.

[F4]

Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction: L-Lipschitz means ∥DF(u)−DF(v)∥2≤L∥u−v∥2 on the stated ball.

[F5]

If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz: for m≥1, a path into Rm continuous on [0,1] and differentiable on (0,1) with derivative norm at most C≥0 is C-Lipschitz there.

Proof

technique · direct
1.1givenF1F2construct

Put h=θ^−θ0 and define R(t)=F(θ0+th)−F(θ0)−tDF(θ0)h for 0≤t≤1. The segment stays in the convex ball. By [F2], R′(t)=(DF(θ0+th)−DF(θ0))h.

2.1step 1.1F3F4F5algebra

If m=0, the path R is identically the zero vector, so ∥R(1)−R(0)∥2=0≤L∥h∥22. If m≥1, [F4] and [F3] give ∥R′(t)∥2≤Lt∥h∥22≤L∥h∥22 for 0<t<1; since R is continuous on [0,1] and differentiable on (0,1), [F5] gives ∥R(1)−R(0)∥2≤L∥h∥22. In either case, R(0)=0 yields the first-order remainder bound ∥F(θ0+h)−F(θ0)−DF(θ0)h∥2≤L∥h∥22.

3.1givenF1step 2.1F6algebra

If L>0, then ∥h∥2≤r≤μ/(2L), so [F1], step 2.1, and the reverse-triangle inequality [F6] yield ∥F(θ^)−F(θ0)∥2≥μ∥h∥2−L∥h∥22≥(μ/2)∥h∥2. If L=0, step 2.1 gives zero remainder, and the same lower bound follows from [F1].

4.1givenstep 3.1F6algebra∎

The fit condition and y=F(θ0)+e give ∥F(θ^)−F(θ0)∥2≤∥F(θ^)−y∥2+∥e∥2≤ε+δ. Combining with step 3.1 gives (μ/2)∥h∥2≤ε+δ, and division by μ/2>0 proves the asserted bound.

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