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An implicit sensitivity evaluated at an approximate state has a local residual bound

Statement

Let R(z,θ) be C2 near an exact solution R(z∗,θ0)=0, with A(z)=Rz(z,θ0) invertible and B(z)=Rθ(z,θ0). The implicit sensitivity at a state z is S(z)=−A(z)−1B(z). Suppose the local inverse Ψ of z↦R(z,θ0), supplied near z∗ by the implicit function theorem, is defined on a convex residual neighborhood containing 0 and r=R(z~,θ0), with ∥DΨ(y)∥≤κ along their segment. Suppose also that the segment from z∗ to z~=Ψ(r) lies in a convex neighborhood where ∥A(z)−1∥≤M, ∥B(z)∥≤B0, and A,B are Lipschitz with constants LA,LB. Then ∥S(z~)−S(z∗)∥≤κ(M2LAB0+MLB)∥r∥. This bounds the implicit sensitivity formula evaluated at an approximate state. It is not by itself a bound for the derivative of a stopped iteration.

Facts & Assumptions

Given: The exact root, local inverse chart, residual, inverse-norm and Lipschitz bounds in the statement.

[F1]

The Euclidean implicit function theorem with derivative formula: if the state Jacobian of a C1 residual is invertible at a zero, there is a local C1 solution branch and its derivative formula; applying the theorem to R(z,θ0)−r=0 gives the local inverse Ψ(r).

[F2]

On a convex open set, a uniform bound ∥Df(z)v∥2≤M∥v∥2 implies ∥f(y)−f(x)∥2≤M∥y−x∥2: on a convex open set, a uniform derivative bound ∥Df(z)v∥≤M∥v∥ implies ∥f(y)−f(x)∥≤M∥y−x∥.

[F3]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): the derivative of a composition is the composition of the derivatives at the corresponding points.

[A1]

The residual segment lies inside the inverse chart and the state segment lies inside the neighborhood where all stated bounds hold.

Proof

technique · first control state error by residual, then perturb the inverse-Jacobian formula
1.1F1F2F3A1given

For 0<t<1, put γ(t)=Ψ(tr). By [F1] and [F3], γ is differentiable and ∥γ′(t)∥=∥DΨ(tr)r∥≤κ∥r∥. For 0<ε<1/2, apply [F2] to γ on the convex open interval (0,1) at ε and 1−ε to obtain ∥Ψ((1−ε)r)−Ψ(εr)∥≤κ(1−2ε)∥r∥. Letting ε↓0 and using continuity of Ψ at 0 and r gives ∥z~−z∗∥≤κ∥r∥.

1.2givenalgebra

Put A=A(z~), A∗=A(z∗), B=B(z~), and B∗=B(z∗). The identities S(z~)−S(z∗)=(A∗−1−A−1)B+A∗−1(B∗−B) and A∗−1−A−1=A−1(A−A∗)A∗−1 follow by expansion and multiplication.

2.1A1step 1.2givenalgebra

The inverse bounds, B bound, and Lipschitz bounds give ∥A∗−1−A−1∥≤M2LA∥z~−z∗∥ and ∥B∗−B∥≤LB∥z~−z∗∥. The triangle and product inequalities then yield ∥S(z~)−S(z∗)∥≤(M2LAB0+MLB)∥z~−z∗∥.

3.1F1step 1.1step 2.1algebra∎

Substitute step 1.1 into step 2.1 to obtain the asserted residual bound. If r=0, local uniqueness gives z~=z∗ and the same inequality reads 0≤0.

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