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Paired force evaluations give a second-order Hessian-vector estimate

Statement

Let V∈C4(U,R) on an open set U⊆Rd, let x∈U, and let v∈Rd satisfy ∥v∥2=1 and x±ϵv∈U for ϵ>0. Assume the segments x+[−ϵ,ϵ]v lie in U. Define

Dϵ(x,v):=∇V(x+ϵv)−∇V(x−ϵv)2ϵ.

If approximate gradient evaluations g^± satisfy ∥g^±−∇V(x±ϵv)∥2≤η, and M4 bounds the third derivatives of each gradient coordinate in the Taylor remainders on these segments, then

∥g^+−g^−2ϵ−∇2V(x)v∥2≤d M46ϵ2+ηϵ.

Thus the deterministic truncation error is O(ϵ2) and the force-evaluation error contributes O(η/ϵ). This is an error bound for the Hessian action on the supplied direction; it is not an operator-norm bound for a full Hessian approximation.

Facts & Assumptions

Given: The smooth potential, admissible line segment, unit direction, finite-difference step, and force discrepancy bound in the Statement.

[F1]

Ck maps and multi-index derivative notation in Euclidean space, The Hessian matrix and critical points of a scalar field, and The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: V∈C4 gives continuous coordinate partials of each gradient coordinate through order three. Applying If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative to those iterated partials through order two gives their total derivatives, whose Jacobian entries are continuous. A total derivative computes every directional derivative, and its matrix is the Jacobian identifies the matrix of D(∇V) with the Jacobian, and The Hessian of a C2 scalar field is symmetric makes its Jacobian entries agree with the Hessian convention Hij=∂i∂jV.

[F2]

By [F1], Ck maps and multi-index derivative notation in Euclidean space, and repeated application of The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), each restriction t↦∂jV(x+tv) is C3 on its open interval domain. The one-dimensional case of Multivariable Taylor formula with a Lagrange remainder along a line segment then gives its coordinatewise degree-two Taylor expansion with Lagrange remainder.

[A1]

For each coordinate j, the third derivative of t↦∂jV(x+tv) is bounded in absolute value by M4 for t∈[−ϵ,ϵ]; this is the remainder bound required for [F2].

Given: The hypotheses of the Statement.

Proof

1.1A1F1F2givenalgebra

Let I be the open interval component of {t∈R:x+tv∈U} containing [−ϵ,ϵ]; it exists because that line segment lies in the open set U. Each gj(t):=∂jV(x+tv) is C3 on the open, convex interval I. Apply [F2] to gj at 0 with displacements ±ϵ. The linear terms are ±ϵ(Hv)j by [F1], the quadratic terms agree, and the coordinate remainders satisfy ∣(r±)j∣≤M4ϵ3/6 by [A1]. Thus ∥r±∥2≤d M4ϵ3/6. Subtracting and dividing by 2ϵ cancels the constant and quadratic terms and gives ∥Dϵ(x,v)−∇2V(x)v∥2≤d M4ϵ2/6.

2.1givenstep 1.1algebra∎

Write e±=g^±−g(x±ϵv). Then ∥e+−e−∥2≤∥e+∥2+∥e−∥2≤2η, so their contribution after division by 2ϵ is at most η/ϵ. Adding this to step 1.1 proves the inequality. A bound on this single direction alone supplies no bound on an operator acting on all directions.

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