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Validated Hessian enclosures and a small gradient residual certify a nearby index-one saddle

Statement

Let $d \geq 2$ , let $V \in C^{4} (U, R)$ on an open $U \subseteq R^{d}$ , and let $x \in U$ . Put $H = \nabla^{2} V (x)$ and $ϵ_{g} = ∥ \nabla V (x) ∥_{2}$ . Suppose a validated ordered eigenvalue enclosure certifies index one with gap $γ > 0$ in the sense of Validated ordered eigenvalue enclosures for an index-one Hessian. If $ϵ_{g} = 0$ , then $x$ is an exact critical point with Hessian index one.

If $ϵ_{g} > 0$ , put $ρ = 2 ϵ_{g} / γ$ . Assume $B_{2} (x, 2 ρ) \subset U$ , the Hessian is $L$ -Lipschitz in the induced Euclidean operator norm there, and $L ρ / γ \leq 1 / 2$ . Then there is exactly one critical point $x_{*}$ in ${\overline{B}}_{2} (x, ρ)$ , it satisfies $∥ x_{*} - x ∥_{2} \leq ρ$ , and its Hessian has exactly one negative eigenvalue and no zero eigenvalues. The reported residual $ϵ_{g}$ , gap $γ$ , conditioning bound $1 / γ$ , Hessian Lipschitz constant $L$ , and radius condition are part of the certificate; a residual or a Ritz value alone is not a certificate.

Facts & Assumptions

Given: The potential, point, gradient residual, interval evidence, gap, and local Hessian-Lipschitz hypotheses in the Statement.

[F1]

Validated ordered eigenvalue enclosures for an index-one Hessian: the validated intervals certify $λ_{1} (H) \leq - γ$ and $λ_{j} (H) \geq γ$ for $j \geq 2$ .

[F2]

Since $V \in C^{4}$ , it is $C^{2}$ , and The Hessian of a $C^{2}$ scalar field is symmetric gives $H^{T} = H$ . In the standard Euclidean inner product this is a self-adjoint endomorphism, so Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis gives an orthonormal eigenbasis.

[F3]

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 define $H$ and the gradient. Since $V \in C^{4}$ , the gradient's coordinate partials exist and are continuous; If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative at each point makes the gradient totally differentiable. Its derivative matrix is continuous and equals the gradient Jacobian by A total derivative computes every directional derivative, and its matrix is the Jacobian, and The Hessian of a $C^{2}$ scalar field is symmetric identifies it with the Hessian. Thus $\nabla V$ is $C^{1}$ , and a critical point is its zero.

[F4]

A small residual gives a nearby root under a uniform inverse and Lipschitz bound: if $D F (x)$ is invertible, has inverse norm at most $β$ , and is $L$ -Lipschitz on a sufficiently large ball, then $∥ F (x) ∥$ small enough gives a unique nearby zero within $2 β ∥ F (x) ∥$ .

[F5]

Courant-Fischer min-max principle for self-adjoint endomorphisms on finite-dimensional real inner product spaces: the largest eigenvalue is the maximum Rayleigh quotient, and the second largest is at most the maximum Rayleigh quotient on the orthogonal complement of a leading eigenvector.

[F6]

The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector: the maximum defining the finite-dimensional Euclidean operator norm exists, and its definition bounds $∥ B w ∥_{2}$ by $∥ B ∥_{o p} ∥ w ∥_{2}$ .

[F7]

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

Given: The data in the Statement.

Proof

1.1F1F2F6givenalgebra

By [F1], the eigenvalues of $H$ are all at least distance $γ$ from zero. Using the orthonormal eigenbasis in [F2], $H$ is invertible and each unit vector is expanded by $H^{- 1}$ by at most $1 / γ$ ; [F6] therefore gives $∥ H^{- 1} ∥_{o p} \leq 1 / γ$ .

2.1F1F3F4givenstep 1.1algebra

If $ϵ_{g} = 0$ , [F3] says $x$ is critical, and [F1] says its Hessian has exactly one negative eigenvalue. Suppose henceforth $ϵ_{g} > 0$ and use the inverse bound $β = 1 / γ$ from step 1.1. The radius in [F4] is then exactly $2 β ϵ_{g} = ρ$ , and its Lipschitz condition is $β L ρ = L ρ / γ \leq 1 / 2$ . Since $B_{2} (x, 2 ρ) \subset U$ , every hypothesis of [F4] holds on the stated ball, so there is exactly one zero $x_{*}$ of $\nabla V$ in ${\overline{B}}_{2} (x, ρ)$ and $∥ x_{*} - x ∥_{2} \leq ρ$ .

3.1F1F2step 2.1givenalgebra

Set $A = - H (x)$ and $E = - (H (x_{*}) - H (x))$ . Then $A$ is symmetric with largest eigenvalue at least $γ$ and second-largest at most $- γ$ . The Lipschitz hypothesis is in the induced Euclidean operator norm, so $∥ E ∥_{o p} \leq L ρ \leq γ / 2$ by step 2.1.

4.1F1F2F5F6F7step 2.1step 3.1algebra∎

Let $q_{1}$ be a unit leading eigenvector of $A$ . The maximum Rayleigh formula in [F5] yields $λ_{1} (A + E) \geq q_{1}^{T} (A + E) q_{1} \geq λ_{1} (A) - ∥ E ∥_{o p} \geq γ / 2 > 0$ by [F6, F7]. The second-eigenvalue min-max formula in [F5], applied to $q_{1}^{⊥}$ , yields $λ_{2} (A + E) \leq λ_{2} (A) + ∥ E ∥_{o p} \leq - γ / 2 < 0$ . Every later eigenvalue is at most the second, so $A + E = - H (x_{*})$ has exactly one positive eigenvalue and no zero eigenvalue. Thus $H (x_{*})$ has exactly one negative eigenvalue and no zero eigenvalues. Together with step 2.1 this proves the stated local certificate; if $ϵ_{g} = 0$ , step 2.1 already proved it.

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Sources

Graeme Henkelman, Blas P. Uberuaga, and Hannes Jónsson, A climbing image nudged elastic band method for finding saddle points and minimum energy paths (standard reference, not scraped)
Xuanyu Liu, Huajie Chen, and Christoph Ortner, Convergence of the Discrete Minimum Energy Path (standard reference, not scraped)