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Unsuccessful-poll accumulation points are stationary under shrinking mesh

Statement

Consider an infinite continuation of the polling and mesh-update rules of Finite-budget deterministic direct search with positive-spanning polling, using its admissible inputs and exact branch rules but with no finite poll-budget cap. Write its fixed finite positive-spanning poll list as $D$ and its forcing function as $ρ$ . Suppose a subsequence of unsuccessful polls indexed by $k_{j}$ has $Δ_{k_{j}} \to 0$ and centers $x_{k_{j}}$ in a compact set $K \subseteq V$ , where $V \subseteq U$ is open. Assume $f : U \to R$ is totally differentiable at every point of $V$ and $\nabla f$ is $L$ -Lipschitz there. Here an accumulation point means a limit of a subsequence in the Euclidean metric. Then every such point is stationary, meaning $\nabla f (\overset{ˉ}{x}) = 0$ . Compactness ensures at least one such accumulation point. This asymptotic claim concerns the untruncated continuation; a finite-budget return is not a stationarity certificate.

Facts & Assumptions

Given: The poll and forcing rules Finite-budget deterministic direct search with positive-spanning polling, positive angle A finite positive-spanning poll set has a positive uniform angle, uniform neighborhood radius A compact set inside an open set has a uniform neighborhood radius, local descent Local descent inequality for a Lipschitz gradient on an open segment, and compact metric sequential compactness In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle.

[F1]

There is $κ_{D} > 0$ with $\max_{i} (- ⟨ u, d_{i} ⟩) \geq κ_{D}$ for every unit vector $u$ (A finite positive-spanning poll set has a positive uniform angle).

[F2]

For sufficiently short displacements from any center in compact $K$ , the segment remains in $V$ (A compact set inside an open set has a uniform neighborhood radius).

[F3]

If the segment $[x, x + h]$ lies in $V$ , then $f (x + h) \leq f (x) + ⟨ \nabla f (x), h ⟩ + \frac{L}{2} ∥ h ∥_{2}^{2}$ (Local descent inequality for a Lipschitz gradient on an open segment).

[F4]

A sequence is a function on the naturals, convergence is the metric limit, and every sequence in compact metric $K$ has a subsequence converging in $K$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, Open cover, subcover, compact metric space, and compact subset of a metric space).

[F5]

On $R^{n}$ with $n \geq 1$ , the Euclidean metric is $d_{2} (x, y) = ∥ x - y ∥_{2}$ ; thus the $L$ -Lipschitz hypothesis gives $∥ \nabla f (u) - \nabla f (v) ∥_{2} \leq L ∥ u - v ∥_{2}$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[F6]

The finite poll list has a largest norm $D_{\max} = \max_{i} ∥ d_{i} ∥_{2}$ (Positive-spanning poll sets).

Proof

1.1F2F6givenalgebra

Use the defined finite maximum $D_{\max}$ from [F6], and choose the uniform radius $δ > 0$ from [F2]. For all sufficiently large $j$ , $Δ_{k_{j}} D_{\max} < δ$ , so every segment from $x_{k_{j}}$ to $x_{k_{j}} + Δ_{k_{j}} d_{i}$ remains in $V$ .

2.1F1F3step 1.1algebra

If $\nabla f (x_{k_{j}}) \neq 0$ , [F1] supplies a poll direction $d_{i}$ with $⟨ \nabla f (x_{k_{j}}), d_{i} ⟩ \leq - κ_{D} ∥ \nabla f (x_{k_{j}}) ∥_{2}$ . The poll is unsuccessful, so its in-domain trial value is at least $f (x_{k_{j}}) - ρ (Δ_{k_{j}})$ . Apply [F3] to this direction and combine the two bounds to obtain $κ_{D} ∥ \nabla f (x_{k_{j}}) ∥_{2} \leq ρ (Δ_{k_{j}}) / Δ_{k_{j}} + (L Δ_{k_{j}} / 2) D_{\max}^{}$ . If the gradient is zero, the same inequality holds.

3.1givenF4F5step 2.1algebra∎

The right side tends to zero by the forcing condition and $Δ_{k_{j}} \to 0$ , while $κ_{D} > 0$ ; hence $∥ \nabla f (x_{k_{j}}) ∥_{2} \to 0$ . For any accumulation subsequence converging to $\overset{ˉ}{x} \in K$ , [F5] gives $∥ \nabla f (x_{k_{j}}) - \nabla f (\overset{ˉ}{x}) ∥_{2} \leq L ∥ x_{k_{j}} - \overset{ˉ}{x} ∥_{2} \to 0$ , so $\nabla f (\overset{ˉ}{x}) = 0$ . By [F4] at least one accumulation subsequence exists.

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Sources

A. R. Conn, K. Scheinberg, and L. N. Vicente, Introduction to Derivative-Free Optimization, §§2.1–2.2 and ch. 8 (standard reference, not scraped)
Tamara G. Kolda, Robert Michael Lewis, and Virginia Torczon, Optimization by Direct Search, §§3.1–3.6 (standard reference, not scraped)