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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Exact gradient line search has a global objective-gap contraction

Statement

Let U⊆Rd be open and convex, let f∈C2(U,R) have an attained global minimum f∗=f(x∗) at x∗∈U, and fix x0∈U. Suppose the initial sublevel set K:={y∈U:f(y)≤f(x0)} is compact and is contained in a convex open set V⊆U on which

mI⪯Hf(y)⪯MI(y∈V)

for constants 0<m≤M. For every current iterate x∈K, the oracle Ideal exact gradient line-search step is well-defined. If g:=∇f(x)≠0 and x+ is its returned point, then

f(x+)−f∗≤(1−mM)(f(x)−f∗).

If g=0, the oracle returns the unique minimizer x=x∗. Repeated exact steps from x0 therefore satisfy

f(xk)−f∗≤(1−mM)k(f(x0)−f∗)

for every generated k≥1. When m=M, the factor is zero for k≥1; no 00 convention is used. This is an objective-gap statement and does not identify an iterate-error bound without an additional curvature estimate.

Facts & Assumptions

Given: U,f,x∗,x0,K,V,m,M as in the Statement, and a current x∈K.

[F1]

Ideal exact gradient line-search step: if g≠0, it returns the unique global minimizer of t↦f(x−tg) over feasible t≥0; if g=0, it returns the stationary point.

[F2]

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: Hf is the Hessian matrix and ∇f is the vector of standard-coordinate partial derivatives.

[F4]

A C2 function is convex exactly when its Hessian is positive semidefinite: a C2 function with positive-semidefinite Hessian on a convex open set is convex there.

[F5]

Multivariable Taylor formula with a Lagrange remainder along a line segment: along a segment in a convex open domain, the second-order Taylor remainder is the Hessian quadratic form at a point on that segment.

[F6]

A Hessian lower bound gives strong convexity and gradient-gap bounds: applied to the restriction of f to V, the lower Hessian bound gives the strong-convexity inequality, uniqueness of an attained minimizer, and ∥∇f(x)∥22≥2m(f(x)−f∗) for points of V.

[F7]

A compact subset of a metric space is closed and bounded: a compact subset of a metric space is closed and bounded.

[F8]

Metric continuity characterisations, with countable choice for the sequential converse: a continuous map between metric spaces has closed preimages of closed sets; this implication uses no choice principle.

[F10]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real-valued function on a nonempty compact metric space attains its minimum.

Proof

1.1F2F3F6givenchoose

If g=0, [F6] gives 0≥2m(f(x)−f∗); since f(x)≥f∗ and m>0, f(x)=f∗, and uniqueness gives x=x∗. Now assume g≠0 and put r(t):=x−tg for t≥0. [F3] gives ϕ′(0)=−∥g∥22<0 for ϕ(t):=f(r(t)). Openness of U and this strict derivative provide a sufficiently small t>0 with r(t)∈U and f(r(t))<f(x)≤f(x0), so S:={t≥0:r(t)∈K} is nonempty.

2.1F4F7F8F9F11step 1.1

By [F4], f is convex on V, so its sublevel set K is convex because K⊂V. By [F7], K is closed and bounded. The affine map r is continuous since ∥r(s)−r(t)∥2=∣s−t∣∥g∥2, so [F8] makes S=r−1(K)∩[0,∞) closed. Boundedness gives c∈Rd and R0>0 with K⊂B(c,R0); hence ∥y∥2≤∥c∥2+R0=:R on K. For t∈S, the triangle inequality [F11] yields t∥g∥2=∥x−r(t)∥2≤∥x∥2+R. Thus S is bounded and compact by [F9].

3.1F1F6F10givenstep 2.1algebra

The continuous restriction of ϕ to the nonempty compact set S attains a minimum t∗∈S by [F10]. Since 0∈S, we have f(r(t∗))≤f(x). Every feasible ray parameter t with r(t)∈U but t∉S has f(r(t))>f(x0)≥f(x)≥f(r(t∗)), so t∗ is a global feasible-ray minimizer. If two distinct parameters minimized, their endpoints y1,y2 would be distinct because g≠0; their midpoint z lies in K by step 2.1. Apply the strong-convexity inequality [F6] from z to each endpoint and add: f(z)≤(f(y1)+f(y2))/2−m∥y1−y2∥22/8, strictly below their common minimum. Thus the minimizer is unique and the output specified by [F1] is well-defined.

4.1F2F3F12givenstep 1.1step 3.1

The small step in step 1.1 strictly decreases f, so the minimizer from step 3.1 has t∗>0. Since r(t∗)∈K⊂U and U is open, t∗ is interior to the feasible ray and ϕ′(t∗)=0. Convexity of V and step 2.1 put the segment [x,r(t∗)] in V. By [F2], [F3], and the Hessian bounds, m∥g∥22≤ϕ′′(s)=gTHf(r(s))g≤M∥g∥22 on this segment, so ϕ′ is strictly increasing. The mean-value theorem [F12] applied to ϕ′ on [0,t∗] gives ∥g∥22=ϕ′(t∗)−ϕ′(0)=t∗ϕ′′(s0)≤t∗M∥g∥22 for some s0∈(0,t∗). Since g≠0, division yields t∗≥1/M.

5.1F5F6givenstep 3.1step 4.1

The candidate t0:=1/M lies in [0,t∗] by step 4.1, so r(t0) lies on the segment from x to r(t∗) and belongs to V. The lower Hessian bound makes ϕ′′(s)>0 there; since ϕ′(t∗)=0, ϕ′(s)<0 for 0≤s<t∗. Hence f(r(t0))≤f(x)≤f(x0) and r(t0)∈K. Taylor's formula [F5] with the upper Hessian bound gives f(r(t0))≤f(x)−M−1∥g∥22+(M/(2M2))∥g∥22=f(x)−∥g∥22/(2M). By the minimization proved in step 3.1, f(x+)≤f(r(t0)).

6.1F1F6givenstep 5.1algebradischarge-induction∎

Subtract f∗ in step 5.1 and use [F6] to get f(x+)−f∗≤f(x)−f∗−∥g∥22/(2M)≤(1−m/M)(f(x)−f∗). The factor is nonnegative because m≤M. Exact steps never increase the objective, so every later iterate remains in K; applying the one-step bound successively proves the stated rate for all generated iterates.

Depends on

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Cited to discharge well-definedness by Ideal exact gradient line-search step.

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