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Uniform pullback smoothness guarantees an Armijo step

Statement

Let M=St⁡(n,r) or Gr⁡(r,n) be compact, let f:M→R be smooth, and let R be a QR or polar retraction with the first-order property of The QR and polar maps are first-order Stiefel and Grassmann retractions. Write f^x(η)=f(Rx(η)). Assume a uniform bound ∥grad⁡f(x)∥≤G and a constant LR≥0 such that, for every x and ∥η∥≤1, ∥grad⁡f^x(η)−grad⁡f^x(0)∥≤LR∥η∥. For a∈(0,1) and a nonstationary iterate, set δ=min⁡{1,(2G)−1,2(1−a)/LR}, where a reciprocal of zero is interpreted as +∞. If the finite backtracking budget tests some tj=βj≤δ, then it accepts a step; the first accepted step obeys t≥βδ. For 0<t≤δ, the bound t∥grad⁡f(x)∥≤1/2 keeps the pullback argument inside its assumed unit tangent ball; an earlier accepted trial may be larger.

Facts & Assumptions

Given: The compact manifold, smooth objective, first-order retraction, constants G,LR,a,β, and finite budget reaching δ.

[F1]

The retraction pullback has derivative at zero equal to the original tangent direction (The QR and polar maps are first-order Stiefel and Grassmann retractions).

[F2]

The Riemannian gradient represents the differential in the chosen metric (The Riemannian gradient is the metric dual of the differential).

[F3]

The finite Armijo search tests tj=βj and accepts only the sufficient-decrease inequality (Finite-budget retraction Armijo backtracking).

Proof

technique · integrate the Lipschitz pullback-gradient bound along the negative-gradient segment
1.1givenalgebra

Since g≠0 and ∥g∥≤G, we have G>0. For 0<t≤δ, η=−tg has norm t∥g∥≤tG≤1/2, so the entire segment sη, 0≤s≤1, lies in the unit ball.

2.1F1F2F3F4givenstep 1.1algebra

By [F1] and [F2], Df^x(0)[η]=⟨g,η⟩=−t∥g∥2. The fundamental theorem of calculus along s↦sη and the assumed Lipschitz bound give f^x(η)≤f(x)−t∥g∥2+∫01LRs∥η∥2 ds=f(x)−t∥g∥2+12LRt2∥g∥2. The integral bound uses [F4] to estimate the inner product of the gradient difference with η. If LR=0 the last error is zero; otherwise t≤2(1−a)/LR makes the right side at most f(x)−at∥g∥2. Thus every tested t≤δ satisfies the Armijo test in [F3].

3.1F3givenstep 2.1algebra∎

Let j be the first tested index with βj≤δ. By step 2.1 this trial passes, so the algorithm stops no later than j. If j=0, then t=1≥βδ because δ≤1. If j>0, minimality gives βj−1>δ, hence βj>βδ. Any earlier accepted grid point is larger still.

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