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Local descent inequality for a Lipschitz gradient on an open segment

Statement

Let U⊆Rn be open, let f:U→R be totally differentiable at every point, and suppose its gradient is L-Lipschitz on U for some L≥0. If x,y∈U and the whole segment [x,y] lies in U, then f(y)≤f(x)+⟨∇f(x),y−x⟩+L2∥y−x∥22.

Facts & Assumptions

Given: Total derivatives and gradients The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn, For a differentiable scalar field, Dvf(a)=⟨∇f(a),v⟩ and the unit direction of steepest ascent is the normalized gradient, the Lipschitz definition and Euclidean metric Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, local continuity of totally differentiable maps Total differentiability gives a local O(∥h∥2) increment bound and therefore continuity, Cauchy–Schwarz Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, derivative rules The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, and the mean-value theorem The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a).

[F1]

The Lipschitz condition is d2(∇f(u),∇f(v))≤Ld2(u,v); for n≥1, d2(a,b)=∥a−b∥2 (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it).

[F5]

Cauchy–Schwarz and norm homogeneity give ∣⟨u,v⟩∣≤∥u∥2∥v∥2 and ∥sv∥2=∣s∣∥v∥2 (Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation).

Proof

1.1givenF3F5construct

Put d=y−x. If d=0, both sides are equal. For d≠0, necessarily n≥1. Define γ(s)=x+sd and I={s∈R:γ(s)∈U}. Since ∥γ(s+h)−γ(s)∥2=∣h∣∥d∥2, the affine path is continuous; openness of U then makes I open, and the segment assumption gives [0,1]⊆I. Put ϕ=f∘γ on I. Total differentiability makes f continuous at every point of γ([0,1]), so ϕ is continuous on [0,1]. For each s∈(0,1), its difference quotient is the directional difference quotient of f at x+sd in direction d, so ϕ′(s)=⟨∇f(x+sd),d⟩.

2.1F1F3F4F5step 1.1algebra

Put a=⟨∇f(x),d⟩, g(s)=ϕ(s)−ϕ(0)−sa, and q(s)=g(s)−L2s2∥d∥22. By [F3] and [F6], q is continuous on [0,1]; by [F4], it is differentiable on (0,1) and q′(s)=⟨∇f(x+sd)−∇f(x),d⟩−Ls∥d∥22. By [F1] and [F5], the inner-product term is at most L∥sd∥2∥d∥2=Ls∥d∥22 for s∈(0,1). Hence q′(s)≤0.

3.1F2step 2.1algebra∎

Apply [F2] to q on [0,1]: for some s∈(0,1), q(1)−q(0)=q′(s)≤0. Since q(0)=0, this says f(y)−f(x)−⟨∇f(x),d⟩≤L2∥d∥22, as required.

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