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A contractive self-consistent map converges and admits residual and inexact-solve bounds

Statement

Let (K,d) be a nonempty complete metric space and let Φ:K→K be a contraction with constant 0≤q<1. In the SCF application, K is a complete invariant subset of the response-map domain from The occupied-projector response map on its strict-gap domain, with d(X,Y)=∥X−Y∥F. Then Φ has a unique fixed point X∗ and exact iterates Xk+1=Φ(Xk) satisfy d(Xk,X∗)≤qkd(X0,X∗). For every X∈K, d(X,X∗)≤d(X,Φ(X))1−q. If computed updates Xk+1∈K obey d(Xk+1,Φ(Xk))≤ηk, then ek+1≤qek+ηk,ek≤qke0+∑j=0k−1qk−1−jηj, where ek=d(Xk,X∗) and X0 is the supplied initial point. Use the convention q0=1, including when q=0.

Facts & Assumptions

Given: A nonempty complete metric space (K,d), a self-map Φ:K→K with d(Φ(X),Φ(Y))≤qd(X,Y) for 0≤q<1, an initial point X0, and inexact updates with errors ηk≥0.

[F1]

A contraction of a nonempty complete metric space has a unique fixed point and its iterates converge (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point).

[F2]

The response map and, in the projector application, the Frobenius residual are defined as in The occupied-projector response map on its strict-gap domain.

[L1]

At every strict occupied gap, that response map selects a unique rank-r projector (A strict occupied gap defines a unique rank-r spectral projector).

[F3]

A contraction with factor q means d(Φ(X),Φ(Y))≤qd(X,Y) for all X,Y∈K (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction).

Proof

1.1givenF1L1

By [F1], there is a unique X∗ with Φ(X∗)=X∗. In the projector application, [L1] supplies the stated map wherever the strict gap hypothesis holds.

2.1step 1.1givenF3induction

For exact iterates, [F3] gives d(Xk+1,X∗)=d(Φ(Xk),Φ(X∗))≤qd(Xk,X∗). Induction from k=0 yields d(Xk,X∗)≤qkd(X0,X∗).

2.2givenstep 1.1F2F3algebra

For any X∈K, [F2] identifies the SCF metric residual with d(X,Φ(X)). The triangle inequality and contraction property give d(X,X∗)≤d(X,Φ(X))+d(Φ(X),Φ(X∗))≤d(X,Φ(X))+qd(X,X∗). Since 1−q>0, rearrangement proves the residual error bound.

2.3givenstep 1.1F3algebra

For an inexact update, the triangle inequality gives ek+1=d(Xk+1,X∗)≤d(Xk+1,Φ(Xk))+d(Φ(Xk),Φ(X∗))≤ηk+qek. This is the one-step recurrence.

3.1step 2.3inductionalgebra∎

Repeated substitution in the recurrence yields ek≤qke0+∑j<kqk−1−jηj: it holds at k=0 with an empty sum, and substituting its kth bound into ek+1≤qek+ηk gives the formula at k+1.

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