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An inexact FAS coarse solve contributes a bounded correction defect

Statement

Consider one two-grid FAS cycle with fixed pre-smoothing result, exact selected coarse root v(u), prolongation P, and a post-smoother that is Lpost-Lipschitz on the relevant corrected states. Suppose a coarse solver returns v^ with residual η=∥FH(v^)−gH(u)∥2 and a separately verified residual-to-solution estimate ∥v^−v(u)∥2≤MHη. Then the inexact-cycle output E^(u) and exact-cycle output E(u) satisfy ∥E^(u)−E(u)∥2≤Lpost∥P∥opMHη. If the exact cycle satisfies the root-error estimate ∥E(u)−u∗∥2≤κ∥u−u∗∥2 on a stated invariant region, then for every iteration input uk in that region whose exact and inexact corrected states satisfy the preceding hypotheses, ∥uk+1−u∗∥2≤κ∥uk−u∗∥2+Lpost∥P∥opMHηk. Invertibility of DFH alone is not asserted to imply the stated residual-to-solution estimate; that bound must be proved under additional coarse-operator hypotheses.

Facts & Assumptions

Given: The two-grid FAS correction formula, the same fine pre-smoothing state in both cycles, a residual-to-solution bound with constant MH, a post-smoother Lipschitz constant Lpost, and (for the iteration inequality) an exact-cycle root-error factor κ.

[F1]

Finite-budget nonlinear FAS two-grid cycle: the coarse difference is prolonged by P(v^−Iu) before post-smoothing.

[F2]

A smooth FAS cycle converges locally under its actual derivative bounds: under its stated hypotheses, the exact cycle has a root-error factor κ=q+Lr/2<1 on the invariant ball.

[A1]

On the chosen region, the exact selected coarse root and the inexact returned value satisfy ∥v^−v(u)∥2≤MH∥FH(v^)−gH(u)∥2.

[A2]

The post-smoother has Lipschitz constant Lpost on the segment joining the exact and inexact corrected states.

Proof

1.1F1A1algebra

Let z=upre+P(v(u)−Iupre) and z^=upre+P(v^−Iupre) be the exact and inexact corrected states. By [F1], subtraction cancels upre and Iupre, so z^−z=P(v^−v(u)). Hence [A1] gives ∥z^−z∥2≤∥P∥op∥v^−v(u)∥2≤∥P∥opMHη.

2.1A2F2step 1.1algebra∎

Applying the same post-smoother to the two corrected states and using [A2] yields ∥E^(u)−E(u)∥2≤Lpost∥P∥opMHη. For an iteration, the triangle inequality and [F2] therefore give ∥uk+1−u∗∥2≤∥E^(uk)−E(uk)∥2+∥E(uk)−u∗∥2≤Lpost∥P∥opMHηk+κ∥uk−u∗∥2.

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