Intertwining Defect Propagation
Abstract
An operator intertwining defect telescopes and propagates with exact norm bounds.
Theorem 1.1 (Intertwining defects telescope exactly).
Proof. Machine-checked in Lean as D5/S3/Observer/Approximation/IntertwiningDefectPropagation.intertwining_defect_telescope (✓ std3). ∎
Source. Repository-derived.
Commentary.
Let T and A be continuous linear endomorphisms of possibly distinct spaces, and let C map the source space to the target. The time-n defect is the sum of the one-step defect transported by the remaining powers of A and the elapsed powers of T.
A noncommutative-ring induction proves exact cancellation. At time zero the finite sum is empty and both sides are zero.
Theorem 1.2 (The propagated defect has a weighted norm bound).
Proof. Machine-checked in Lean as D5/S3/Observer/Approximation/IntertwiningDefectPropagation.norm_intertwining_defect_le (✓ std3). ∎
Source. Repository-derived.
Commentary.
The triangle inequality, the operator-norm composition bound, and the norm bound for powers turn the exact telescope into the finite weighted sum stated in the source corollary.
No finite-dimensional, completeness, inner-product, or nontrivial carrier assumption is used.
Theorem 1.3 (Uniform norm bounds give linear propagation).
Proof. Machine-checked in Lean as D5/S3/Observer/Approximation/IntertwiningDefectPropagation.uniform_norm_intertwining_defect_le (✓ std3). ∎
Source. Repository-derived.
Commentary.
If both operator norms are at most L, every summand is at most L to the power n minus one times the one-step defect norm. There are exactly n summands.
The proof does not need L less than one. Natural subtraction is truncated, so at n equal to zero its exponent is zero while the leading factor n makes the right side zero.
Theorem 1.4 (The bound on A is necessary).
Proof. Machine-checked in Lean as D5/S3/Observer/Approximation/IntertwiningDefectPropagation.left_norm_bound_is_necessary (✓ std3). ∎
Source. Repository-derived.
Commentary.
On the one-dimensional real space take A as multiplication by two, C as the identity, T as zero, L as one, and n as three. The bound on T holds, but the claimed conclusion without the bound on A is false.
Theorem 1.5 (The bound on T is necessary).
Proof. Machine-checked in Lean as D5/S3/Observer/Approximation/IntertwiningDefectPropagation.right_norm_bound_is_necessary (✓ std3). ∎
Source. Repository-derived.
Commentary.
The symmetric one-dimensional counterexample takes A as zero, C as the identity, T as multiplication by two, L as one, and n as three. The bound on A holds, but the conclusion without the bound on T is false.
References
- Truth anchor:
D5/S3/Observer/Approximation/IntertwiningDefectPropagation.intertwining_defect_telescope - Truth anchor:
D5/S3/Observer/Approximation/IntertwiningDefectPropagation.left_norm_bound_is_necessary - Truth anchor:
D5/S3/Observer/Approximation/IntertwiningDefectPropagation.norm_intertwining_defect_le - Truth anchor:
D5/S3/Observer/Approximation/IntertwiningDefectPropagation.right_norm_bound_is_necessary - Truth anchor:
D5/S3/Observer/Approximation/IntertwiningDefectPropagation.uniform_norm_intertwining_defect_le