Manuscript En09-A@1 · En09-A.tex · lemma · eafec9de

Lemma: Complete-channel telescoping

Complete-channel stability and refinement independence › Telescoping on compatible complete interfaces · lines 1368–1396 · proof included or linked

En09-A@1#lem:telescoping, En09-A@1#eq:telescoping

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Let $G_{1}, \dots, G_{N}$ and $G_{1}, \dots, G_{N}$ be compatible channels, with corresponding factors on the same comparison interfaces. Then

d_{⋄} (G_{N} \dots G_{1}, G_{N} \dots G_{1}) \leq min {1, j = 1 \sum^{N} d_{⋄} (G_{j}, G_{j})} .

Proof. Subtract the two compositions and replace one factor at a time:

G_{N} \dots G_{1} - G_{N} \dots G_{1} = j = 1 \sum^{N} G_{N} \dots G_{j + 1} (G_{j} - G_{j}) G_{j - 1} \dots G_{1} .

Channels have diamond norm one. Submultiplicativity therefore bounds each term by $G_{j} - G_{j}_{⋄}$ . The triangle inequality and division by two prove the sum bound. The distance between channels is at most one.

Displayed formulas in this passage

En09-A@1#eq:telescoping · line 1373
$d_{⋄} (G_{N} \dots G_{1}, G_{N} \dots G_{1}) \leq min {1, j = 1 \sum^{N} d_{⋄} (G_{j}, G_{j})} .$
unlabeled · line 1384
$G_{N} \dots G_{1} - G_{N} \dots G_{1} = j = 1 \sum^{N} G_{N} \dots G_{j + 1} (G_{j} - G_{j}) G_{j - 1} \dots G_{1} .$

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section: Telescoping on compatible complete interfaces