Manuscript XXII-D@1 · XXII-D.tex · lemma · 17889d18

Lemma: Complete-channel telescoping

Complete-channel transport and error assembly › Telescoping on actual compatible interfaces · lines 828–863 · proof included or linked

XXII-D@1#lem:telescoping, XXII-D@1#eq:telescoping

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Let

Γ = D G_{k} \dots G_{1} E, Γ = D G_{k} \dots G_{1} E

be compatible channel compositions on matching step interfaces. Suppose

d_{⋄} (E, E) \leq α, d_{⋄} (G_{j}, G_{j}) \leq e_{j}, d_{⋄} (D, D) \leq β .

Then

d_{⋄} (Γ, Γ) \leq min {1, α + j = 1 \sum^{k} e_{j} + β} .

Proof. Insert intermediate compositions replacing one factor at a time. Each difference contains one changed factor, preceded and followed by channels. Since channels have diamond norm one, each term is bounded by the diamond norm of that factor difference. The triangle inequality and division by two give the sum. The half-diamond distance between two channels is at most one.

Displayed formulas in this passage

unlabeled · line 831
$Γ = D G_{k} \dots G_{1} E, Γ = D G_{k} \dots G_{1} E$
unlabeled · line 842
$d_{⋄} (E, E) \leq α, d_{⋄} (G_{j}, G_{j}) \leq e_{j}, d_{⋄} (D, D) \leq β .$
XXII-D@1#eq:telescoping · line 848
$d_{⋄} (Γ, Γ) \leq min {1, α + j = 1 \sum^{k} e_{j} + β} .$

Lemma: Complete-channel telescoping

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