Manuscript XXII-F@1 · XXII-F.tex · lemma · 67821705

Lemma: Compatible complete-channel telescoping

Complete-channel approximation corollary · lines 1102–1132 · proof included or linked

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

Suppose

Γ = D_{k} G_{k} \dots G_{1} E_{k}, Γ = D_{k} G_{k} \dots G_{1} E_{k}

are compatible compositions with matched intermediate interfaces, and

d_{⋄} (E_{k}, E_{k}) \leq α, d_{⋄} (G_{j}, G_{j}) \leq e_{j}, d_{⋄} (D_{k}, D_{k}) \leq β .

Then

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

Proof. Insert and subtract intermediate compositions replacing one factor at a time. Each difference contains one changed factor, with channels before and after it. Their diamond norms are one, so each contribution is bounded by its factor difference. The triangle inequality gives the sum. The half-diamond distance between two channels is at most one.

Displayed formulas in this passage

unlabeled · line 1105
$Γ = D_{k} G_{k} \dots G_{1} E_{k}, Γ = D_{k} G_{k} \dots G_{1} E_{k}$
unlabeled · line 1112
$d_{⋄} (E_{k}, E_{k}) \leq α, d_{⋄} (G_{j}, G_{j}) \leq e_{j}, d_{⋄} (D_{k}, D_{k}) \leq β .$
XXII-F@1#eq:telescoping · line 1118
$d_{⋄} (Γ, Γ) \leq min {1, α + j = 1 \sum^{k} e_{j} + β} .$

Lemma: Compatible complete-channel telescoping

Displayed formulas in this passage

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