Manuscript VII-F@1 · VII-F.tex · lemma · 2d81c03f

Lemma: Location of the active region

An explicit seam-compatible slicing family › Three single-valued functions on the cone · lines 399–434 · proof included or linked

VII-F@1#lem:active-region

For all $x \in C_{μ}$ ,

h (x) - k (x) \geq \frac{11}{30} ρ (x) .

Consequently:

if $s \leq 0$ , then
$F_{s} = s + min {2, h};$
if $s > 0$ , the branch $s + k$ can dominate $h$ only where
$ρ \leq \frac{30 s}{11} \leq \frac{30}{11} < 3;$
for all $s \in [- 1, 1]$ and all $ρ \geq 3$ ,
$F_{s} = 2 + s .$

Proof. Since $min {ξ, η} \leq (ξ + η) /2$ ,

h - k \geq (\frac{2}{3} - \frac{3}{10}) ρ = \frac{11}{30} ρ .

This proves the first two assertions. For $ρ \geq 3$ , $h \geq 2$ , so the outer minimum in (eq:family) equals $2$ .

Displayed formulas in this passage

unlabeled · line 402
$h (x) - k (x) \geq \frac{11}{30} ρ (x) .$
unlabeled · line 408
$F_{s} = s + min {2, h};$
unlabeled · line 412
$ρ \leq \frac{30 s}{11} \leq \frac{30}{11} < 3;$
unlabeled · line 416
$F_{s} = 2 + s .$
unlabeled · line 424
$h - k \geq (\frac{2}{3} - \frac{3}{10}) ρ = \frac{11}{30} ρ .$