Manuscript Em02-A@6 · Em02-A-v6.tex · proposition · cb2f0a9a

Proposition: Failure of automatic internal-core localization

Characteristic chart and exact translated equations · lines 1900–1927 · proof included or linked

Em02-A@6#prop:counterexample

Status: Proved for displayed mathematical data.

The core conditions do not force $r = O (δ)$ . Simultaneous crossed cones on the $5 z$ seed do not force $q_{R} = O (δ)$ .

Proof. Take $u = v = ξ = 0$ and fixed $r = ϵ ℓ$ with $0 < ϵ < β$ . This lies on the $47$ and $5 z$ scalar supports while $r / δ$ is unbounded. A fixed past-null $r$ gives the analogous $5 a$ example.

For the second assertion, fix small $u_{0} > 0$ and put

v = \frac{β - α}{β} u_{0}, ∣ ξ ∣^{2} = \frac{4 α ( β - α )}{β} u_{0} .

Then $q_{1} = q_{2} = 0$ , but $q_{R} = 4 β u_{0} > 0$ . Choose $0 < γ < β$ and set

υ = \frac{β - γ}{β} u_{0}, ∣ ω ∣^{2} = \frac{4 γ ( β - γ )}{β} u_{0} .

Both internal vectors are future null, giving a $5 z$ seed point. It remains fixed as $δ ↓ 0$ .

Displayed formulas in this passage

unlabeled · line 1915
$v = \frac{β - α}{β} u_{0}, ∣ ξ ∣^{2} = \frac{4 α ( β - α )}{β} u_{0} .$
unlabeled · line 1921
$υ = \frac{β - γ}{β} u_{0}, ∣ ω ∣^{2} = \frac{4 γ ( β - γ )}{β} u_{0} .$

Nearby in this section

section: Characteristic chart and exact translated equations