Manuscript IX-A-ST@1 · IX-A-ST.tex · lemma · 6e99d1b2

Lemma: Positive advanced-Cauchy moment

A nonzero boundary witness and failure of regular convergence · lines 968–1008 · proof included or linked

IX-A-ST@1#lem:positive, IX-A-ST@1#eq:positive

Let $f \in D (V; R)$ be nonzero. Define

u_{f} = \partial_{t} (t f), ϕ_{f} = u_{f}, a = f d ϕ_{f}, b = G_{A} a .

Then

\int_{Σ} b_{0} (0, x) d^{3} x = \int_{V} [\partial_{t} (t f)]^{2} d^{4} x =: E_{f} > 0.

The initial trace is compactly supported inside $Σ$ .

Proof. All spatial supports are uniformly bounded on the finite time interval in question. Put

A_{0} (t) = \int_{R^{3}} a_{0} (t, x) d^{3} x, B_{0} (t) = \int_{R^{3}} b_{0} (t, x) d^{3} x .

Integration of $□_{η} b_{0} = a_{0}$ gives $- B_{0}^{''} = A_{0}$ . The advanced solution has zero data at sufficiently late times. Consequently,

B_{0} (0) = - \int_{0}^{\infty} t A_{0} (t) d t = - \int_{V} t f \partial_{t} u_{f} d^{4} x = \int_{V} u_{f}^{2} d^{4} x .

There are no integration boundary terms because $f$ is compact inside $V$ . If $u_{f} = 0$ , then $t f$ is independent of $t$ ; compact time support implies $f = 0$ , a contradiction. Support saturation places the entire trace support inside the initial ball.

Displayed formulas in this passage

IX-A-ST@1#eq:positive · line 977
$\int_{Σ} b_{0} (0, x) d^{3} x = \int_{V} [\partial_{t} (t f)]^{2} d^{4} x =: E_{f} > 0.$
unlabeled · line 971
$u_{f} = \partial_{t} (t f), ϕ_{f} = u_{f}, a = f d ϕ_{f}, b = G_{A} a .$
unlabeled · line 989
$A_{0} (t) = \int_{R^{3}} a_{0} (t, x) d^{3} x, B_{0} (t) = \int_{R^{3}} b_{0} (t, x) d^{3} x .$
unlabeled · line 997
$B_{0} (0) = - \int_{0}^{\infty} t A_{0} (t) d t = - \int_{V} t f \partial_{t} u_{f} d^{4} x = \int_{V} u_{f}^{2} d^{4} x .$

Lemma: Positive advanced-Cauchy moment

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