Manuscript VII-B@1 · VII-B.tex · lemma · be632835

Lemma: A nonzero boundary trace for every nonzero smearing

Exclusion of the first geometric insertion · lines 878–922 · proof included or linked

VII-B@1#lem:witness, VII-B@1#eq:phi-witness, VII-B@1#eq:moment, VII-B@1#eq:positive-moment

Let $0 \neq = f \in C_{c}^{\infty} ({t > 0}; R)$ . There exists $ϕ_{f} \in C_{c}^{\infty} (M; R)$ such that, for

a_{μ} = f \partial_{μ} ϕ_{f}, b_{μ} = E_{A} a_{μ},

the trace $b_{0} ∣_{Σ_{0}}$ is not identically zero. One explicit choice is

ϕ_{f} = - \partial_{t} (t f) .

Proof. Let

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

On every bounded time interval, the spatial supports of $a_{0}$ and $b_{0}$ are contained in a common compact set. Integrating $□ b_{0} = a_{0}$ over space therefore yields

B^{''} (t) = F (t) .

Since $b_{0}$ is advanced, $B = B^{'} = 0$ for sufficiently large positive $t$ . Consequently,

B (0) = \int_{0}^{\infty} tF (t) d t = \int_{M} t f \partial_{t} ϕ_{f} d^{4} x .

Set $u = \partial_{t} (t f)$ , so that $ϕ_{f} = - u$ . Integration by parts gives

B (0) = \int_{M} u^{2} d^{4} x .

This number is strictly positive. Indeed, $u = 0$ would imply that $t f$ is independent of $t$ ; compact time support would then force $f = 0$ . Hence $b_{0} ∣_{Σ_{0}}$ cannot vanish identically.

Displayed formulas in this passage

VII-B@1#eq:phi-witness · line 887
$ϕ_{f} = - \partial_{t} (t f) .$
VII-B@1#eq:moment · line 907
$B (0) = \int_{0}^{\infty} tF (t) d t = \int_{M} t f \partial_{t} ϕ_{f} d^{4} x .$
VII-B@1#eq:positive-moment · line 914
$B (0) = \int_{M} u^{2} d^{4} x .$
unlabeled · line 882
$a_{μ} = f \partial_{μ} ϕ_{f}, b_{μ} = E_{A} a_{μ},$
unlabeled · line 895
$F (t) = \int_{R^{3}} a_{0} (t, x) d^{3} x, B (t) = \int_{R^{3}} b_{0} (t, x) d^{3} x .$
unlabeled · line 902
$B^{''} (t) = F (t) .$

Lemma: A nonzero boundary trace for every nonzero smearing

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