Proposition

Corollary 35 ([Strict compact-magnetic numerical extension]). Let the original exterior satisfy the numerical geometric and asymptotic hypotheses, and let D_m be an arbitrary smooth compactly supported closed two-form. With α=i_E_g closed, …

Corollary 35 ([Strict compact-magnetic numerical extension]). Let the original exterior satisfy the numerical geometric and asymptotic hypotheses, and let Dm be an arbitrary smooth compactly supported closed two-form. With α=iEg closed, use the electromagnetic convention Ig(v)=cn2(|E|g2+|Dm|g2+2Dm(v,E)), μm+Jm(v)=μ+J(v)-Ig(v). Assume also that μm,|Jm|g are integrable and that the ADM and charge flux limits exist with the stated normalizations. Assume the ADM vector is future timelike, the fixed componentwise trapping inequalities are strict, and for some c>0 and 0<β<(1,q) , μm-|Jm|g≥c r-n-β . Then, with the original full-cut infimum a and the original normalized electric charge Q , m≥s|Q|+1-s22XA (0<s<1). Equivalently, m≥|Q| , and when XA>|Q| one also has m≥(XA+Q2/XA)/2 . No magnetic equality classification or non-timelike extension is asserted here.

Prove 1Disprove 0Cited as proposition 2Chain graphAdd Your Proof

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Apply the rest preparation and strict preparation in Proposition 11 ([Preparation at rest]) and Lemma 9 ([The decaying Neumann problem and strictification]), retaining the two closed forms. The magnetic support stays in the compact original region. There the strictly positive margin absorbs the arbitrarily small rescaling errors. The gluing and bending annuli can be taken outside that support, so on those annuli and on the end the electric estimates are unchanged. This yields prepared data with charge Q , energies E_ *,j →m , and full-cut infima a_ *,j ≥(1-o_j(1))a . For each fixed s , apply Proposition 34 ([Prepared numerical bound]) and then pass to this limit. The right side is increasing in the area variable, so the lower comparison of infima is sufficient. This proves. The elementary optimization is given in Proposition 13 ([Reduction of the numerical theorem]). The argument does not require the component electric fluxes to have one sign.