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minimizers for the hartree-fock-bogoliubov theory of neutron stars ...

minimizers for the hartree-fock-bogoliubov theory of neutron stars ...

290 LENZMANN and LEWIN

290 LENZMANN and LEWIN For III, this can be seen as follows (we only consider the term in III that contains (X γ ) x , the other being treated similarly): |〈α R ,ζ R (x)ζ R (y)(X γ ) x α〉| = |Tr(α ∗ R K1/2 K −1/2 X ζR γ αζ R )| ‖K 1/2 α R ‖ S2 ‖K −1/2 X ζR γ ‖ S2 ‖αζ R ‖ S2 , (5.20) using that ‖A‖ S∞ ‖A‖ S2 holds. Next, by the Hardy-Kato inequality, we have that ‖K −1/2 X f ‖ S2 C‖K 1/2 f ‖ S2 . Therefore we find (using that αα ∗ γ and γ 2 γ and ζR 2 1) the following bound: |(5.20)| C‖K 1/2 α R ‖ S2 ‖K 1/2 ζ R γ ‖ S2 ‖αζ R ‖ S2 √ C Tr(Kζ R α R ζR 2α∗ R ζ R) √ Tr(Kζ R γ 2 ζ R ) √ Tr(Kζ R α R αR ∗ ζ R) CTr(Kζ R γζ R ) 3/2 δ(R)( ∫ ) ζ 2 R (x)ρ γ (x) dx + Tr(Tζ R γζ R ) , R 3 where δ(R) → 0 as R →∞. The proof is the same for IV. Finally, we estimate I as follows (we again only consider the term with T x ): |〈α R , [T x ,ζ R (x)]αζ R (y)〉| ‖[T x ,ζ R ]‖‖α R ‖ S2 ‖αζ R ‖ S2 C R Tr(ζ Rγζ R ) C ∫ ζ 2 R R (x)ρ γ (x) dx, R 3 where we used the commutator estimate ‖[T x ,f(x)]‖ C‖∇f ‖ L ∞ combined with the fact that αα ∗ γ . In summary, we have proved the following estimate: |α R (x,y)| 〈α R , (T x + T y )α R 〉−κ ∫∫R 2 dx dy 3 ×R |x − y| ∫∫ 3 ( ∫ − 2μ |α R (x,y)| 2 dx dy δ(R) ζ 2 R (x)ρ γ (x) dx + Tr(Tζ R γζ R )) . R 3 ×R 3 R 3 (5.21) Next, we turn to the left-hand side and derive a lower bound as follows. Recall from Appendix C the lower bound 〈 ( α, T x + T y − κ 1 |x − y| ) 〉 α 2(β − m)〈α, α〉 (5.22) for all α ∈ H 1/2 (R 3 , C q ) ∧ H 1/2 (R 3 , C q ), provided that κ

MINIMIZERS FOR THE HFB-THEORY 291 we deduce that 2μ 0 small, and hence |α R (x,y)| 〈α R , (T x + T y )α R 〉−(κ + ɛ − ɛ) ∫∫R 2 dx dy 3 ×R |x − y| 3 ∫∫ − 2μ |α R (x,y)| 2 |α R (x,y)| dx dy ɛ R 3 ×R ∫∫R 2 dx dy, (5.23) 3 3 ×R |x − y| 3 for some ɛ>0 sufficiently small. Therefore, we arrive at the following estimate: ζR ∫∫R 2(x)ζ R 2(y)|α(x,y)|2 dx dy 3 ×R |x − y| 3 δ(R)( ∫ ) ζ 2 R (x)ρ γ (x) dx + Tr(Tζ R γζ R ) R 3 (5.24) with δ(R) → 0 as R →∞. Step 3: Combining (5.16) and (5.24). We define the sequence {R n } ∞ n=1 of radii given by R n = 4 n ,forn 1. From (5.16) we conclude for all n large enough that ∫ Tr(Tζ Rn+1 γζ Rn+1 ) + ζ 2 R n+1 (x) 2 ρ γ (x) dx R 3 C ∫∫ (R n+1 ) + C ζ 2 2 R n+1 (x) |α(x,y)|2 dx dy. (5.25) R 3 ×R |x − y| 3 √ Next, we let χ Rn = 1 − ζR 2 n so that ζR 2 n + χR 2 n ≡ 1 holds. Therefore we can split and estimate the pairing term in the inequality above as follows: ζ ∫∫R 2 R n+1 (x) |α(x,y)|2 dx dy = ζ 3 ×R |x − y| ∫∫R 2 R n+1 (x)χ 2 R n (y) |α(x,y)|2 dx dy 3 3 ×R |x − y| 3 + ζ ∫∫R 2 R n+1 (x)ζ 2 R n (y) |α(x,y)|2 dx dy =: I + II. 3 ×R |x − y| 3 Since R n = (1/4)R n+1 , we deduce from the support properties of ζ Rn+1 (x) and χ Rn (y) that I C R n ∫R 3 ζ 2 R n+1 (x)ρ γ (x) dx, using ρ αα ∗(x) = ∫ |α(x,y)| 2 dy ρ R 3 γ (x) thanks to αα ∗ γ . Further, we notice that ζ Rn+1 ζ Rn and thus ∫∫ II ζ 2 R n (x)ζ 2 R n (y) |α(x,y)|2 |x − y| where we used inequality (5.24) in the last step. dx dy δ(R n )( ∫ ) ζ 2 R n (x)ρ γ + Tr(Tζ Rn γζ Rn ) ,

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