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

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

294 LENZMANN and LEWIN

294 LENZMANN and LEWIN Next, let ɛ>0 be small enough and consider a real function ψ ∈ span(ϕ i , i = 1,...,K) ⊥ . We introduce the following test state τ ɛ = (n 1 − ɛ)|ϕ 1 〉〈ϕ 1 |+ and compute F (τ ɛ ) = F (τ) − κ √ ∫∫ ɛ R 3 ×R 3 K∑ n i |ϕ i 〉〈ϕ i |+ɛ|ψ〉〈ψ| i=2 a(x,y)ψ(x)ψ(y) |x − y| dx dy + O(ɛ), (6.3) where a := √ τ(1 − τ) ≠ 0 is a (real) finite-rank symmetric operator acting on L 2 (R 3 ). Hence we get a contradiction once we can prove the following. LEMMA 6.2 There exists ψ ∈ span(ϕ i ,i= 1,...,K) ⊥ , ψ = ψ, such that ∫∫ R 3 ×R 3 a(x,y)ψ(x)ψ(y) |x − y| Proof Assume, on the contrary, that ∫∫ R 3 ×R 3 dx dy = 4π K∑ i=1 a(x,y)ψ(x)ψ(y) |x − y| √ |̂ϕ i ψ(k)| ni (1 − n i ) ∫R 2 dk > 0. |k| 3 2 dx dy = 0 for all ψ ∈ span(ϕ i ,i= 1,...,K) ⊥ . This means that the (real) symmetric Hilbert- Schmidt operator R with kernel a(x,y)|x − y| −1 vanishes when it is restricted to span(ϕ i ,i= 1,...,K) ⊥ .AsR 0, this implies that we must have K a(x,y) |x − y| = ∑ i,j=1 a ij ϕ i (x)ϕ j (y) (6.4) where (a ij ) is a real nonnegative symmetric matrix. Multiplying (6.4) by |x − y| and taking x = y we deduce that a(x,x) = ∑ K √ i=1 ni (1 − n i )ϕ i (x) 2 = 0 for every x ∈ R 3 . This implies that a = √ τ(1 − τ) = 0, which is a contradiction with τ 2 ≠ τ. The proof of Theorem 3 is now complete.

MINIMIZERS FOR THE HFB-THEORY 295 7. Proof of Theorem 1 In this section, we present the proof of Theorem 1. By Lemma 4.1, we note that Theorem 1(ii) is an immediate consequence of Theorem 1(i). Hence it suffices to prove Theorem 1(i), where we argue by contradiction as follows. We suppose that 0 0. To prepare the proof of Lemma 7.1, we first establish the following fact, which is an adaptation of a similar result in [34, Lemma I.1] to the fractional Sobolev space H 1/2 (R 3 ). An essential ingredient is the localization formula in Lemma A.1. LEMMA 7.2 Let {f n } be a bounded sequence in H 1/2 (R 3 ), and suppose that {|f n | 2 } vanishes in the sense that ( ) lim sup sup |f n (x)| n→∞ y∈R ∫|x−y|R 2 dx = 0 for all R>0. 3 Then ‖f n ‖ L p → 0 for 2

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