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

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

## 298 LENZMANN and LEWIN

298 LENZMANN and LEWIN Proof We do not detail the proof which is an adaptation of ideas of Lions [34], where one introduces another sequence of concentration functions Q 2 n (R) = sup ρ y∈R ∫|x−y|R √ √ Kγ n K (x) dx (7.13) 3 in addition to the sequence {Q 1 n }∞ n=1 defined in (7.2). We refer, for instance, to [15] where a similar argument has been detailed. The rest of the proof follows well-known ideas of Lions [34] coupled with the fact that our minimizing sequence {γ n } is bounded in X, hence (up to a subsequence) converges strongly locally in the trace class. The proof is the same in X. As we have already mentioned, our model given by E(γ,α) is invariant under spatial translations. Thus we may assume that (in Lemma 7.3) the sequence of translations is given by which we do for all the rest of the proof. y n = 0 for all n 1, Step 3: Splitting of the energy. The next step shows that we must have I(λ) = I(λ 1 ) + I(λ − λ 1 ) and I(λ 1 ) has a minimizer. (7.14) To streamline our notation with Lemma 7.3, we introduce a new radial cutoff function 0 ˜χ 1 such that ˜χ ≡ 1 for |x| 7/5 and ˜χ ≡ 0 for |x| 8/5. Given the sequence {R n } from Lemma 7.3, we define the functions χ n = ˜χ(·/R n ) and = √ 1 − χn 2. Likewise, we define {(γ n 1,α1 n )} n∈N and {(γn 2,α2 n )} n∈N by ζ Rn (γ 1 n ,α1 n ):= χ n(γ n ,α n )χ n and (γ 2 n ,α2 n ):= ζ R n (γ n ,α n )ζ Rn . Note that, since (γ n ,α n ) ∈ K, one easily verifies that (γn i,αi n ) ∈ K for i = 1, 2. Furthermore, we have γn 1 → γ 1 in S 1 , by Lemma 7.3, which in particular implies that Tr(γn 1) = ∫ ρ γn (x)χn 2(x) dx → λ1 . Next, we prove that E(γ n ,α n ) E(γ 1 n ,α1 n ) + E(γ 2 n ,α2 n ) + o(1). (7.15) To deal with the kinetic energy, we use Lemma A.1 and Tr(γ n ) = λ to find that Tr(Tγ n ) Tr(Tγ 1 n ) + Tr(Tγ2 n ) − CλR−2 n . (7.16)

MINIMIZERS FOR THE HFB-THEORY 299 The next step in the proof of (7.15) is to separate the direct term, which can be done as follows: ∫∫ ργ 1 D(ρ γn ,ρ γn ) = D(ρ γ 1 n ,ρ γ 1 n ) + D(ρ γ 2 n ,ρ γ 2 n ) + 2 n (x)ρ γ 2 n (y) dx dy. |x − y| Here, we note that ∫∫ χn (x) 2 ρ γn (x)ζ n (y) 2 ρ γn (y) ∫ = |x|6R n /5 ∫ + ∫ + |x − y| ∫ |y|7R n /5 6R n /5|x|8R n /5 6R n /5|x|8R n /5 Therefore, ∫∫ χn (x) 2 ρ γn (x)ζ n (y) 2 ρ γn (y) |x − y| dx dy χ n (x) 2 ρ γn (x)ζ n (y) 2 ρ γn (y) dx dy |x − y| ∫ χ n (x) 2 ρ γn (x)ζ n (y) 2 ρ γn (y) dx dy |y|9R n /5 |x − y| ∫ χ n (x) 2 ρ γn (x)ζ n (y) 2 ρ γn (y) dx dy. |x − y| 7R n /5|y|9R n /5 dx dy 10λ2 R n + C||ρ γn || 2 L 6/5 (B(0,9R n /5)\B(0,6R n /5)) . Furthermore, by (7.12), we have that ρ γn 1 B(0,2Rn )\B(0,R n ) → 0 in L 1 (R 3 ). Since this is also a bounded sequence in L p (B(0, 2R n ) \ B(0,R n )) for all 1 p 3/2, we infer by interpolation that it also tends to 0 in L p (B(0, 2R n ) \ B(0,R n )) for 1

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