Annales Henri Poincaré - Volume 2 by Vincent Rivasseau (Chief Editor)

By Vincent Rivasseau (Chief Editor)

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1. Set f (x) := xK1 (x). The proof is immediate from the observation that limx→0 f (x) = 1 and that f (x) < 0 for positive x. 2. Let g(u, v) := K1 (R|u − v|)S(u, v)/|u − v|. The function g fulfills g(u, v) = g(−u, −v). Changing the variable in the integral v := −s yields the desired result. We will now turn to the main goal of this section, namely proving that the function ϕL has its maximum at zero. We will start with the massless case and extend the result to the general massive case. In the latter it is enough for us to assume R ≥ 1.

48) β Furthermore, N (ψ, (Jβ2 Iβ )ψ) ≥ −Zα (ψ, ( i=1 β χ2 (xi ) Z 2α χ21 (xi ) + 2 )ψ) + |xi − R| |xi + R| 2R (49) ZαN Z 2α ≥− + . R 2R Here we have dropped the inter-electronic potential and used (25). We obtain Eb = (ψ, Hψ) − Es ≥ −N (ψ, Lψ) − ZαN Z 2α + ≥ 0, R 2R (50) where the last inequality can be written as, Z˜ 2 α ˜ − R (ψ, Lψ) ≥ 0 − Zα 2 N (51) with Z˜ := Z/N . At this point we will bound the localization error in two different ways yielding, through equation (51), the conditions of the theorem.

5) with parameter ζ, to conclude fixed. Insert Γ MH MH ¯ (λ, β, ζ, α) < Eλ,β,ζ,α [Γ] Eext MH ¯ ¯ = Eλ,β, ¯ [Γ] − (ζ − ζ) ζ,α 1 ρ¯ |x| MH ¯ α) − (ζ − ζ) ¯ = Eext (λ, β, ζ, 1 ρ¯. 33) ¯ Together with the same argument, where ζ and ζ¯ are exchanged, we get, for ζ > ζ, − MH ¯ α) 1 (λ, β, ζ, E MH (λ, β, ζ, α) − Eext <− ρ < ext ¯ |x| ζ −ζ 1 ρ¯. 22), where the ρn are now the unique densities of the minimizers for ζ¯n → ζ, that the Hartree energy is differentiable in ζ, and MH ∂Eext ∂ζ = −A. 22), we show that MH ∂Eext ∂α = R.

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