By Gruson L., Skiti M.

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D1 d1 41 n . 1 we obtain (252) (Tn log 1 2 cn ) ≤ (1 + or (1)). 1 we find by easy calculations (253) Tn 1 1 = o( n ). d1 r2 Hence at this step we can continue the proof of the even case to get the estimate for the case k = 1. Now let show the adaptations to do to get the case k > 1. Focusing on two steps, we follow the proof in the even case up to formula 190 that we recall (254) ∂ β ϕλ,σ (x) = 2∂ β log( 1 1 n ) + oδ,Θ,C (1)( |β| ) for all multi-index β such that |β| ≤ . di 2 d min Hence from this we obtain (255) (Tn ϕλ,σ )2 = 4(Tn log( 1 2 1 2 )) + oδ,Θ,C (1)((Tn ) ).

Letting → 0 we are done. Now let us prove the first one. Writting n = 2k + 1 and recalling we are working in conformal normal coordinates around x, up to errors terms we can suppose we are on flat 1 space and that we have to compute (−∆) 2 (−∆)k H. First, reasoning as in the even case we have the following estimate for (−∆)k H(r) (235) Now we recall a well known that we will use to continue transform that we denote by (236) (−∆)k H(r) = O(r2−2k ). 3) our analysis. Given f ∈ L1 (Rn ) radial, it is well known that its Fourier fˆ is still radial and verifies the following formula ∞ n−2 fˆ(r) = 2πr− 2 n f (s)J n−2 (2πrs)s 2 ds, 2 0 where J n−2 is the Bessel function of first kind and of order 2 following asymptotics at 0 (237) J n−2 (t) = t n−2 2 2 n−2 2 .

1. 12) there exists a sequence ρl → 1 and ul such that the following holds : Pgn u + ρl Qng = ρl κPgn enul in M. 4 with Ql = ρl Qng and Q g α ul is bounded in C for every α ∈ (0, 1). Hence up to a subsequence it converges uniformly to a solution of (3). 1 is proved. 1 As said in the introduction, the condition Pgn non-negative is only required to make the exposition clear. Indeed if Pgn has some negative eigenvalues the arguments change in the following way. To obtain Moser-Trudinger type inequality and its improvement we impose the additional condition u ˆ ≤ C where u ˆ is the component of u in the direct sum of the negative eigenspaces.

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