By Tarantello G.

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R∞ Consequently 1/2 R∞ |fn+p (x) − fn1/2 (x)|2 µ(dx) n+p = k=n+1 dνk (xk ) − 2 dµk n+p k=n+1 dνk (xk ) dµk R∞ = 2 1 − 1/2 + 1 µ(dx) n+p dνk (xk ) dµk k=n+1 1/2 µk (dxk ) R n+p =2 1− H(µk , νk ) . 6) k=n+1 On the other hand we know by assumption that ∞ H(µ, ν) = H(µk , νk ) > 0, k=1 or, equivalently, that ∞ − log H(µ, ν) = − log[H(µk , νk )] < +∞. k=1 22 Chapter 1 Consequently, for any ε > 0 there exists nε ∈ N such that if n > nε and p ∈ N, we have n+p − log[H(µk , νk )] < ε. 6) if n > nε we have R∞ fn |2 dµ ≤ 2(1 − e−ε) .

This fact can be easily proved, using convolutions, when dim H < ∞. Let in fact ρ ∈ Cb∞ (H) be nonnegative, with compact support and such that H ρ(x)dx = 1. Then it is easy to see that setting ϕt (x) = t−n ρ H x−y t ϕ(y)dy, t > 0, we have ϕt ∈ Cb∞ (H) and limt→0 ϕ − ϕt 0 =0. If dim H = ∞, it was proved in 1954 by J. Kurtzweil [152] that there exists a sequence (ϕε ) ⊂ U Cb (H) ∩ C ∞ (H) such that limε→0 supx∈H |ϕ(x) − ϕε (x)| = 0. However in 1973 A. S. Nemirowski and S. M. Semenov [176] proved that U Cb2 (H) is not dense in U Cb (H), whereas U Cb1,1 (H) is.

1. 4). Let us prove (ii). Let ϕ ∈ U Cb (H), t > 0, and ϕt = Vt/2 Ut ϕ. Then ϕt is Lipschitz 2 continuous. 4 the mapping x → ϕt (x) + |x|t is convex. 6, since the mapping x → Ut ϕ(x) − |x|2 t is is concave too. 1. Finally, we prove (iii). 5). We have |ϕ(x) − ϕt (x)| = |ϕ(x) − Vt/2 Ut ϕ(x)| ≤ |ϕ(x) − Vt/2 ϕ(x)| + |Vt/2 ϕ(x) − Vt/2 Ut ϕ(x)|. 6) and the fact that Vt is a contraction, we ﬁnd 1 α |ϕ(x) − ϕt (x)| ≤ Cα [ϕ]α2−α (t/2) 2−α + ϕ − Ut/2 ϕ 0 . 5) follows. 6). We ﬁrst remark that for any ε > 0 and x ∈ H, there exists zε,x ∈ H such that |zε,x − y|2 |zε,x − x|2 ϕt (x) < inf ϕ(y) + − + ε.

### A quantization property for blow up solutions of singular Liouville-type equations by Tarantello G.

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