Page 136 - Textos de Matemática Vol. 47

126 M. G. TEMIDO
Proof. We first assume > 0. When = 0, the proof of (2.2) for (i,j) in In,m\{(0, 1), (1, 0)} can be seen in [2]. By now we also assume that {un,m} is a sequence of normalized levels, that is
nm(1 (un,m))!⌧, n,m!+1.
Consider (k, `) = sup wi,j and let ↵ be a real number such that
i k,j `
0<↵< 1 (1,0)  1 (1,1).
1+ (1,0) 1+ (1,1)
Take also p := [n↵], q = [m↵], := (1,1) and ⇤ := (1,0).
We split the sum in (2.2) into eight terms. Namely, with A1 := {i 2 N : 1ip},A2 :={i2N:p<in},B1 :={j2N:1jq}, B2 :={j2N:q<j<m},weconsiderthesubsetsA1⇥B1,A1⇥B2,
A2 ⇥B1, A2 ⇥B2, A1 ⇥{0}, A2 ⇥{0}, {0}⇥{B1} and {0}⇥{B2}.
In this proof K, K1, K2,... denote suitable constants.
Take into account that 1 (x) ⇠ (x)/x, x!+1, from nm(1 (un,m))!
⌧, n, m ! +1,!we deduce exp u2n,m ⇠Kun,m,
2 n m that is un,m ⇠ p2 ln(n m),
K >0,
and 2ln(nm) ⇠1, n,m!+1, u 2n , m
! +1. Considering un,m = x/an,m +
Xp Xq u2n,m nm |ri,j ⇢n,m|exp 1+wi,j
i=1 j=1
 2n1+↵m1+↵ exp⇣ u2n,m ⌘
1+
⇣ ⇣u2 ⌘⌘2  2n1+↵m1+↵ exp n,m 1+
2
Kn1+↵ 2 m1+↵ 2 (u )2 1+ 1+ n,m 1+
1 1+ 1+
K (nm)1+↵ 2 (ln(nm)) 1 !0, n,m!+1,
n, m bn,m, x 2 R, similarly, we obtain
a = (2 ln nm)1/2 and b n,m n,m
a 1 ln(4⇡ ln(n m))/2. (2.3) n,m n,m
= a
For the term concerning A1 ⇥ B1, we have s!uccessively