Page 139 - Textos de Matemática Vol. 47

STONGLY DEPENDENT GAUSSIAN RANDOM FIELDS 129 Likewise, we deal with the sums related to A2 ⇥ B2. Indeed, since
(p,q)ln(nm)  k, with k > 0, we get
n m Xn Xm r i , j e x p u 2n , m ! i=p+1 j=q+1 ln(n m) 1 + wi,j
!! 2
n2m2 u2 1+ (p,q) ln(n m)
exp n,m
ln(nm) 2 nm

K n2m2 ln(nm) k+↵2ln(nm)
:i>p,j>q i,j ln(n m) i=p+1 j=q+1 nm n m ;
⇥ 8< Xn Xm r + :i=p+1j=q+1 i,j ln(ij) ln(ij)
9= ln(nm) ;

⇥ 8< s u p | r l n ( i j ) | + Xn Xm l n i j 1 9=
✓ ◆ ↵2 ln(n m) 1 ln(n m) (n m)2
deduce
i 1 u2n,m !
nm
Xp i=1
|ri,0 ⇢n,m|exp
1+wi,0
exp⇣ klnln(nm)⌘ ⇢ ⇣k+↵2 ln(nm)⌘⇥ o(1)+

Z 1Z 1 exp 2kln(nm) ln(nm) 0 0

|lnxy|dxdy
K2
K3{o(1)+o(1)}=o(1), n,m!+1.
k+↵2 ln(n m)
Let us consider now the sums associated with A1 ⇥ {0} and A2 ⇥ {0}. From nowonwetake 0.Duetothefactthatsupwi,0  sup wi,j = ⇤,we
i 1,j 0
 2n1+↵m
Kn1+↵ 2⇤m1 2⇤ (ln(nm)) 1⇤
1+
K(nm)1+↵ 2 ⇤ ln(nm)!0, n,m!+1.
!! 2 ⇤ u2 1+
exp
n,m 2
1+ 1+ 1+