A Proof of Remark 3.1 Suppose Assumption 2.2 is satisfied, lim$_{t}$$_{→}$$_{1}$ Ξ( t ) = ∞ and conditions (3.3) and (3.4) hold. Then L’Hˆ opital rule will give (notice that due to (3.4) and continuity of σ$_{z}$ ( t ) in the vicinity of 1 we have lim$_{t}$$_{→}$$_{1}$ ( 1 + σ 2 $_{z}$( t ) < 2) Since by L’Hˆ opital rule we have it follows that where f ( t ) = ∫ t 0 1 Σ$_{z}$( s )+ σ $^{2}$− $_{s}$ds and lim$_{t}$$_{→}$$_{1}$ f ( t ) = ∞ . Since lim$_{x}$$_{→∞}$ log( x ) x α = 0, for any α > 0 we need to show that for some α > 0 to establish (3.6). Consider any α ∈ (0 , 1) and denote by then for t ≥ t ∗ we have where c is some positive constant. Due to this expression and since lim$_{t}$$_{→}$$_{1}$ f ( t ) = ∞ , α < 1 and, 30 f 1 − $^{α}$( t ) = (1 − α ) ∫ t t ∗ 1 g ( s ) ds + c 0 < g ( t ) = f $^{α}$( t ) f $^{′}$( t ) , lim sup t → 1 f $^{α}$( t ) f $^{′}$( t ) < ∞ (A.14) 0 ≤ lim t → $_{1}$λ $^{2}$( t )Ξ( t ) log log (Ξ( t )) ≤ 1 2 lim t → 1 ( 1 + σ 2 $_{z}$( t ) lim t → 1 ( Σ$_{z}$( t ) + σ 2 − t ) log log ( λ − $^{2}$( t ) = 1 2 lim t → 1 ( 1 + σ 2 $_{z}$( t ) lim t → 1 ( Σ$_{z}$( t ) + σ 2 − t ) log (∫ t 0 1 Σ$_{z}$( s ) + σ 2 − s ds = 1 2 lim t → 1 1 + σ 2 $_{z}$( t ) lim t → 1 log( f ( t )) f $^{′}$( t ) , lim t → $_{1}$λ $^{2}$( t )Ξ( t ) = 0 , lim t → $_{1}$λ $^{2}$( t )Ξ( t ) log log (Ξ( t )) = 1 2 lim t → 1 ( 1 + σ 2 $_{z}$( t ) lim t → 1 ( Σ$_{z}$( t ) + σ 2 − t ) log log (Ξ( t )) .