“…Because H B−1 (e B ) = 0, H B−1 (x) − H B−1 (e B ) = H B−1 (ξ )(x − e B ) and x ranges onto (H B,1 (d − 1), H B,2 (d − 1)) := (x 1,B , x 2,B ), when B → ln d, (x 1,B , x 2,B ) → (x 1,ln d , x 2,ln d ), it follows |H B−1 (ξ )| 1/μ > 0 independent on B. Moreover H B (x) − H B (e B ) = (1/2)H B (ξ )(x − e B ) 2 = −(1/2ξ )(x − e B )2 . Thus there exists m > 0 such thatH B (x) − H B e B m 2 x − e B 2 m 2 μ 2 H 2 B−1 (x).Therefore, near ln d,taking x = L B,i (d − √ 1 − λ 2 u * 2 ), one derive ψ i,u * (λ) 2 mμ d − √ 1 − λ 2 u * 2 − e B = 2 mμ √ 1 − u * 2 − √ 1 − λ 2 u * 2 4 mμ √ 1 − λ 2 .…”