џ 1. Introduction - Issues of Analysis

32 A. Prusinska, A. Tret'yakovwhere‖ω(h, u 1 , u 2 , θ)‖ ≤supx∈U ε (x 0 )On account of R and ‖u i ‖, i = 1, 2, we have∥∥∥F (p+1) (x)[h + u 2 + θ(s 1 − s 2 )] p ∥∥ .‖h + u 2 + θ(s 1 − s 2 )‖ ≤ 4 · ‖h‖,and from assumption, ‖h‖ ≤ ε 2. By the above,Moreover,‖ω(h, u 1 , u 2 , θ)‖ ≤ C · ‖h‖ p ≤ 4 p · C · ε2 · ‖h‖p−1 . (10)F (p) (x 0 )[h + u 2 + θ(s 1 − s 2 )] p−1 =∑p−1Cp−1F i (p) (x 0 )[h] p−1−i [u 2 + θ(s 1 − s 2 )] i =i=0∑p−1F (p) (x 0 )[h] p−1 +i=1C i p−1F (p) (x 0 )[h] p−1−i [u 2 + θ(s 1 − s 2 )] i , (11)where‖u 2 + θ(s 1 − s 2 )‖ ≤ 3 · ‖h‖/R. (12)Taking into account the choice of R,∥ ∑p−1∥∥∥∥ C i∥p−1F (p) (x 0 )[h] p−1−i [u 2 + θ(s 1 − s 2 )] i ≤i=1∥≤ ∥F (p) ∑p−1(x 0 ) ∥ · Cp−1‖h‖ i p−1−i (3 · ‖h‖) i /R i ≤i=1∥≤ ∥F (p) (x 0 ) ∥ · ‖h‖ p−1 · 4 p−1 /R ≤ 4 p · (p − 1)! · ε2 · C · ‖h‖p−1 . (13)And nally, applying (9)(13) in (8) we obtain∥ F (s 11) − F (s 2 ) −(p − 1)! F (p) (x 0 )[h] p−1 · [s 1 − s 2 ]∥ ≤≤ 4 p · ε · C · ‖h‖ p−1 · ‖s 1 − s 2 ‖.