џ 1. Introduction - Issues of Analysis

The theorem on existence of singular solutions 31[] −11Ψ(x) = x −(p − 1)! F (p) (x 0 )[h] p−1 [F (x)], x ∈ U ε (x 0 ).Let us rst prove that there exists a number q, 0 < q < 1, such thath(Ψ(s 1 ), Ψ(s 2 )) ≤ q · ‖s 1 − s 2 ‖, (7)for any s 1 , s 2 ∈ U ε (x 0 ), such that s 1 = x 0 + h + u 1 , s 2 = x 0 + h + u 2 ,where ‖u i ‖ ≤ ‖h‖R , i = 1, 2 and {R = max 21,ε·C · ∥1(p−1)!∥F (p) (x 0 ) ∥ } .By Lemmas 2 and 3, we haveh(Ψ(s 1 ), Ψ(s 2 )) = inf {‖z 1 − z 2 ‖ : z i ∈ Ψ(s i ), i = 1, 2} === inf{‖z 1 − z 2 ‖ :1(p − 1)! F (p) (x 0 )[h] p−1 [z i ] =1(p − 1)! F (p) (x 0 )[h] p−1 [s i ] − F (s i ), i = 1, 2} ≤≤ inf{‖z‖ :1(p − 1)! F (p) (x 0 )[h] p−1 [z]1=(p − 1)! F (p) (x 0 )[h] p−1 [s 1 − s 2 ] − F (s 1 ) + F (s 2 )} ≤(p − 1)!≤‖h‖ p−1 C 1 ·∥ F (s 11) − F (s 2 ) −(p − 1)! F (p) (x 0 )[h] p−1 [s 1 − s 2 ]∥ .Further, taking into account Lemma 4, we have≤ supθ∈[0,1]∥ F (s 11) − F (s 2 ) −(p − 1)! F (p) (x 0 )[h] p−1 [s 1 − s 2 ]∥ ≤∥ F ′ (s 2 + θ(s 1 − s 2 )) −From the Taylor expansion, we obtain1(p − 1)! F (p) (x 0 )[h] p−1 ∥ ∥∥∥· ‖s 1 − s 2 ‖ . (8)F ′ (s 2 + θ(s 1 − s 2 )) = F ′ (x 0 ) + . . . ++ 1(p−1)! F (p) (x 0 )[s 2 − x 0 + θ(s 1 − s 2 )] p−1 + ω(h, u 1 , u 2 , θ) == 1(p−1)! F (p) (x 0 )[s 2 − x 0 + θ(s 1 − s 2 )] p−1 + ω(h, u 1 , u 2 , θ), (9)