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Controllability of Single-valued Chapter | 7  219


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             there exists a positive number δ such that EjðΠ 2 zÞj # δ , which means that
             Π 2 ðzÞAB δ 0. Now, to prove that Π 2 maps bounded sets into equicontinuous
                    0
             sets in C .Let t 1 ; t 2 AJ; t 2 . t 1 , and consider B q as a bounded set as in Step
                    b
             1. Let zAB q , then if E . 0 and E # t 1 # t 2 , one has
                     EjðΠ 2 zÞðt 2 Þ 2 ðΠ 2 zÞðt 1 Þj # 3E   Ð    t 1 2E ½S ðt 2 2sÞ2S ðt 1 2sފðΠ 1 zÞðsÞds   2
                                     2
                                               0
                                                      0

                                           0
                                                                     2
                                          Ð
                                             t 1  0   0
                                           t 1 2ε  ½S ðt 2 2sÞ2S ðt 1 2sފðΠ 1 zÞðsÞds
                                      1 3E
                                                          2

                                          Ð
                                             t 2  0
                                      1 3E   S ðt 2 2sÞðΠ 1 zÞðsÞds
                                           t 1
                                      # 3ðt 1 2 EÞδ  Ð t 1 2E  jS ðt 2 2 sÞ 2 S ðt 1 2 sÞj ds
                                                                  2
                                                            0
                                                    0
                                                0
                                      1 3Eδ  Ð t 1 2ε  jS ðt 2 2 sÞ 2 S ðt 1 2 sÞj 2
                                            t 1
                                                       0
                                               0
                                                         2
                                      1 3ðt 2 2 t 1 Þδ  Ð  t 2  jS ðt 2 2 sÞj ds
                                                   0
                                                t 1
             as t 2 -t 1 and ε sufficiently small, the right-hand side of the above inequality
             tends to zero. By the Arzela Ascoli theorem it suffices to show that Π 2
             maps B q into a precompact set in H. Let 0 , t , b be fixed and let ε be a
             real number satisfying 0 , ε , t. For zAB q ,
                                        ð t2E
                                     0       0
                           ðΠ 2E zÞðtÞ 5 S ðEÞ  S ðt 2 s 2 EÞðΠ 1 zÞðsÞds
                                         0
                   0
             since S ðtÞ is a compact operator for t . 0, the set Z E ðtÞ 5 fðΠ 2E zÞðtÞ: zAB q g
             is precompact in H for every E; 0 , E , t. Furthermore
                    EjðΠ 2 zÞðtÞ 2 ðΠ 2E zÞðtÞj # 2E   Ð    t2E  S ðt2sÞ½ðΠ 1 zÞðsÞ2ðΠ 1E zÞðsފds   2
                                   2
                                            0

                                         0
                                    1 2E   Ð    t  S ðt2sÞðΠ 1 zÞðsÞds   2
                                            0
                                         t2ε

                                          2     Ð  t2E             2
                                    # 2Oϕ O 1 ðt 2 EÞ  EjðΠ 1 zÞðsÞ 2 ðΠ 1E zÞðsÞj ds
                                         A L     0
                                    1 2Oϕ O 1 E  Ð  t  EjðΠ 1 zÞðsÞj ds
                                                       2
                                          2
                                         A L  t2ε
                                    # 2Oϕ O 1 ðt2εÞ Ξ 1 2Oϕ O 1 ε δ
                                                2
                                                          2
                                                        2
                                          2
                                         A L           A L
             so ZðtÞ 5 fðΠ 2 zÞðtÞ: zAB q g is precompact in H. Hence Π 2 is completely
             continuous.
                                                       ^
                Step 3: To prove that there exists an open set UCCðJ; HÞ with z=2λΠðzÞ
                               ^
             for λAð0; 1Þ and yA@U. Let λAð0; 1Þ and
                                     λ  ð t             ð t
                  zðtÞ 5λðΠzÞðtÞ5λφð0Þ1  ðt2sÞ α21 BðsÞuðsÞds1λ  fðs;z s 1y s Þds
                                    ΓðαÞ 0               0
                        λ  ð t      ð  s
                      1     ðt2sÞ α21  σðτ;z τ 1y τ ÞdWðτÞ ds
                       ΓðαÞ 0       2N
                               2
                                      λ  ð  s
                      1λ S ðt2sÞ φð0Þ1    ðs2τÞ α21 BðτÞuðτÞdτ
                        Ð
                         t 0
                         0
                               4
                                     ΓðαÞ 0
                                                                    3
                                                  τ
                                         s


                                                 ð
                                        ð
                      1  Ð  s fðτ;z τ 1y τ Þdτ1  1  ðs2τÞ α21  σðη;z η 1y η ÞdWðηÞ dτ 5 ds:
                        0
                                     ΓðαÞ 0       2N
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