Generalized Ordinary Differential Equations in Abstract Spaces and Applications. Группа авторов

Generalized Ordinary Differential Equations in Abstract Spaces and Applications

target="_blank" rel="nofollow" href="#fb3_img_img_25c40c62-77b8-5594-805c-6252b7a22d4f.png" alt="epsilon greater-than 0"/>, there is a division d equals left-parenthesis t Subscript i Baseline right-parenthesis element-of upper D Subscript left-bracket a comma b right-bracket

d equals left-parenthesis t Subscript i Baseline right-parenthesis element-of upper D Subscript left-bracket a comma b right-bracket

for which

parallel-to f Subscript k Baseline left-parenthesis t right-parenthesis minus f Subscript k Baseline left-parenthesis s right-parenthesis parallel-to less-than StartFraction epsilon Over 3 EndFraction comma

for every k element-of double-struck upper N and t Subscript i minus 1 Baseline less-than s less-than t less-than t Subscript i , i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue .

Take tau Subscript i Baseline element-of left-parenthesis t Subscript i minus 1 Baseline comma t Subscript i Baseline right-parenthesis . Because the sequence left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N converges pointwisely to f 0 , we have f Subscript k Baseline left-parenthesis t Subscript i Baseline right-parenthesis right-arrow f 0 left-parenthesis t Subscript i Baseline right-parenthesis and also , for i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue . Thus, for every epsilon greater-than 0 , there is k 0 element-of double-struck upper N such that, whenever k greater-than k 0 , we have parallel-to f Subscript k Baseline left-parenthesis t Subscript i Baseline right-parenthesis minus f 0 left-parenthesis t Subscript i Baseline right-parenthesis parallel-to less-than StartFraction epsilon Over 3 EndFraction and for i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue period

Take an arbitrary t element-of left-bracket a comma b right-bracket . Then, either t equals t Subscript i for some , or t element-of left-parenthesis t Subscript i minus 1 Baseline comma t Subscript i Baseline right-parenthesis for some . In the former case, parallel-to f Subscript k Baseline left-parenthesis t right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to less-than StartFraction epsilon Over 3 EndFraction period The other case yields

parallel-to f Subscript k Baseline left-parenthesis t right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f Subscript k Baseline left-parenthesis t right-parenthesis minus f Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis minus f 0 left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f 0 left-parenthesis tau Subscript i Baseline right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to less-than epsilon period

Then, parallel-to f Subscript k Baseline minus f 0 parallel-to less-than epsilon and, therefore, f Subscript k Baseline right-arrow f 0 uniformly on left-bracket a comma b right-bracket .

Lemma 1.14: Let be a sequence in . The following assertions hold:

1 if the sequence of functions converges uniformly to as on , then , for , and , for ;

2 if the sequence of functions converges pointwisely to as on and , for , and , for , where , then the sequence converges uniformly to as .

Proof. We start by proving left-parenthesis normal i right-parenthesis . By hypothesis, the sequence left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N converges uniformly to f 0 . Then, Moore–Osgood theorem (see, e.g., [19]) implies

limit Underscript k right-arrow infinity Endscripts limit Underscript s right-arrow t Superscript minus Baseline Endscripts f Subscript k Baseline left-parenthesis s right-parenthesis equals limit Underscript s right-arrow t Superscript minus Baseline Endscripts limit Underscript k right-arrow infinity Endscripts f Subscript k Baseline left-parenthesis s right-parenthesis comma t element-of left-parenthesis a comma b right-bracket period

Therefore, f Subscript k Baseline left-parenthesis t Superscript minus Baseline right-parenthesis right-arrow f 0 left-parenthesis t Superscript minus Baseline right-parenthesis , for t element-of left-parenthesis a comma b right-bracket . In a similar way, one can show that , for every t element-of left-bracket a comma b right-parenthesis .

Now, we prove left-parenthesis i i right-parenthesis . It suffices to show that Скачать книгу