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

Generalized Ordinary Differential Equations in Abstract Spaces and Applications

right-brace"/> is an equiregulated set. Indeed, by Lemma 1.13, f Subscript k

converges uniformly to f 0

Since the function f 0 is regulated, its lateral limits exist. Then, for every t 0 element-of left-bracket a comma b right-bracket and every epsilon greater-than 0 , there is delta greater-than 0 such that

StartLayout 1st Row 1st Column parallel-to f 0 left-parenthesis t 0 Superscript minus Baseline right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to 2nd Column less-than epsilon comma for t 0 minus delta less-than-or-slanted-equals t less-than t 0 comma 2nd Row 1st Column parallel-to f 0 left-parenthesis t right-parenthesis minus f 0 left-parenthesis t 0 Superscript plus Baseline right-parenthesis parallel-to 2nd Column less-than epsilon comma for t 0 less-than t less-than-or-slanted-equals t 0 plus delta comma EndLayout

for every t element-of left-bracket a comma b right-bracket .

But the hypotheses say that we can find n 0 element-of double-struck upper N such that, for n greater-than-or-slanted-equals n 0 , we have

parallel-to f Subscript n Baseline left-parenthesis t 0 minus delta right-parenthesis minus f 0 left-parenthesis t 0 minus delta right-parenthesis parallel-to less-than epsilon comma parallel-to f Subscript n Baseline left-parenthesis t 0 right-parenthesis minus f 0 left-parenthesis t 0 right-parenthesis parallel-to less-than epsilon comma parallel-to f Subscript n Baseline left-parenthesis t 0 Superscript plus Baseline right-parenthesis minus f 0 left-parenthesis t 0 Superscript plus Baseline right-parenthesis parallel-to less-than epsilon comma

parallel-to f Subscript n Baseline left-parenthesis t 0 plus delta right-parenthesis minus f 0 left-parenthesis t 0 plus delta right-parenthesis parallel-to less-than epsilon and parallel-to f Subscript n Baseline left-parenthesis t 0 Superscript minus Baseline right-parenthesis minus f 0 left-parenthesis t 0 Superscript minus Baseline right-parenthesis parallel-to less-than epsilon period

When t element-of left-bracket a comma b right-bracket satisfies t 0 minus delta less-than-or-slanted-equals t less-than t 0 , we have, for every n greater-than-or-slanted-equals n 0 ,

parallel-to f Subscript n Baseline left-parenthesis t 0 Superscript minus Baseline right-parenthesis minus f Subscript n Baseline left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f Subscript n Baseline left-parenthesis t 0 Superscript minus Baseline right-parenthesis minus f 0 left-parenthesis t 0 Superscript minus Baseline right-parenthesis parallel-to plus parallel-to f 0 left-parenthesis t 0 Superscript minus Baseline right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to plus parallel-to f 0 left-parenthesis t right-parenthesis minus f Subscript n Baseline left-parenthesis t right-parenthesis parallel-to less-than 3 epsilon period

On the other hand, when t element-of left-bracket a comma b right-bracket satisfies t 0 less-than t less-than-or-slanted-equals t 0 plus delta , we get, for n greater-than-or-slanted-equals n 0 ,

parallel-to f Subscript n Baseline left-parenthesis t right-parenthesis minus f Subscript n Baseline left-parenthesis t 0 Superscript plus Baseline right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f Subscript n Baseline left-parenthesis t right-parenthesis minus f 0 left-parenthesis t right-parenthesis parallel-to plus parallel-to f 0 left-parenthesis t right-parenthesis minus f 0 left-parenthesis t 0 Superscript plus Baseline right-parenthesis parallel-to plus parallel-to f Subscript n Baseline left-parenthesis t 0 Superscript plus Baseline right-parenthesis minus f 0 left-parenthesis t 0 Superscript plus Baseline right-parenthesis parallel-to less-than 3 epsilon period

But this yields the fact that left-brace f Subscript n Baseline right-brace Subscript n element-of double-struck upper N is an equiregulated sequence, and the proof is complete.

The next lemma guarantees that, if a sequence of functions left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N is bounded by an equiregulated sequence of functions, then is also equiregulated.

Lemma 1.15: Let be a sequence of functions in . Suppose, for each , the function satisfies

(1.2) parallel-to f Subscript k Baseline left-parenthesis s 2 right-parenthesis minus f Subscript k Baseline left-parenthesis s 1 right-parenthesis parallel-to less-than-or-slanted-equals StartAbsoluteValue h Subscript k Baseline left-parenthesis s 2 right-parenthesis minus h Subscript k Baseline left-parenthesis s 1 right-parenthesis EndAbsoluteValue comma

for every , where for each and the sequence is equiregulated. Then, the sequence is equiregulated.

Proof. Let epsilon greater-than 0 be given. Since the sequence left-brace h Subscript k Baseline right-brace Subscript k element-of double-struck upper N is equiregulated, it follows from Theorem 1.11 that 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 such that StartAbsoluteValue h Subscript k Baseline left-parenthesis t Superscript prime Baseline right-parenthesis minus h Subscript k Baseline left-parenthesis t right-parenthesis EndAbsoluteValue less-than-or-slanted-equals epsilon comma for every k element-of double-struck upper N and Скачать книгу