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C. generate, create datamatrix 2d barcode none with .net projects Windows Forms 1 + . ,. Proof For = 0 an .NET DataMatrix d = 1 these estimates follow from De nition 3.10.

2 and Theorem 3.10.6(iv).

For intermediate values of we interpolate between these two extreme values by using Lemma 3.9.8 as follows.

By that lemma and Theorem 3.10.6(iv), we have for all x X and t > 0 (recall that At x D (A) for t > 0) ( A) At x 2C.

At x 1 . ( A)At x + ( )At x X X 1 2C M0 (M1 /t + M0 ) e t x. X . C1 (1 + t )e t x. X . . Strongly continuous semigroups To get the second inequality we argue essentially in the same way but replace Theorem 3.10.6(iv) by De nition 3.

10.2: ( A) ( A) 1 x 2C. ( A) 1 x 1 . ( A)( A) 1 x X X C1 1 1 + . /. x. X C2 1 1 + . x. X .. Lemma 3.10.9 enabl es us to add the following conclusion to Theorem 3.

8.2 in the case of an analytic semigroup: Theorem 3.10.

10 Let A be an analytic semigroup on X , and de ne the spaces X , R as above. Let s R, xs X , 1 < p , and f p L loc ([s, ); X ), with < 1 1/ p. Then the function x given by (3.

8.2) is a strong solution of (3.8.

1) in X . Proof This follows from (3.8.

2), Lemma 3.10.9, and H lder s inequality.

o Let us end this section with a perturbation result. The feedback transform studied in 7 leads to a perturbation of the original semigroup of the system, so that the generator A of this semigroup is replaced by A + T for some operator T . In the analytic case we are able to allow a fairly large class of perturbations T without destroying the analyticity of the perturbed semigroup.

Theorem 3.10.11 Let A be the generator of an analytic semigroup on the Banach space X , and de ne the fractional spaces X , R, as in Section 3.

9. If T B(X ; X ) for some , R with < 1, then the operator (A + T T ). X generates an analytic semigroup A X 1 on X 1 . For [ 1, + 1], T T T the spaces X are invariant under A X 1 , and the restriction A X of A X 1 to T X is an analytic semigroup on X . The generator of A X is (A + T ). X +1 if [ visual .net DataMatrix 1, ], and it is the part of A + T in X if ( , + 1].9 Moreover, T if 0 [ 1, + 1] (so that A.

X is an analytic s emigroup on X ) and if we let T X , R, be the analogues of the spaces X with A replaced by A + T , then T X = X for all [ 1, + 1]. Proof We begin by studying the special case where = 1 and 0 < 1. By Theorem 3.

10.6, every in some half-plane C+ belongs to the resolvent set of . See De nition 3.14.12 and Theorem 3.14.14. 3.10 Analytic semigroups and sectorial operators A, and for all C+ , ( A) 1 C . (3.10.9).

We claim that A + T has the same property on some other half-plane C+ . For all C+ we may write A T in the form A T = ( A)(1 ( A) 1 T ). Here A maps X 1 one-to-one onto X , so to show that A T is invertible it suf ces to show that 1 ( A) 1 T is invertible in B(X 1 ).

Fix some (A). Then ( A) T B(X 1 ), and ( A) 1 T. B(X 1 ). = ( A) ( A) 1 ( A) T ( A) ( A) 1. B(X 1 ). B(X 1 ) B(X 1 ) ..
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