By V.P. Maslov

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**Example text**

0 Definition 3. the generator. Theorem 1. A is called the generator of the semigroup or simply (a) The subset DA is dense in X and A is a closed operator. (b) DA - X and A is a bounded operator iff the semigroup U(t) is uniformly* continuous in t. This means that U(t) - U(T) I - 0 as t - T; in other words, U(t)X - U(T)X uniformly in x, IXP <. 1. T11e reader should avoid confusion with uniformity in t. 36 (c) U(t)DA C. DA for all t, and A commutes with U(t); AU(t)X = U(t)AX, For X E p(A), (A - A)-1 also commutes with U(t).

0 Tao Both limits exist obviously when -ImA > w. -iatU(t)xdt- x 0 0 for an arbitrary x E X. The operator x a is bounded for -ImA > w (its norm does not exceed therefore Mfoe(w + Hence for x E DA ifine-1ltU(t)Axdt - fim 1-t fine-i't[U(T + t) - U(t)]xdt = 0 o Ta0 ifine-iatU(t)Axdt -x, 0 just as above. Thus (4) is proved. RA(A) is bounded and defined everywhere, therefore closed. So A - A is closed as well and the proof of (a) is complete. To prove (5) we observe that n A E p(A), x E X dan 1RA(A)x - (-1)nn(RA(A))n+ix, (9) (RX(A))-1 (indeed, the resolvent of a closed operator A is a holomorphic function on It may be represented by the Neumann series p(A).

Hence we may pass to the limit as n -* m in (15),, obtaining x = X. kim - iuR -111 I4 (16) . Since R = ImRA does not depend on A (Lemma 2), (16) yields x e R (R denotes It remains to show that (a) implies the closure of R). Thus we obtain (a). N = Ker RA - (0). Just another application of Lemma 2 completes the proof We may assume that w = 0, as in the proof of Theorem 1. Then after this. M/IImAt for A lying in the lower half-plane. Thus the family of uRAu operators -iPR_ip, u E (0,'), is uniformly bounded.