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Sample text
C/ is also unique. e. a/ D f 1 af . a/ depends continuously on a > 0. 10) twice to find aaf Q 00 . a/ Q : On the other hand, aaf Q 00 . a/. a/ D a1=˛ . 10) we have 1 2 f 00 . a ˛ / for all a; > 0: ˛ to get f 00 . 1/ and integrate twice. 1/ C f. ˛ 1/ with some integration constant C 2 R. Since f 2 BF is non-trivial, we know that 0 < ˛ 6 1. a a1=˛ /. 14. We have ¹stable lawsº SD ID. Proof. 13. If g is (weakly) stable, f . / WD log g. / D ˇ ˛ C . Thus, for 0 < c < 1, f. 1 c/ is a Bernstein function, and we get that gc .
And f2 . ˇ /= ˇ 3 Bernstein functions 21 are again completely monotone. Since ˛ C ˇ 6 1, 7! ˛Cˇ 1 is completely monotone. 6, h0 is completely monotone. 7 says that on the set BF the notions of pointwise convergence, locally uniform convergence, and even convergence in the space C 1 coincide. 8. 1 fn . / D f . 0; 1/. 1 fn . k/ . 0; 1/. 0; / In both formulae we may replace lim infn by lim supn . Proof. 7(ii) we know that f 2 BF. e f and . 1 ! 0; 1/. 1 fn0 . / D f 0 . 0; 1/. k/ . By the mean value theorem, jfn .
This proves at once that n ! 0; 1/ as n ! 1. 0; / at all continuity points > 0 of . If such that j ! 0; j/ For a sequence of arbitrary j ! 0 we find continuity points ıj ; Áj , j 2 N, of such that 0 < ıj 6 j 6 Áj and ıj ; Áj ! 0. 0;Áj / Since both sides of the inequality coincide, the claim follows. 0C/ we find for each R > 1 a D lim f . 0 D lim lim fn . dt / ! 1 ŒR;1/ Letting R ! 1 n ŒRj ; 1/ ! 0: we get : That we do not need to restrict ourselves to continuity points Rj follows with a similar argument as for the coefficient b.