Read e-book online Analytic Functions Błażejewko 1982: Proceedings of a PDF

By Boris Nikolaevič Apanasov (auth.), Julian Ławrynowicz (eds.)

ISBN-10: 3540127127

ISBN-13: 9783540127123

ISBN-10: 3540386971

ISBN-13: 9783540386971

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Read or Download Analytic Functions Błażejewko 1982: Proceedings of a Conference held in Błażejewko, Poland, August 19–27, 1982 PDF

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Additional resources for Analytic Functions Błażejewko 1982: Proceedings of a Conference held in Błażejewko, Poland, August 19–27, 1982

Example text

That u(x+lllxle )' s we have [ ~ Then, if ~(x+lllxle) lu(xl-u(x+lllxle) III xl s] p luk(xl-uk(x+lllxle s ) [lllxl s =[ ~ k~l[ u(x) > min ui(x+lllxle) =u(x+lllxle s )' 1~~q s lu(x)-uk(x+lllxle s )I p III xl ] q ]p = l~(x)-~(x+lllxles) lllxl I p ] 52 Petru Caraman and, as above, in the hypothesis (20), we obtain (19) in this case as well. Finally, since uE21, '1t Hence, since each we deduce by (19) that is an arbi trary function of Uk' q k Z cap (E ,E ,D) , 1 k=l p 0 as desired. The same argument still holds for (iii'), max ( u 1 (x) , ••• , u (i') q but with u(x) (x) ] • is trivial since u =0 is an admissible function; arguing as for (ii), we also establish (ii').

In this case, we precise that, if there are components with u Do n E1 =~, such that the restriction such that 01 Do it is enough to consider only admissible functions n E0 = I"lI YJ, ul = 0 and if there are components D1 Do then it is enough to consider only admissible = 1. In these two cases, VUIDo = vUI01 = 0, 01 so that, if we eliminate from the open set D all the components of functions such that uI these two kinds, the value of LEMMA 3. open D = (UD ) U (UD ) U (UD 1 ) , k k mm q q introduced above, while If ~ cap D p (E ,E ,D) o 1 remains unchanged.

And any 0 m we conclude that {(So(zm)u,u)U} In particular, U,WEU. Choosing A=l and A=i, suc- 29 Positive Definiteness and Holomorphy cessively, we conclude that is a Cauchy-sequence for any weakly complete, that weakly in U,WEU. So (Zm) , U to an operator for any S(~)=So(~) ~EDo' S(z)EB(u:u), s(u) (z)EQ, ZED, S(·) ZED. n(u:u) Thus - we conclude that is, in fact, in S(·)EH(D;B(U:U». ace). ~ ~:Q+n. and clearly UEU, the function Finally, since also and the proof is complete. 2) should be modified by using any ho1omorphic cover map Riemann mapping converges lim (So(zm)u,u)U = (S(z)u,u)U The last theorem can be extended somewhat by letting domain in B(u:U), S(')E(D;B(U:U» Moreover, since for any unit vector s (u) (z) is ho1omorphic in This implies, since every Hilbert space is being an element of n:n+Q We shall not pursue this point here.

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Analytic Functions Błażejewko 1982: Proceedings of a Conference held in Błażejewko, Poland, August 19–27, 1982 by Boris Nikolaevič Apanasov (auth.), Julian Ławrynowicz (eds.)


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