By Harthong J.
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1 J where 1/p+1/q=1. As in the proof of SCHMIDT's inequality of classical complex analysis we replace the domain G by a certain ball with the radius R=(IGI/IS 1 1)1/3 and the centre x. That means that G and BR(x) have the same volume. c~' , 3 ZQ \'"2-2'( dr ~ -R - 3-lq • whence the assertion follows. This proposition yields the following 2 3 g Corollary Let u 4< CH(G). itioD The operator TG:L H(G)----+L ~ p. q. q<~3 • and it holds the estimate -p ITG~[l l ] '(diQ",Gl ,,", 'I,H ~~ ~-Jp ~+%j"~ ~"~+" I.
E Q e> with respeot to G" and the density 37 a e an(a. 9) = bn(a. e) follows for eaoh of 5i in 51 direotion 815 ~ 51 . Makin. e oonsideration as for the proof of the maximum modulus theorem we let u(x) = v(x) for all x E G. Remark The zeros of·an H-relular funotion are not neoessarily lated [Fae5]·. See. for instanoe • a(x)' = (lege 1 -x1e,,) + (lege 2 -x2 e,,) . • 1 5 12 1 5 13 Let Theore. iso- (IBIBBSTRASS' Theore. ) (u( 1»i E R be a sequenoe of H-valued funotions in G. 3... af& at. 4 t , Let I be an arbitrary oompaot set oontained in G.
Af& at. 4 t , Let I be an arbitrary oompaot set oontained in G. By the help of CAUCHY's integral formula we get the followin, estimate: C£ max lu(i)(y)-u(j)(y)' .!. d (I( aGllflt-Z. are harmonic functions. 3, suoh that .!. YE 1 rx=YT a Klx-y C max with u 3 =~ k=e uke k on oompaot subsets of G. in partioular. Du(i) --+ Du . 2 •... in G • we finally obtain Beoause Du(i) u £ 'H(G). The last step is possible as the domain G oan be exhausted by compact sets. =" = • 38 Corollary 1 5 14 Let (u u(i). (i) )i t:.