Introduction to Several Complex Variables: Lectures by by Lipman Bers, Marion S. Weiner, Joan Landman

By Lipman Bers, Marion S. Weiner, Joan Landman

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Proof of Theorem 8. · iii) implies i). )CD. J J J J That UI:. is bounded follows from the fact that g assumes its maximum J over E on II {I t I =1}. and Ua,; j is bounqed. ii) implies iv). If Dis bounded, we may choose If Dis unbounded, choose '(z) =max ~(z) (-log~(z),

We now extend -this definition as follows: : Definition 18. Let D be open in«:, 4>: D- B.. Then ,P is said to be subharmonic if: i) - 00 ~ cP < co ' 4> f . 00 ii) r,l is upper semi -continuous; i. e. lim sup 9 (p' ) < 9 (p) pr- p - - iii) for any domain DCC D; if h is harmonic in D and continuous on 0 0 ii00 , then h > ~ on SD implies h > t/> in D , - 0 - 0 Property 1 (Mean Value Property, 1). Let {I z-z0 I < r} C D, ~ sub- harmonic in D. Then 211' f/1 (z ) < -21 0 - '11 S (z + reiB) df/ • tP 0 0 Property 2 (Mean Value Property, II).

But IJ is a closed set X€~.. X€1J and ~-&C B • therefore min AB(x) > 0; hence il C B . 0 0 X€1J (b) Now, let x, y € B . We must show that the lirie segment joining 0 them belongs to B . ,; (t), 0 0 0 < t < 1 , lying in B • joining x and y; - - 0 ~ (0 )=x, +(1 )=y. For t sufficiently small# the line segment (x# (/) (t) )CB . As t- 1 there cannot exist at such 0 that for all t < t the line segment 0 but the segment (x,~{t 0 Hence B is convex. ~ (t) '~B, Y- 0 )CB 0 because this would violate (a). 37 In the case r/J ~ c 2, it suffices to show that if B0 is any component of .

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