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HW9ds - Key to Homework 9 Thansk to Da Zheng for providing...

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Key to Homework 9, Thansk to Da Zheng for providing the tex-file December 7, 2011 1. Page 204, #20 Let { f α } be a normal family of holomorphic functions on a domain U . Prove that { f 0 α } is a normal family. Proof. Pick any sequence { f 0 n } j { f 0 α } , then by the given condtion, for the sequence { f n } , there is a subsequence which converges uniformly on compact sets of U , say { f n k } . By Corollary 3.5.2, { f 0 n k } converges uniformly on compact subset of U . Hence, for any sequence of { f 0 n } , we can find a subsequence which converges uniformly on compact subsets, which implies that { f 0 α } is a normal family by definition. 2 2. Page 205, #24 Let Ω j C be a bounded domain and let { f j } be a sequence of holomorphic functions on Ω. Assume that Z Ω | f j ( z ) | 2 dxdy < C < where C does not depend on j . Prove that { f j } is a normal family. Proof. First we note that, as the hint given in the problem, you can use the hint in Problem 8, chapter 4 (Page 146), which you need to show that, if f is holomorphic on D ( Q, ), then | F ( Q ) | 2 1 π 2 Z D ( Q, ) | F ( z ) | 2 dxdy. We will follow the hint given on Page 146 in proving the above. Below is the proof of problem #24 1
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Solution : We will show that { f j } is unformly bounded on every compact sub- sets of Ω, then the Montel theorem in the book will imply that { f j } is nor- mal on Ω. To do so, let K Ω be compact. By the Lebesgue number lemma (see Munkers: Topology P. 175-176, or a better proof can be found at http://mathblather.blogspot.com/2011/07/lebesgue-number-lemma-and-corollary.html), there exists r K > 0 such that for each z K , D ( z, r K )) Ω. Now fix Q K . by the Cauchy integral formula, for every 0 r r K , f 2 j ( Q ) = 1 2 πi I ∂D ( Q,r ) f 2 j ( ζ ) ζ - Q So, if we parameterize ∂D ( Q, r ) as re , where θ [0 , 2 π ]. | f j ( Q ) | 2 = 1 2 πi I ∂D ( Q,r ) f 2 j ( ζ ) ζ - Q 1 2 π I ∂D ( Q,r ) f 2 j ( ζ ) ζ - Q | | = 1 2 π Z 2 π 0 | f j ( Q + re ) | 2 r rdθ = 1 2 π Z 2 π 0 | f j ( Q + re ) | 2 Now, use the Fubini theorem (and the polar coordinates) as well as the above inequality, Z D ( Q,r K ) | f j ( x, y ) | 2 dxdy = Z r K 0 Z 2 π 0 | f j ( Q + re ) | 2 rdθdr 2 π Z r K 0 | f j ( Q ) | 2 rdr 2 π r 2 K 2 | f j ( Q ) | 2 . Thus, by the assumption, C > Z Ω | f j ( z ) | 2 dxdy Z D ( Q,r K ) | f j ( x, y ) | 2 dxdy = πr 2 K | f j ( Q ) | 2 . Hence, for every Q K , | f j ( Q ) | 2 C πr 2 K which proves our claim. 2 3.( Marty’s Theorem ) 2
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Let F be a family of holomorphic functions on a region U on C . Prove that F is normal (in the general sense) if and only if for every compact subset K of U , there is a constant C K such that f # ( z ) C K for all Z K and f ∈ F , where f # ( z ) := | f 0 ( z ) | 1 + | f ( z ) | 2 Proof. ” = ” Suppose that we have F as a normal family, but it does not satisfy the Marty’s Criterion , i.e. there exists a compact subset K , a sequence of points z n K , and a sequence of functions { f n } j F , such that f # n ( z n ) := | f 0 ( z n ) | 1 + | f ( z n ) | 2 n, for each n However, F being a normal family implies that either { f n } has a uniformly convergent subsequence on K , say { f n k } , or uniformly divergent subsequence on K , say { f n l } .
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