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Unformatted text preview: Serge Ballif MATH 502 Homework 4 February 15, 2008 (1) Consider a continuous family { f t ,t ∈ ( 1 , 1) } of functions that are continuous on the closed unit disc U and holomorphic on the open disc U . (In other words, t 7→ f t is a continuous map from ( 1 , 1) to the Banach space C ( U ) .) Suppose that f has a single, simple zero in U , say z . Show that for  t  sufficiently small the function f t also has a single, simple zero in U , say z t and that the mapping t 7→ z t is continuous. What can you say in the case of a multiple zero? Define ε = min { f ( z )  :  z  = 1 } (the minimum exists on a compact set). Since the map t 7→ f t is continuous, we know that for each u ∈ U there exists δ > such that  t s  < δ implies  f t ( u ) f s ( u )  < ε . We are concerned with the case where s = 0, and u = z . For  t  < δ we have  f t ( z ) f ( z )  < ε = min { f ( z )  :  z  = 1 } . Hence, we can invoke Rouch´ es theorem to conclude that f and f t (  t  < δ ) have the same number of zeros inside the unit disc; namely, they both have a single simple zero....
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This note was uploaded on 04/01/2008 for the course MATH 502 taught by Professor Johnridener during the Summer '08 term at Pennsylvania State University, University Park.
 Summer '08
 JOHNRIDENER
 Math

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