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### m621-hurwitz-theorem

Course: M 621, Fall 2009
School: UMass (Amherst)
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Word Count: 723

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MATH 621 COMPLEX ANALYSIS, NOTES ON A THEOREM OF WEIERSTRASS FARSHID HAJIR MARCH 1, 2007 16 : 46 Here is some clari cation for the proof of Theorem 5.3 from Chapter 2 of SteinShekarchi. Theorem 0.1. If f1 , f2 , f3 , . . . is a sequence of functions holomorphic on an open set C which converge to a function f on and if this convergence is uniform on all &lt;a href=&quot;/keyword/compact-subset/&quot;...

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MATH 621 COMPLEX ANALYSIS, NOTES ON A THEOREM OF WEIERSTRASS FARSHID HAJIR MARCH 1, 2007 16 : 46 Here is some clari cation for the proof of Theorem 5.3 from Chapter 2 of SteinShekarchi. Theorem 0.1. If f1 , f2 , f3 , . . . is a sequence of functions holomorphic on an open set C which converge to a function f on and if this convergence is uniform on all <a href="/keyword/compact-subset/" >compact subset</a> s of , then (1) the function f is holomorphic on , and (2) the sequence of derivatives fn converges uniformly to f on all <a href="/keyword/compact-subset/" >compact subset</a> s of . Proof of (2). . This theorem was apparently rst given by Weierstrass, not Hurwitz, as I mistakenly said in class. The proof of (1) using Morera from the book or lecture is straightforward. Here is a fussy elaboration of the proof given in SteinShekarchi. Let be a <a href="/keyword/compact-subset/" >compact subset</a> of . Since is compact and is open, there exists a chain of sets , with open and compact. (Easy veri cation left to the reader). Now there exists &gt; 0 such that , where := {z | D (z) }. We will now show that fn f uniformly on , which is all we need do since . Claim. If F (z) is holomorphic on (or even just on ), then 1 sup |F (z)| sup |F ( )|. z Note that although and are not compact, they are both contained in the compact set , hence the two sup s above are well-de ned real numbers. To prove the claim, we apply Cauchy s formula just as in the book and lecture: For all z , D (z) , so C (z) , giving us |F (z)| = verifying the claim. 1 1 2 i C (z) F ( ) dz z 1 2 sup |F (z)| 2 2 z 1 sup |F (z)|, z 2 FARSHID HAJIR MARCH 1, 2007 16 : 46 1 Now suppose &gt; 0 is given. Since fn f uniformly on integer N such that there exists an |fn (z) f (z)| &lt; /2 for all z and all n N. Then, for n N and z , we apply the claim to F (z) := fn (z) f (z) and nd that 1 |F (z)| = |fn (z) f (z)| sup |fn (z) f (z)| z 1 sup |fn (z) f (z)| z 1 &lt; . 2 We have shown that fn f uniformly on . The most common application of the above theorem is contained in: Corollary 0.2. If the functions e0 (z), e1 (z), e2 (z), . . . are holomorphic on the open set and the series n 0 en (z) converges to a function f on and it does so uniformly on <a href="/keyword/compact-subset/" >compact subset</a> s of , then f is holomorphic on and f (z) = n 0 en (z) for z . Later, after we prove the Maximum Modulus and Argument Principles, we ll be able to obtain the following theorems. Theorem 0.3. If en (z) are holomorphic on |z| &lt; R and f (z) = n en (z) converges uniformly on the circles Cr (0) for all 0 &lt; r &lt; R, then f is holomorphic on |z| &lt; R. The proof follows from the Corollary once we invoke the maximum modulus principle. Theorem 0.4 (Hurwitz). If fn are holomorphic and non-vanishing on and converge uniformly to f on <a href="/keyword/compact-subset/" >compact subset</a> s of , then f is either identically zero or non-vanishing on . We ll be able to deduce this Hurwitz from Weierstrass after we prove the Argument Principle. Alternative Hint for Exercise 2 from Chapter 2 of Stein-Shekarchi. To evaluate I = 0 sin(x)/xdx, we can note that I= 1 lim 2i 0 eix dx + x eix dx . x Now we integrate eiz /z over the indented semi-circle, consisting of a radius R semicircle in the upper half plane and the real ax...

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