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math185f09-hw3sol

# math185f09-hw3sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

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MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 3 SOLUTIONS For a real-valued function of two real variables, u : Ω R R , we say that u is twice continuously differentiable if all second-order partial derivatives u xx , u yy , u xy , u yx exist and are continuous on Ω R . The set of all twice continuously differentiable functions on Ω R is denoted C 2 R ). 1. We mentioned Tauberian theorems in class. Here is an example of an easy one (easy relative to other Tauberian theorems). Let n =0 a n z n be a power series with radius of convergence 1 and suppose lim n →∞ na n = 0 . (a) Show that lim m →∞ m n =0 n | a n | m = 0 . ( Hint: Problem 4 (a), Problem Set 3 , Math 104 , Spring 2009.) Solution. Let ε > 0 be given. Since lim n →∞ na n = 0, there exists N 1 N such that n | a n | < ε/ 2 whenever n > N 1 . Now by the Archimedean property, there exists N 2 N such that | a 1 | + 2 | a 2 | + · · · + N 1 | a N 1 | N 2 < ε 2 . Hence | a 1 | + 2 | a 2 | + · · · + N 1 | a N 1 | m < ε 2 whenever m > N 2 . Now for m > max { N 1 , N 2 } , m n =0 n | a n | m = N 1 n =0 n | a n | m + m n = N 1 +1 n | a n | m < ε 2 + m - N 1 m ε 2 < ε. Hence lim m →∞ m n =0 n | a n | m = 0 . (b) Define a function f by f ( z ) = X n =0 a n z n for all | z | < 1 . Let x be a real variable and suppose the following left limit exists lim x 1 - f ( x ) = A. Show that the series n =0 a n converges to A . Date : October 18, 2009 (Version 1.0). 1

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Solution. Let x R and 0 x < 1. We have X m n =0 a n - f ( x ) = X m n =0 a n (1 - x n ) - X n = m +1 a n x n (1 - x ) X m n =0 | a n | (1 + x + · · · + x n - 1 ) + X n = m +1 | a n | x n < m (1 - x ) m n =0 n | a n | m + X n = m +1 n | a n | x n n . Let ε > 0. Since lim n →∞ na n = 0, there exists M 1 N such that m | a m | < ε/ 2 whenever m > M 1 . By (a), there exists M 2 N such that m n =0 n | a n | m < ε 2 for m > M 2 . Hence when m > max { M 1 , M 2 } , X m n =0 a n - f ( x ) < m (1 - x ) ε 2 + ε 2 m X n = m +1 x n = ε 2 m (1 - x ) + 1 m x m +1 1 - x < ε 2 m (1 - x ) + 1 m (1 - x ) . Then for 1 - x = 1 /m and m > max { M 1 , M 2 } , we get X m n =0 a n - f 1 - 1 m < ε, which implies that lim m →∞ X m n =0 a n = lim m →∞ f 1 - 1 m . Since lim x 1 - f ( x ) = A exists, we must have lim m →∞ f 1 - 1 m = A and so we get X n =0 a n = A. 2. Recall that C is both a real vector space of dimension 2 and a complex vector space of dimen- sion 1. A function ϕ : C C is called R -linear if ϕ is a linear transformation of real vector spaces, ie. ϕ ( λ 1 z 1 + λ 2 z 2 ) = λ 1 ϕ ( z 1 ) + λ 2 ϕ ( z 2 ) for all λ 1 , λ 2 R and z 1 , z 2 C . (2.1) It is called C -linear if ϕ is a linear transformation of complex vector spaces, ie. ϕ ( λ 1 z 1 + λ 2 z 2 ) = λ 1 ϕ ( z 1 ) + λ 2 ϕ ( z 2 ) for all λ 1 , λ 2 C and z 1 , z 2 C . (2.2) (a) Prove that if ϕ is C -linear, then it is R -linear. Give an example to show that the converse is false. Solution. This is obvious since R C and so (2.1) is a special case of (2.2). For a counterexample to the converse, consider the complex conjugate function, ϕ : C C , 2
ϕ ( z ) = z . For λ 1 , λ 2 R , ϕ ( λ 1 z 1 + λ 2 z 2 ) = λ 1 z 1 + λ 2 z 2 = ¯ λ 1 ¯ z 1 + ¯ λ 2 ¯ z 2 = λ 1 ¯ z 1 + λ 2 ¯ z 2 = λ 1 ϕ ( z 1 ) + λ 2 ϕ ( z 2 ) and so ϕ is R -linear. However, for λ 1 = i , z 1 = 1, λ 2 = z 2 = 0, we see that ϕ ( i ) = - i 6 = i = (1) and so it is not C -linear.

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