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Unformatted text preview: not hold? ± 1 Problem. Let ( x n ) and ( y n ) be bounded sequences of nonnegative real numbers. Let x * = lim sup ( x n ). Prove that if y * = lim ( y n ) exists, then lim sup ( x n y n ) = x * y * . Solution Set z * = lim sup n →∞ x n y n . Choose ε > 0. Then there exists an N > 0 such that n > N = ⇒ y *ε < y n < y * + ε. Therefore, if we set u m = sup n ≥ m x n , v m = sup n ≥ m x n y n , then for m > N we have v m = sup n ≥ m x n y n ≥ ( y *ε ) sup n ≥ m x n = ( y *ε ) u m . Hence z * = lim m →∞ v m ≥ ( y *ε ) lim m →∞ u m = ( y *ε ) x * . Since ε is arbitrary, we conclude that z * ≥ y * x * . Also, the preceding problem implies that z * ≤ x * y * (why?), so we conclude that we have z * = x * y * . ± 2...
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 Summer '07
 Staff
 Math, sup xn, sup xn yn, lim wm

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