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practice3 - not hold ± 1 Problem Let x n and y n be...

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MATH 4317 Real Analysis I SOME RECOMMENDED PROBLEMS WITH SOLUTIONS Here are a few practice problems with solutions. Try to work these WITHOUT looking at the solutions! After you write your own solution, you can compare to my solution. Your solution does not need to be identical—there are often many ways to solve a problem—but it does need to be CORRECT. Problem. Let ( x n ) and ( y n ) be bounded sequences of nonnegative real numbers. Let x * = lim sup ( x n ). Prove that lim sup n →∞ x n y n lim sup n →∞ x n lim sup n →∞ y n . Solution Set x * = lim sup n →∞ x n , y * = lim sup n →∞ y n , z * = lim sup n →∞ x n y n . We want to show that z * x * y * . For m N define u m = sup n m x n , v m = sup n m y n , w m = sup n m x n y n . By one of the equivalent characterizations of limsup, we then have that x * = lim m →∞ u m , y * = lim m →∞ v m , z * = lim m →∞ w m . We have the following inequalities for each m (WHY?): w m u m v m . Hence z * = lim m →∞ w m lim m →∞ u m v m = lim m →∞ u m lim m →∞ v m = x * y * . Question: Must we actually have equality, or does there exist an example where equality does
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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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