MATH 321 Spring 2015 Homework 1 Solutions

# MATH 321 Spring 2015 Homework 1 Solutions - Homework 1...

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Homework 1 Solutions - Math 321,Spring 2015 1a Pointwise convergence: for a fix x [0 , ) If x = 0 , f n ( x ) = 0 for all n so it converges to 0. If x 6 = 0 , we have lim n →∞ nx 1 + n 2 x 2 = lim n →∞ x n 1 n 2 + x 2 = 0 0 + x 2 = 0 The convergence is not uniform on [0 , ): If f n converges uniformly to f then f must be identically zero (by above and uniqueness of limit). i.e. sup x [0 , ) nx 1 + n 2 x 2 0 , n → ∞ That is for any > 0 there is N such that for all n N, | nx 1+ n 2 x 2 | ≤ for every x > 0 (a single works for all x ). However, we have f n (1 /n ) = 1 / 2 for all n so the sequence cannot converges uniformly on this domain. (when speaking about uniform convergence, don’t forget to state the domain where the sequence converges uniformly on). Observe that any interval containing 0 or has 0 as a limit point will contain 1 n for all n sufficiently large and f n (1 /n ) = 1 / 2 so the sequence cannot converge to 0 uniformly on this domain. Now consider intervals of the form [ a, ) , a > 0 then for all x in this interval nx 1 + n 2 x 2 nx n 2 x 2 = 1 nx 1 an where a/n 0 , n → ∞ independent of x. In the same way, the sequence also converges uniformly on ( -∞ , a ] , a < 0 and hence also on any intervals that are subsets of them. In summary, the sequence converges uniformly on any interval not

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