hw1-sol - Modern Analysis Homework 1 Possible solutions 1...

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Unformatted text preview: Modern Analysis, Homework 1 Possible solutions 1. Consider ε = 1, and let f = lim n →∞ f n . By uniform convergence, there exists N ∈ N such that n > N implies k f n- f k u ≤ 1. Since each f n is bounded, this shows that f is bounded and that k f n k u ≤ k f k u + 1 if n > N . Set M = max max n =1 ,...,N k f n k u , k f k u + 1 . Then k f n k u ≤ M for all n ∈ N . 2. (a) Let f = lim n →∞ f n , g = lim n →∞ g n , and ε > 0. By uniform convergence, there exists N f ,N g ∈ N such that if n > N f and x ∈ E , then | f n ( x )- f ( x ) | < ε/ 2, and if n > N g and x ∈ E , then | g n ( x )- g ( x ) | < ε/ 2. It follows that if n > max( N f ,N g ) and if x ∈ E , then | ( f n ( x )+ g n ( x ))- ( f ( x )- g ( x )) | ≤ | f n ( x )- f ( x ) | + | g n ( x )- g ( x ) | < ε . (b) As in (a), let f = lim n →∞ f n , g = lim n →∞ g n , and ε > 0. By problem 1, we know that { g n } is uniformly bounded, ie there exists G > 0 such that for all n ∈ N and all x ∈ E , we have | g n ( x ) | < G . By the uniform convergence of { f n } , there exists N f ∈ N such that n > N f and x ∈ E implies that | f n ( x )- f ( x ) | < ε 2 G . By the uniform convergence of { g n } , we may find N g ∈ N such that n > N g and x ∈ E implies that | g n ( x )- g ( x ) | < ε 2(sup x ∈ E | f ( x ) | +1) . It follows that if n > max( N f ,N g ) and x ∈ E , then we have | f n ( x ) g n ( x )- f ( x ) g ( x ) | < | f n ( x )- f ( x ) | · | g n ( x ) | + | f ( x ) | · | g n ( x )- g ( x ) | < ε. 3. Many examples are possible; here’s one: let’s define two functions f,g : (0 , ∞ ) → R by f ( x ) = x if x ≤ 1 /n 1 /n if x > 1 /n and simply g n ( x ) = 1 /x for all n ∈ N . Since k f n k u ≤ 1 /n for all n ∈ N , it follows that f n converges uniformly to the zero function (and of course g n converges uniformly to 1 /x ). Then f n ( x ) g n ( x ) = 1 if x ≤ 1 /n 1 nx if x > 1 /n This function converges pointwise to the zero function, but does not do so uniformly: for a given n ∈ N , we have f n (1 /n ) g n (1 /n ) = 1. 4. (a) First note (say, by looking at the graph) that { x } is discontinuous precisely on Z . Similarly, for a given n ∈ N , { nx } is discontinuous precisely on 1 n Z (integral multiples of 1 n )). In fact, if p ∈ Z , then lim x → p/n- { nx } = 1 while lim x → p/n + { nx } = 0. For n ∈ N , let g n ( x ) = { nx } n 2 . Then | g n ( x ) | ≤ 1 /n 2 for all x ∈ R , so by the Weierstrass M-test, since ∑ 1 n 2 converges, it follows that the sequence f n ( x ) def = 1 ∑ n k =1 g k ( x ) converges uniformly to the function f . Also each f n ( x ) is clearly Riemann-integrable on any bounded domain [ a,b ], so by uniform convergence, we get that f too is Riemann-integrable and Z b a f ( x ) dx = lim n →∞ Z b a f n ( x ) dx = lim n →∞ Z b a n X k =1...
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hw1-sol - Modern Analysis Homework 1 Possible solutions 1...

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