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Pitt MATH 1530 - HOMEWORK

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Chapter 5 1 a Let fk be a sequence of functions from A ⊂ n to m. Suppose there are constants mk such that ||fk(x) – f(x)|| ≤ mk for all x ∈ A and such that mk → 0. Prove that fk → f uniformly. For ε > 0, let K be so that k ≥ K ⇒ mk < ε. Then k ≥ K implies ||fk(x) – f(x)|| < ε for all x ∈ A. Hence, fk converges uniformly to f. b If mk → m ∈  and ||fk(x) – fl(x)|| ≤ |mk – ml| for all x ∈ A, show that fk converges uniformly. ||fk(x) – fl(x)|| ≤ |mk – ml| ≤ |m – m| ≤ ε, ∀ε > 0. Therefore, fk converges uniformly. 4 Let fn: [1,2] →  be defined by fn(x) = x/(1+x)n. a Prove that ∑∞=1)(nnxf is convergent for x ∈ [1,2]. By the ratio test, 1)1/(1)1/()1/(lim1<+=+++∞→xxxxxnnn. Hence, the series converges. b Is it uniformly convergent? Yes. ∑∑∑∞+==∞=+=+−+111)1/()1/()1/(knnknnnnxxxxxx. Then 0)1()1(1)1(1>+−+=++nnxxnxxdxd. The maximum error on [1,2] is achieved at x = 2. Since the series converges at x = 2, the error can be made smaller than any ε > 0 by choosing k large enough to achieve uniformity. c Is ()∑∫∫∑∞∞=121211)()( dxxfdxxfnn Since the series converges uniformly by part b, ()∑∫∫∑∞∞=121211)()( dxxfdxxfnn follows from Corollary 5.3.2. 5 Suppose that fk → f uniformly, where fk: A ⊂ n → ; gk → g uniformly, where gk: A → ; there is a constant M1 such that ||g(x)|| ≤ M1 for all x; and there is aconstant M2 such that ||f(x)|| ≤ M2 for all x. Show that fkgk → fg uniformly. Find a counterexample if M1 or M2 does not exist. Are M1 and M2 necessary for pointwise convergence? ||fk(x)gk(x) – f(x)g(x)|| = ||fk(x)gk(x) – fk(x)g(x) + fk(x)g(x) – f(x)g(x)|| ≤ ||fk(x)|| ||gk(x) – g(x)|| + ||g(x)|| ||fk(x) – f(x)||. Let ε > 0 and x ∈ A. Since fk(x) converges uniformly f(x), there exists N1 such that k ≥ N1 ⇒ ||fk(x)|| < ||f(x)|| + 1 < 1 + M2 and there exists N2 such that k ≥ N2 ⇒ ||fk(x) – f(x)|| < ε2M1 . There exists N3 such that k ≥ N3 ⇒ ||gk(x) – g(x)|| < ε2(1+M2) . Let k ≥ max{N1,N2,N3}. Then ||fk(x)gk(x) – f(x)g(x)|| < (1+M2)ε2(1+M2) + M1ε2M1 = ε, ∀x ∈ A. If M1 or M2 does not exist, then it need not be uniformly convergent. For example, fk(x) = x → x converges uniformly on  and gk(x) = 1/k → 0 converges uniformly on . But fk(x)gk(x) = x/k → 0 converges nonuniformly on . However, for pointwise convergence M1 or M2 are not necessary. In this case 1 + M2 can be replaced by 1+ ||f(x)|| and M1 can be replaced by ||g(x)|| in the proof for uniform convergence. 9 Suppose that the functions gk are continuous and ∑∞=1kkg converges uniformly on A ⊂ n. If xk → x0 in A, prove that ∑∑∞=∞=→101)()(nnnknxgxg as k → 0. Since the functions gk are continuous and since ∑∞=1)(nknxg converges uniformly, by Corollary 5.1.5, g(x) = ∑∞=1)(nnxg is continuous. Therefore, xk → x0 implies that g(xk) → g(x0) by the definition of continuity. 13 In Theorem 5.3.3, show that fk → f uniformly. Let ε >0 and M > 0 such that |x – x0| ≤ M for all x ∈ (a,b). Then let k be such that |fn’(x) – fm’(x)| < ε and |fn(x0) – fm(x0)|< ε, ∀m,n ≥ k. By the mean value theorem there exists t ∈ (a,b) for each x ∈ (a,b) and n,m ≥ k such that |fn(x) – fm(x)| ≤ |fn(x) – fm(x) – (fn(x0) – fm(x0))| + |fn(x0) – fm(x0)| = |fn’(t) – fm’(t)| |x – x0| + |fn(x0) – fm(x0)| ≤ Mε + ε = (1 + M)ε Therefore, fk is a converges uniformly to f.19 Prove that 312sinxnnxn∑∞= defines a continuous function on all or . Let x0 ∈ , there exists a > 0 ∈  such that x0 ∈ [-a,a]. On [-a,a], 32sinxnnx ≤ a3/n2. Since ∑∞=123nna converges on [-a,a] by the Weierstrass M test 312sinxnnxn∑∞= converges uniformly on [-a,a] and therefore 312sinxnnxn∑∞= is continuous on [-a,a]. Since a is an arbitrary positive number the series is continuous on all of . 28 Let fn(x) = x/n. Is fn uniformly convergent on [0,396]? On ? Let n ∈ [0,396] and ∈ . If x > n then |x/n – 0| = x/k > 1. Therefore, fn is nonuniformly on  and on


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