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Real Analysis { Project Exercise 1. (3+5+7=15 marks) For x 2 [0; =2] we de ne fn(x) = nx 1 + n sin(x) : 1. Find the pointwise limit of (fn), i.e. fund the function f : [0; =2] ! R such that fn(x) ! f(x) for all x 2 [0; =2]. Solution: 2. Prove that fn does not converge uniformly to f on [0; =2]. 3. Prove that for all a 2 (0; =2], fn ! f uniformly on [a; =2] (Hint: sin is non-decreasing on [a; =2], so sin(x)  sin(a) if x 2 [a; =2]). Solution: Exercise 2. (3+3+6+8=20 marks) 1. Let f : [0;1) ! R be continuous such that limx!+1 f(x) = 0. Prove that f is bounded on [0;1). Solution: 2. Deduce that for any a > 0 there exists C 2 R such that for all n  1 ne􀀀na  Ce􀀀na 2 : P 3. Deduce that nena converges for any a > 0. Solution: 4. Prove that the series P e􀀀nx+cos(nx) is, continuous and differentiable (with a continuous derivative) on (a;1) for any a > 0. Solution: 1. Let (fn) be a sequence of functions that are continuous on [a; b] and differentiable on (a; b). Use the Lipschitz estimate to prove that jfn(x)􀀀fp(x)􀀀(fn(c)􀀀fp(c))j  jb􀀀aj supy2(a;b) jf0n (y)􀀀fp 0 (y)j for all x 2 [a; b] and all n; p 2 N (make explicit the function on which you use the Lipschitz estimate). Solution: 2. Deduce that jfn(x) fp(x) jfn(c)fp(c)j + jbaj sup y2(a;b) n p jf0 (y f0 (y)j: Solution: 3. Prove Proposition 1 (hint: uniform Cauchy criterion for sequences of functions). Solution: Exercise 3. (MTH3140 only) (4+2+4=10 marks) We want to prove the following proposition, mentioned in the lecture notes. Proposition 1 Let (fn) be and continuous on an interval [a; b], and differentiable on (a; b). Let c 2 [a; b]. Assume that (fn(c)) converges and that (f0n ) converges uniformly on (a; b). Then (fn) converges uniformly on [a; b].

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