Real Analysis Math 125A, Fall 2012 Sample Final Questions. x 1+x y. x y x. 2 (1+x 2 )(1+y 2 ) x y

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1 . Define f : R R by Real Analysis Math 25A, Fall 202 Sample Final Questions f() = Show that f is continuous on R. Is f uniformly continuous on R? To simplify the inequalities a bit, we write 3 + = For,y R, we have f() f(y) = y + + y 2 +y 2 y + + y 2 +y 2. Using the inequality 2 y 2 +y 2, we get + y 2 +y 2 = y +y 2 2 y (+ 2 )(+y 2 ) [ ] + y y (+ 2 )(+y 2 ) [ ] y 2 y 2 (+ 2 )(+y 2 ) ( 2 +y + ) y It follows that y f() f(y) 2 y for all,y R. Therefore f is Lipschitz continuous on R, which implies that it is uniformly continuous (take δ = ǫ/2).

2 2. Does there eist a differentiable function f : R R such that f (0) = 0 but f () for all 0? No such function eists. We have f (0) = lim 0 [ f() f(0) The mean value theorem implies that for for every 0, there is some ξ strictly between 0 and (so ξ 0) such that ]. f() f(0) = f (ξ). Since limits preserve inequalities, it follows that [ ] f() f(0) lim, 0 so we cannot have f (0) = 0. 2

3 3. (a) Write out the Taylor polynomial P 2 () of order two at = 0 for the function +. and give an epression for the remainder R 2 () in Taylor s formula + = P2 ()+R 2 () < <. (b) Show that the limit eists and find its value. lim 0 [ ] +/2 + 2 (a) The function and its derivatives are given by f() = +, f(0) =, f () = 2 (+) /2, f (0) = 2, f () = 4 (+) 3/2, f (0) = 4, f () = 3 8 (+) 5/2. The Taylor polynomial and remainder are P 2 () = 2 k=0 where ξ is between 0 and, which gives k! f(k) (0) k, R 2 () = 3! f (ξ) 3, + = (+ξ) 5/2 3 (b) For this part, we only need the Taylor polynomial of order one, + = (+ξ) 3/2 2 where ξ is between 0 and. Since ξ 0 as 0, it follows that [ ] +/2 + lim = 0 8 lim ξ 0 (+ξ) 3/2 =

4 4. (a) Suppose f n : A R is uniformly continuous on A for every n N and f n f uniformly on A. Prove that f is uniformly continuous on A. (b) Does the result in (a) remain true if f n f pointwise instead of uniformly? (a) Let ǫ > 0. Since f n f converges uniformly on A there eists N N such that f n () f() < ǫ 3 for all A and n > N. Choose some n > N. Since f n is uniformly continuous, there eists δ > 0 such that f n () f n (y) < ǫ 3 for all,y A with y < δ. Then, for all,y A with y < δ, we have f() f(y) f() f n () + f n () f n (y) + f n (y) f() < ǫ, which implies that f is uniformly continuous on A. (b) The result does not remain true if f n f pointwise. For eample, consider f n : [0,] R defined by f n () = n. Then f n is uniformly continuous on [0,] because it is a continuous function on a compact interval, but f n f pointwise where { 0 if 0 <, f() = if =. The limit f is not even continuous on [0,]. 4

5 5. Define f n : [0, ) R by f n () = sin(n) +n. (a) Show that f n converges pointwise on [0, ) and find the pointwise limit f. (b) Show that f n f uniformly on [a, ) for every a > 0. (c) Show that f n does not converge uniformly to f on [0, ). (a) If > 0, then f n () +n 0 as n so f n () 0. Also, f n (0) = 0 for every n, so f n (0) 0. Thus, f n 0 pointwise on [0, ). (b) We have f n () +na < na for all a <, so given ǫ > 0 take N = /a and then f n () < ǫ for all n > N, meaning that f n 0 uniformly on [a, ). (c) If (f n ) converges uniformly on [0, ), then it must converge to the pointwise-limit 0. Let n = π/(2n). Then f n ( n ) = +π/2. Therefore, if 0 < ǫ 0 /(+π/2), there eists [0, ) such that f n () ǫ 0, which means that f n does not converge uniformly to 0 on [0, ). 5

6 y Figure : Plot of the function f n () = sin(n)/(+n) on [0,] for n = 20 (green), n = 00 (red), and n = 500 (blue). Remark. The non-uniform convergence of the sequence near = 0 is illustrated in the figure. We can also write the proof in terms of the sup-norm. Let f a = sup f() [a, ) denote the sup-norm of f on [a, ). If a > 0, then f n a na 0 as n, so f n 0 uniformly on [a, ). If a = 0, then f n 0 +π/2 for every n N, so (f n ) does not converge uniformly to 0 on [0, ). 6

7 6. Suppose that f() = n= sinn n 3, g() = (a) Prove that f,g : R R are continuous. n= (b) Prove that f : R R is differentiable and f = g. cosn n 2. (a) Since sinn n 3 n 3, cosn n 2 n 2, n= n= n 3 < n 2 <, the Weierstrass M-test implies that both series converge uniformly(and absolutely) on R. Each term in the series is continuous, and the uniform limit of continuous functions is continuous, so f, g are continuous on R. (b) The series for g is the term-by-term derivative of the series for f. Since the series for g converges uniformly, the theorem for the differentiation of sequences implies that f is differentiable and f = g. 7

8 7. Let P = {2,3,5,7,,...} be the set of prime numbers. (a) Find the radius of convergence R of the power series f() = p = p P (b) Show that 0 f() 2 for all 0 <. (a) We write the series as f() = a n n n=2 where Then a n = { if n is prime, 0 if n isn t prime. a n n n for every n = 2,3,4,... Therefore, if < the series converges by comparison with the convergent geometric series n. Furthermore, if >, the terms in the series do not approach 0. So the radius of convergence of the series is R =. (b) As in (a), and using the sum of the geometric series, we have for 0 < that 0 p P p n = 2 n = 2, n=2 n=0 which proves the result. 8

9 8. Let (X,d) be a metric space. (a) Define the open ball B r () of radius r > 0 and center X. (b) Define an open set A X. (c) Show that the open ball B r () X is an open set. (a) The open ball is defined by B r () = {y X : d(,y) < r}. (b) A set A X is open if for every A there eists r > 0 such that B r () A. (c) Suppose that y B r (). We have to show that B r () contains an open ball B s (y) for some s > 0. Choose s = r d(,y) > 0. (Draw a picture!) If z B s (y), then by the triangle inequality d(,z) d(,y)+d(y,z) < d(,y)+s = r, meaning thatz B r (). Thus, B s (y) B r (), which proves theresult. 9