Question 1 (a) ( =⇒ ) Since fA (x) = inf{|x − y| : y ∈ A} = 0, by the definition of inf, for every n ∈ N there exists xn ∈ A such that |x − xn | < n1 . Let ε > 0 be given and choose N such that N > 1ε (by the Archimedean property.) Then for the sequence {xn }, n ≥ N implies |x − xn | < 1 1 ≤ < ε. n N Hence {xn } is convergent, xn ∈ A for all n and lim xn = x. n→∞ ( ⇐= by contradiction) Since |x − y| ≥ 0, inf |x − y| ≥ 0 for all x, y ∈ R. Suppose fA (x) = inf{|x − y| : y ∈ A} = ε > 0 for some x. Then for any sequence {xn } such that all xn ∈ A, |xn − x| ≥ inf{|x − y| : y ∈ A} = ε and hence cannot converge to x. (b) fA (x) = inf{|x − z| : z ∈ A} = inf{|x − y + y − z| : z ∈ A} ≤ inf{|x − y| + |y − z| : z ∈ A} ≤ |x − y| + inf{|y − z| : z ∈ A} = |x − y| + fA (y) where we have assumed and hence fA (x) − fA (y) ≤ |x − y|. Interchanging the variables x and y leads to the inequality fA (y) − fA (x) ≤ |x − y|, and hence |fA (x) − fA (y)| ≤ |x − y|, as desired. We furnish the proof with the following details: Inequality on line 3. Let S and T be two non-empty sets, both bounded from below in R. For any s ∈ S and t ∈ T , inf{|s + t| : s ∈ S, t ∈ T } ≤ |s + t| ≤ |s| + |t| and hence inf{|s + t| : s ∈ S, t ∈ T } provides a lower bound on |s| + |t|. Hence inf{|s + t| : s ∈ S, t ∈ T } ≤ inf{|s| + |t| : s ∈ S, t ∈ T } and the inequality on line 3 is a consequence of this result. Inequality on line 4. Let U be a non-empty set, bounded from below in R. For any constant c and any u ∈ U , inf{c + u : u ∈ U } ≤ c + u inf{c + u : u ∈ U } − c ≤ u and hence inf{c + u : u ∈ U } provides a lower bound on U . Hence inf{c + u : u ∈ U } − c ≤ inf U inf{c + u : u ∈ U } ≤ c + inf U 1 and the inequality on line 4 is a consequence of this result. Now, let ε > 0 be given and choose δ = ε. Then |fA (x) − fA (y)| ≤ |x − y| < δ = ε. Hence fA (x) is uniformly continuous on R. (c) (i) Since • fA (x) is continuous, • fA (x) + fB (x) is continuous and • fA (x) + fB (x) ̸= 0 ∀x ∈ R, fA (x) is continuous on R. fA (x) + fB (x) Question 2 (a) (i) 1 α xn+1 = xn + 2 xn √ √ 1 α = xn − 2 α + + α 2 xn √ 2 √ xn α = 1− + α. 2 xn √ 2 √ xn α Since xn > α > 0, we have that > 0 and 1 − > 0; and so 2 xn √ 2 √ √ xn α xn+1 = 1− + α > α. 2 xn √ √ This result combined with x1 > α together imply that xn > α for all n ∈ N, by the Principle of Mathematical Induction (PMI). (ii) We show that {xn } is a decreasing sequence. Firstly, √ 1 √ α 5√ 2 α+ √ = α < 2 α = x1 , x2 = 2 4 2 α 2 √ where the inequality is satisfied since α > 0. Now assume xk−1 − xk > 0 for some k > 1, then α α 1 xk−1 − xk + − xk − xk+1 = 2 xk−1 xk 1 α(xk − xk−1 ) = xk−1 − xk + . 2 xk−1 xk √ √ α α Since xn > α for all n, we have that 0 < < 1, and in particular 0 < = xn xk−1 xk √ √ α α < 1. Then xk−1 xk α(xk − xk−1 ) 1 xk−1 − xk + xk − xk+1 = 2 xk−1 xk 1 > (xk−1 − xk + xk − xk−1 ) = 0, 2 where the inequality is satisfied under the assumption that xk − xk−1 < 0. Hence this result together with x2 − x1 < 0 imply that {xn } is a decreasing sequence (by PMI). √ Since {xn } is decreasing and bounded from below by α, it converges by the Monotone Convergence Theorem to a number, say x. Letting n → ∞ on both sides of the recursion relation, α 1 x+ x= 2 x 2 x =α √ x= α since the negative root contradicts our previous assumptions. (b) (i) an+1 2 + 2/n 2 2 · 4 · 6 · . . . · (2n) 2n + 2 . Then = = → . By the Ratio 1 · 4 · 7 · . . . · (3n + 1) an 3n + 4 3 + 4/n 3 n+1 ∞ P (−1) 2 · 4 · 6 · . . . · (2n) is absolutely convergent. Test, 1 · 4 · 7 · . . . · (3n + 1) n=1 Let an = (ii) √ √ n+1− n = √ 1 1 1 √ ≥ √ √ = √ √ . By the Comparn+1+ n 2n + n ( 2 + 1) n ∞ √ P √ 1 ison Test with the 12 -series, ( n + 1 − n) diverges. However, an = √ √ > n+1+ n n=1 1 1 1 √ √ = an+1 and lim an = lim √ = √ = lim √ p n→∞ n→∞ n→∞ n+2+ n+1 n+1+ n n( 1 + 1/n + 1) 1 1 p lim √ · = 0 · 12 = 0. Furthermore all an > 0. Hence by the Altern→∞ n lim 1 + 1/n + 1 n→∞ ∞ √ P √ nating Series Test, (−1)n+1 ( n + 1 − n) is conditionally convergent. Let an = n=1 3 Question 3 (a) Consider the closed interval [0, m]. Then sin(nx) + x n+x 1+x ≤ n+x 1+m → 0. ≤ n |fn (x)| = Hence {fn (x)} converges uniformly to 0 on [0, m], and converges pointwise to 0 on [0, ∞) since there is a closed interval [0, m] containing any M > 0. Now consider the closed interval [0, ∞). Let ε = 41 and define the sequence {xn }∞ n=2 = n ∈ [0, ∞). Then for all n ≥ 2 we have sin(n2 ) + n 2n n−1 ≥ 2n 1 1 = − 2 2n 1 1 1 ≥ − = = ε. 2 4 4 |fn (xn ) − 0| = We have shown that there exists a sequence {xn } with all xn ∈ [0, ∞) such that for n ≥ 2, |fn (xn ) − 0| ≥ ε. So, {fn (x)} does not converge uniformly on [0, ∞). (b) By the Archimedean property, for any positive real number v satisfying 0 < v < 1, there exists N ∈ N such that N > v1 . Hence, for all n ≥ N , gn (v) = 1 since n1 ≤ N1 < v < 1. Thus gn (x) = 1 on [v, 1] for all n ≥ N and hence {gn (x)} converges uniformly to 1 on [v, 1], and converges pointwise to 1 on (0, 1] since for any w such that 0 < w < 1 there is a closed interval [ w2 , 1] that contains it. 1 However, by choosing ε = 1 and defining the sequence {xn } = n+1 , |gn (xn )−1| = |0−1| = 1 = ε. Hence there exists a sequence {xn } with all xn ∈ (0, 1] and |gn (xn ) − 1| ≥ ε; and so {gn (x)} is not uniformly convergent on [0, 1]. Question 4 (a) Note that |nk (x · ln(x + 1))2n | ≤ nk (1 · ln 2)2n = nk (ln 2)2n for all x ∈ [0, 1]. By the Ratio Test, k (n + 1)k (ln 2)2n+1 1 = 1+ ln 2 → 1 · ln 2 = ln 2 < 1. nk (ln 2)2n n 4 Hence ∞ P ∞ P nk (ln 2)2n converges and n=1 nk (x · ln(x + 1))2n converges uniformly over [0, 1] n=1 by the Weierstrass M -test. (b) First, we show for a fixed x ∈ [v, π − v] that the sequence of numbers {an } = 1 − e−x/n is decreasing. x x − 1 − exp − an − an+1 = 1 − exp − n n+1 x x = exp − − exp − n+1 n x x −x/n exp =e − −1 n n+1 x −x/n −1 =e exp n(n + 1) x x −x/n ≥e 1+ − 1 = e−x/n · ≥ 0, n(n + 1) n(n + 1) where the inequalities are satisfied because e−x/n > 0 and x n(n+1) ≥ 0 over [v, π − v]. Next, we show that the sequence of functions {fn (x)} = 1 − e−x/n converges uniformly to 0 over [v, π − v]. For all x ∈ [v, π − v], |fn (x)| = 1 − e−x/n ≤ 1 − e−π/n π π = → 0. ≤1− 1− n n n P Finally, we show that the partial sums sn = sin(kx) are uniformly bounded over k=1 [v, π − v] for all n ∈ N. For all x ∈ [v, π − v], sin(nx + 1) sin(nx/2) sin(x/2) 1 1 ≤ ≤ . | sin(x/2)| sin(v/2) |sn | = By the above results, the series of functions ∞ P (1 − e−x/n ) sin(nx) satisfies the conditions n=1 of Dirichlet’s Test for series of functions and hence converges uniformly on the closed interval [v, π − v]. It follows that it also converges pointwise on [v, π − v]. 5 (c) By the Ratio Test, an+1 (n + 1)3 ((n + 1)!)3 (3n)! = · 3 an (3n + 3)! n (n!)3 3 (n + 1)3 1 = 1+ n (3n + 3)(3n + 2)(3n + 1) (1 + n1 )6 16 1 = → = . 3 2 1 3·3·3 27 (3 + n )(3 + n )(3 + n ) Hence the radius of convergence is 27. Question 5 (a) Consider the sequence of partial sums: sn = n k X x k=1 1 xk+1 − k k+1 x x2 x2 x3 xn+1 − + − + ··· − 1 2 2 3 n+1 n+1 x . =x− n+1 = We consider two cases. Case 1. |x| ≤ 1. For |x| ≤ 1, x − lim n→∞ xn+1 xn+1 ≤ sn ≤ x + and n+1 n+1 xn+1 1 = lim lim |xn+1 | ≤ 0 · 1 = 0. n→∞ n + 1 n→∞ n+1 Hence by the Squeeze Theorem, lim sn exists and equals x. n→∞ Case 2. |x| > 1. Since |x| > 1, there exists N ∈ N such that |x| > 1 + 6 1 N (by the Archimedean property.) Then for all n ≥ N we have |sn | = x − ≥ > = = > = xn+1 n+1 xn+1 − |x| n+1 (1 + N1 )n+1 − |x| n+1 (1 + N1 )n+1 N + 1 N + 2 n · · ... − |x| N +1 N +2 N +3 n+1 (1 + N1 )N +1 1 + N1 1 + N1 1 + N1 · · . . . − |x| N +1 1 + N1+1 1 + N1+2 1 + n1 !n−N (1 + N1 )N +1 1 + N1 − |x| N +1 1 + N1+1 !n (1 + N1+1 )N 1 + N1 − |x|, N 1 + N1+1 1+1/N where the first term forms a geometric progression with common ratio 1+1/(N > 1. +1) This term diverges and by the Comparison Test, |sn | diverges. Hence sn diverges (by the contrapositive of: sn converges =⇒ |sn | converges.) n ∞ P x xn+1 Combining these two results gives us that the series of functions − conn n+1 n=1 verges pointwise over D = [−1, 1]. It is not true, however, that "∞ # ∞ X xn+1 d d d xn xn+1 d X xn − = lim sn = lim sn = − n→∞ dx dx n=1 n n+1 dx n→∞ dx n n+1 n=1 for all x ∈ D. From the LHS, d d lim sn = (x) = 1 n→∞ dx dx for all x ∈ D. From the RHS, d sn = lim (1 − xn ) n→∞ dx n→∞ lim which is 0 for x = 1 and does not exist for x = −1. (b) 2x + nx3 −x 1 + nx2 x = 1 + nx2 |fn (x) − x| = 7 1 1 which has maxima at x = ± √ . So for all x ∈ R, |fn (x) − x| ≤ √ → 0. Hence, n 2 n {fn (x)} converges uniformly on R to f (x) = x. However, (1 + nx2 )(2 + 3nx2 ) − (2nx)(2x + nx3 ) (1 + nx2 )2 (1)(2) =2 fn′ (0) = 12 fn′ (x) = for all n. Hence lim n→∞ fn′ (0) ′ ′ = 2 ̸= 1 = f (0), and it is not true that f (x) = lim n→∞ d (fn (x)) dx for all x ∈ R. (c) Let ε > 0 be given. Since {gn (x)} converges pointwise on [0, 1], there exists N1 , N2 such that n ≥ N1 implies |gn (0) − K| < ε and n ≥ N2 implies |gn (1) − K| < ε. Let N = max{N1 , N2 }. Then for all n ≥ N , we have that gn (0) > K − ε and gn (1) < K + ε, and K − ε < gn (0) ≤ gn (x) ≤ gn (1) < K + ε for all x ∈ [0, 1], under the assumption that gn (x) is monotone increasing on [0, 1]. Hence |gn (x) − K| < ε for all n ≥ N and x ∈ [0, 1], i.e., the convergence of {gn (x)} is uniform on [0, 1]. 8