Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.


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1 Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms: Let (V, ) be a normed vector space and let u, v V. Prove that u v u v. Proof. (i): B the usual triangle inequalit for metrics we have d(x, z) d(, z) (d(x, ) + d(, z)) d(, z) = d(x, ) and also d(, z) d(x, z) (d(, x) + d(x, z)) d(x, z) = d(, x) = d(x, ). Therefore, d(x, z) d(, z) d(x, ). (ii): B the usual triangle inequalit for norms we have u v = (u v) + v v u v + v v = u v. Likewise, we have v u v u = ( 1)(u v) = 1 u v = u v. Therefore, u v u v. (2) Suppose that ϕ : [0, ) [0, ) satisfies ϕ(0) = 0, ϕ(r) > 0 for all r > 0, and for all a, b [0, ): i. ϕ(a + b) ϕ(a) + ϕ(b) (i.e., ϕ is subadditive), ii. if a b then ϕ(a) ϕ(b) (i.e., ϕ is monotone increasing). Let (X, d) be a metric space and let D : X X R be defined b D(x, ) := ϕ(d(x, )). Prove that D is a metric on X. Proof. We must check the axioms. Since d is and ϕ are nonnegative, D is nonnegative. We have D(x, x) = ϕ(d(x, x)) = ϕ(0) = 0 for all x X, and for x we have d(x, ) > 0, hence D(x, ) = ϕ(d(x, )) > 0 b our assumption on ϕ. For each x, we have D(x, ) = ϕ(d(x, )) = ϕ(d(, x)) = D(, x), so that D is smmetric. It remains to show that D satisfies the triangle inequalit: let x,, z X. Then b the triangle inequalit for d we have d(x, z) d(x, )+d(, z). Appling monotonicit of ϕ, followed b subadditivit shows that D(x, z) = ϕ(d(x, z)) ϕ(d(x, ) + d(, z)) ϕ(d(x, )) + ϕ(d(, z)) = D(x, ) + D(, z), so the triangle inequalit holds for D. (3) Let (X, d) be a metric space. (i) Prove D 1 (x, ) := d(x, )/(1 + d(x, )) is a metric on X. (ii) Let α > 0. Prove D 2 (x, ) := α d(x, ) is a metric on X. (iii) Prove D 3 (x, ) := min{d(x, ), 1} is a metric on X. (iv) Prove D 4 (x, ) := d(x, ) 1/2 is a metric on X. Proof. In each case D i = ϕ i d for an appropriate function ϕ i. It therefore suffices to show that each ϕ i satisfies the assumptions of problem (2). (i): Let ϕ 1 : [0, ) [0, ) be the function ϕ 1 (x) = x/(1 + x). Then clearl ϕ(0) = 0 and ϕ(x) > 0 for x > 0. If x, [0, ) then ϕ 1 (x) + ϕ 1 () x/(1 + x + ) + /(1 + x + ) = ϕ 1 (x + ), so ϕ 1 is subadditive. 1
2 2 (ii): Let ϕ 2 (x) = α x. Then ϕ 2 (0) = 0 and for x > 0 we have ϕ 2 (x) = α x > 0 since α > 0. Also, ϕ 2 is monotone increasing since α > 0, and ϕ 2 is subadditive (and in fact ϕ 2 (x + ) = ϕ 2 (x) + ϕ 2 ()). (iii): Let ϕ 3 (x) = min{x, 1}. Then ϕ 3 (0) = 0 and ϕ 3 (x) > 0 for x > 0. If x then ϕ 3 (x) = min{x, 1} min{, 1} = ϕ 3 (), so ϕ 3 is monotones increasing. Let x, 0. If x + 1 then we have ϕ 3 (x + ) = x + = ϕ 3 (x) + ϕ 3 (). If either x 1 or 1 then either ϕ 3 (x) = 1 or ϕ 3 () = 1, hence ϕ 3 (x + ) = 1 ϕ 3 (x) + ϕ 3 (). If x 1 and 1 and x + 1 then ϕ 3 (x + ) = 1 x + = ϕ 3 (x) + ϕ 3 (). These three cases exhaust the possibilities and in all cases we have shown ϕ 3 (x + ) ϕ 3 (x) + ϕ 3 (), so ϕ 3 is subadditive. (iv): Let ϕ 4 (x) = x 1/2. Once again it is clear that ϕ 4 (0) = 0 and ϕ 4 (x) > 0 for all x > 0. Also, ϕ 4 is monotone increasing: suppose toward a contradiction that there exist 0 x < with x 1/2 1/2. Since the map r r 2 is monotone increasing on [0, ), this implies that x, a contradiction. It remains to show that ϕ 4 is subadditive on [0, ). Since ϕ 4 is monotone increasing, for x, [0, ) we have ϕ 4 (x+) = (x+) 1/2 (x+2x 1/2 1/2 +) 1/2 = [(x 1/2 + 1/2 ) 2 ] 1/2 = x 1/2 + 1/2 = ϕ 4 (x)+ϕ 4 (). (4) Let (X 1, d 1 ), (X 2, d 2 ),... be a sequence of metric spaces. Let X = n N X n, i.e., X is the set of all sequences x = (x 1, x 2,... ) with x n X n for all n N. Prove that the function d : X X R defined b is a metric on X. d(x, ) = n=1 2 n d n (x n, n ) 1 + d n (x n, n ) Proof. For each n let d n : X n X n R be the function d n(x n, n ) = 2 n d n(x n, n). B 1+d n(x n, n) problem (3).(i) and (3).(ii), for each n the function d n is a metric on X n. Moreover, for each n N we have d n(a, b) 2 n for all a, b X n. Therefore, d(x, ) = n=1 d n(x n, n ) n=1 2 n 1 for all x, X. We have d(x, x) = 0 since d n(x n, x n ) = 0 for all n. Also, if x, X are distinct, then there is some n such that x n n, hence d(x, ) d n(x n, n ) > 0. We also have d(x, ) = n=1 d n(x n, n ) = n=1 d n( n, x n ) = d(, x), so d is smmetric. It remains to verif the triangle inequalit for d. Let x,, z X. Then for each n we have d n(x n, z n ) d n(x n, n ) + d n( n, z n ), and hence d i(x i, z i ) (d i(x i, i ) + d i( i, z i )) = d i(x i, i ) + d i( i, z i ). Since the left hand side converges to d(x, z) and the right hand side converges to d(x, ) + d(, z), this completes the proof. (5) We sa that a subset A of a metric space X is bounded if there is some x 0 X and some constant C < such that d(a, x 0 ) < C for all a A. Prove that a finite union of bounded sets is again bounded. Proof. Let A 1,..., A n be bounded subsets of X. Then for each 1 i n we can find x i X and a C i > 0 such that d(a, x i ) < C i for all a A i. Let C = max{c i + d(x i, x 1 ) : 1 i n}.
3 If a n A i then a A i for some 1 i n and hence b the triangle inequalit, d(a, x 1 ) d(a, x i ) + d(x i, x 1 ) < C i + d(x i, x 1 ) C. This shows that n A i is bounded. (6) Let A be a nonempt subset of a metric space X. The diameter of A, denoted diam(a), is defined b diam(a) = sup{d(a, b) : a, b A}. Prove that A is bounded if and onl if diam(a) is finite. Proof. Suppose first that A is bounded and fix some x 0 X and C > 0 such that d(a, x 0 ) < C for all a A. Then for an a, b A b the triangle inequalit we have d(a, b) d(a, x 0 ) + d(x 0, b) = d(a, x 0 ) + d(b, x 0 ) < C + C = 2C. Since a, b A were arbitrar this shows that diam(a) 2C. Conversel, suppose that diam(a) = D <. B hpothesis A is nonempt, so fix some a 0 A. Then for all a A we have d(a, a 0 ) diam(a) = D <, so A is bounded. Additional problems for honors section (next page): Definition 0.1. Let I R be an interval. A function ϕ : I R is said to be concave if for all t [0, 1] and x, I. ϕ(tx + (1 t)) tϕ(x) + (1 t)ϕ() (7) Let ϕ : [0, ) [0, ) be a concave function. For each a, b [0, ) with a < b, define the function t a,b : (b, ) [0, 1] b t a,b (z) := z b z a, so that b = t a,b (z)a + (1 t a,b (z))z for all z (b, ). (i) Prove that ϕ(b) t a,b (z)ϕ(a) for all z (b, ). (ii) Let z in part (i) to conclude that ϕ is monotone increasing. Proof. (i): Fix z (b, ). Since t a,b (z) [0, 1] and ϕ(z) 0 we have (1 t a,b (z))ϕ(z) 0. Therefore, since ϕ is convex we have ϕ(b) = ϕ(t a,b (z)a + (1 t a,b (z))z) t a,b (z)ϕ(a) + (1 t a,b (z))ϕ(z) t a,b (z)ϕ(a). (ii): Given a, b [0, ) with a < b, part (i) shows that ϕ(b) t a,b (z)ϕ(a) for all z (b, ). We have 1 t a,b (z) = b a 0 as z, so t z a a,b(z) 1 as z and hence ϕ(b) lim z t a,b (z)ϕ(a) = ϕ(a). (8) Suppose that ϕ : [0, ) [0, ) is a function with ϕ(0) = 0. Consider the following statements: (i) ϕ is twice differentiable with ϕ 0. (ii) ϕ is differentiable and ϕ is monotone decreasing. (iii) For all x,, z [0, ) with x < < z, ϕ() ϕ(x) x ϕ(z) ϕ(). z (iv) ϕ is concave. (v) The function ϕ(x)/x is monotone decreasing on (0, ). 3
4 4 (vi) ϕ is subadditive, i.e., for all x, [0, ). Prove that (i) (ii) (iii) (iv) (v) (vi). ϕ(x + ) ϕ(x) + ϕ() Proof. (i) (ii): Assume ϕ is twice differentiable with ϕ 0. Suppose toward a contradiction that ϕ is not monotone decreasing. Then there is some 0 x < such that ϕ (x) < ϕ (). Appling the Mean Value Theorem to ϕ, we obtain some c (x, ) such that ϕ (c) = ϕ () ϕ (x) > 0, a contradiction. x (ii) (iii): Assume ϕ is differentiable and ϕ is monotone decreasing. Given x,, z [0, ) with x < < z, b the Mean Value Theorem we can find some c (x, ) and some d (, z) such that ϕ (c) = ϕ() ϕ(x) and ϕ (d) = ϕ(z) ϕ(). Then c < d, so as ϕ is monotone decreasing x z we see that ϕ() ϕ(x) = ϕ (c) ϕ ϕ(z) ϕ() (d) =. x z (iii) (iv): Assume that (iii) holds. Fix an x, z [0, ) and 0 t 1. We must show that ϕ(tx + (1 t)z) tϕ(x) + (1 t)ϕ(z). If we let s = 1 t then this is equivalent to the inequalit ϕ((1 s)x + sz) (1 s)ϕ(x) + sϕ(z). Therefore, after switching the roles of x and z and of t and s if necessar, we ma assume without loss of generalit that x z. If x = z or if t = 0 or t = 1 then the inequalit is trivial, so we ma assume that x < z and 0 < t < 1. Then we let = tx + (1 t)z, so that x < < z. B (iii), we have ϕ() ϕ(x) x ϕ(z) ϕ(). z Since x and z are positive, the inequalit still holds after multipling both sides b ( x)(z ). Doing this and rearranging terms shows that ϕ()(z x) ϕ(z)( x) + ϕ(x)(z ). Since x = (1 t)(z x) and (z ) = t(z x), after dividing b z x we obtain the inequalit ϕ() (1 t)ϕ(z) + tϕ(x), as we set out to show. (iv) (v): Assume ϕ is concave. Given 0 < x <, let t = x, so that 0 < t < 1. Then x = t + (1 t)0, so appling convexit and the fact that ϕ(0) = 0 we have ϕ(x) = ϕ(t + (1 t)0) tϕ() + (1 t)ϕ(0) = x ϕ(), and hence ϕ(x)/x ϕ()/. (v) (vi): Assume that ϕ(x)/x is monotone decreasing on (0, ). Given x, [0, ) we must show that ϕ(x + ) ϕ(x) + ϕ(). This is clear when either x = 0 or = 0, so assume that x > 0 and > 0. Then our assumption implies that ϕ(x)/x ϕ(x + )/(x + ) and ϕ()/ ϕ(x + )/(x + ), so that ϕ(x) + ϕ() = x ϕ(x) x so the proof is complete. + ϕ() ϕ(x + ) ϕ(x + ) x + x + x + = ϕ(x + ),
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