HOMEWORK ASSIGNMENT 6

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1 HOMEWORK ASSIGNMENT 6 DUE 15 MARCH, ) Suppose f, g : A R are uniformly continuous on A. Show that f + g is uniformly continuous on A. Solution First we note: In order to show that f + g is uniformly continuous on A, we must show that for every ε > 0, there exists δ > 0 such that for any x, y A with x y < δ, (f + g)(y) (f + g)(x) < ε. Fix ε > 0. Since f is uniformly continuous, there exists δ f > 0 such that for any x, y A with x y < δ f, f(x) f(y) < ε/2. Since g is uniformly continuous, there exists δ g > 0 such that for any x, y A with x y < δ g, g(x) g(y) < ε/2. Let δ = min{δ f, δ g }. Then if x, y A are such that x y < δ, then x y < δ f, so f(x) f(y) < ε/2, and x y < δ g, so g(x) g(y) < ε/2. Therefore (f + g)(y) (f + g)(x) = f(x) f(y) + g(x) g(y) f(x) f(y) + g(x) g(y) < ε/2 + ε/2 = ε. 1

2 2 DUE 15 MARCH, ) Let f : [0, 1] R be a continuous function. For each 0 x 1, let F (x) = max 0 t x f(t). How do we know F is well-defined (that is, how do we know this maximum exists)? Show that F is continuous. Solution We know the value F (x) = max 0 t x f(t) exists, since f is continuous on the compact set [0, t], and it must therefore have a maximum on that set. We show that F is continuous. We observe that F is non-decreasing on [0, 1]. To see this, suppose that 0 x < y 1. Then [0, x] [0, y], so the maximum of f on the set [0, y] is at least as large as the maximum of f on the subset [0, x]. This means F (x) F (y) when 0 x < y 1. Fix ε > 0. Since f is continuous on the compact set [0, 1], f is uniformly continuous, and there exists δ > 0 such that if x, y [0, 1] and x y < δ, f(x) f(y) < ε. We will show that if x, y [0, 1] and x y < δ, F (x) F (y) < ε. To that end, fix x, y [0, 1] with x y < δ. Assume without loss of generality that x y. Since F is non-decreasing, F (x) F (y). This means F (y) F (x) = F (y) F (x). We need to show that F (y) F (x) < ε. Fix some s [0, y] such that F (y) = f(s) (that is, since F (y) is the maximum value of f on [0, y], we can pick an s [0, y] where this maximum is attained). We consider two cases. Either s [0, x], in which case F (x) = max f(t) f(s) = F (y). 0 t x In this case, F (x) F (y) F (x), and F (x) = F (y). In the second case, s (x, y]. This means that s x = s x y x < δ. By our choice of δ, f(s) f(x) < ε. Since F (x) = max 0 t x f(t) f(x), we deduce that F (y) = f(s) < f(x) + ε F (x) + ε, so that F (y) F (x) < ε. In either case, F (y) F (x) < ε, as desired.

3 HOMEWORK ASSIGNMENT 6 3 3) If f : A B is uniformly continuous and g : B R is uniformly continuous, then g f : A R is uniformly continuous. Solution We want to show that for every ε > 0, there exists δ > 0 such that for any x, y A with x y < δ, g(f(x)) g(f(y)) < ε. Fix ε > 0. Since g is uniformly continuous on B, there exists η > 0 such that for any w, z B with w z < η, g(w) g(z) < ε. Since f is uniformly continuous on A, there exists δ > 0 such that for any x, y A with x y < δ, f(x) f(y) < η. We claim that this δ is the one we want. To that end, fix x, y A with x y < δ. Let w = f(x) and z = f(y). Then by our choice of δ, By our choice of η, w z = f(x) f(y) < η. g(f(x)) g(f(y)) = g(w) g(z) < ε.

4 4 DUE 15 MARCH, ) Suppose that I, J are disjoint, closed intervals, and f is a function which is defined on I J, uniformly continuous on I, and uniformly continuous on J. Show that f is uniformly continuous on I J. Give an example to show that this conclusion may not hold if I, J are not assumed to be closed. Solution If either I = or J =, the result is trivial. We may therefore assume that neither interval is empty. We want to show that for any ε > 0, there exists δ > 0 such that for any x, y I J with x y < δ, f(x) f(y) < ε. The first step of the proof is to establish the following: There exists a positive number δ 0 such that for any x I and y J, x y δ 0. We note that since the sets are disjoint intervals, it must be that either x < y for every x I and y J or x < y for every x J and y I. That is, one of the intervals lies entirely to the left of the other. By switching I and J if necessary, we may assume that I lies entirely to the left of J. This means that any member of J is an upper bound for I, and sup I is finite. Since I is closed and the supremum of a set lies in the closure of that set whenever the supremum is finite, we deduce that sup I I = I. Similarly, inf J J = J. Therefore sup I < inf J. Let δ 0 = inf J sup I > 0. Then for any x I and y J, since x < y, x y = y x. Moreover, since x sup I and inf J y, y x inf J sup I = δ 0. This proves the claim at the beginning of the paragraph. Next, fix ε > 0. Since f is uniformly continuous on I, there exists δ I > 0 such that for any x, y I with x y < δ I, f(x) f(y) < ε. Since f is uniformly continuous on J, there exists δ J > 0 such that for any x, y J with x y < δ J, f(x) f(y) < ε. Let δ = min{δ 0, δ I, δ J }. Fix x, y I J. Since x y < δ δ 0, either x, y I or x, y J. That is, it is impossible to have one of x, y in I and the other in J. This is because if x I and y J, x y δ 0, while we know that x y < δ 0. If x, y I, then since x y < δ δ I, f(x) f(y) < ε. If x, y J, then since x y < δ J, f(x) f(y) < ε. In either case, f(x) f(y) < ε. For the counterexample, we recall the first fact used in the previous proof: That there is a positive distance δ 0 between the sets I and J. For the counterexample, we wish to use some I, J which are not separated by a positive distance. For our first counterexample, let I = [ 1, 0] and J = (0, 1]. Let f(x) = 0 if x I and f(x) = 1 if x J. Then f is uniformly continuous on I, since it is constant on I. Also, f is uniformly continuous on J, since it is constant on J. But the function f is not even continuous on I J = [ 1, 1], since there is a jump discontinuity from 0 to 1 at x = 0. For our second counterexample, we will provide a function f which is uniformly continuous on I and J, and continuous on I J, but not uniformly continuous. Let I = ( 1, 0) and J = (0, 1). Let f(x) = 0 for all x I and f(x) = 1 for all x J. Then I J = ( 1, 0) (0, 1). We claim that f is continuous on I J, but not uniformly continuous. We first show continuity. Fix x ( 1, 0) (0, 1). Fix ε > 0 and let δ = x. Then if y I J and

5 HOMEWORK ASSIGNMENT 6 5 y x < x, we consider two cases. If x < 0, x = x, and y x < x implies y < 0. Therefore y ( 1, 0), and f(y) f(x) = 0 0 = 0 < ε. In the second case, x > 0, x = x, and y x = x y < x implies x y < x, which in turn implies that y > 0. Therefore y (0, 1) and f(y) f(x) = 1 1 = 0 < ε. In either case, if y x < δ = x, f(y) f(x) < ε. This shows that f is continuous. We next show that f is not uniformly continuous on I J. To obtain a contradiction, assume f is uniformly continuous on I J. Then there exists δ > 0 such that for any x, y I J with x y < δ, f(x) f(y) < 1. Let x = δ/3 and y = δ/3. Then x, y I J and x y = δ/3 ( δ/3) = 2δ/3 < δ. However which is not less than 1. f(x) f(y) = 1 0 = 1,

6 6 DUE 15 MARCH, ) Suppose f : R R is a function such that (1) lim x 0 f(x) = 1 and (2) f(x + y) = f(x)f(y) for any x, y R. Complete the following steps. (i) Show that f is continuous on R. Hint: Use lim y x f(y) = lim y 0 f(x + y) to check continuity. (ii) Show that f(x) > 0 for all x. Hint: We know by continuity that f(0) = 1 > 0. First prove that f(x) > 0 for every x > 0 by contradiction. Assume there exists a positive number x such that f(x) 0. Then let s = inf{x > 0 : f(x) 0}. Show that s > 0 and then consider the relationship between f(s) and f( s ). Then use property (2) to 2 show that f( x) = f(x) 1 for every x R, and then use this to deduce that f(x) > 0 for every negative x. (iii) Show that for every real number r, for every real number x, f(rx) = [f(x)] r. Hint: First prove this when r is a natural number using property (2). Then prove this when r is an integer. Extend this to the case when r is a rational number. Last, use continuity to prove the result for all real numbers. (iv) Show that if f(1) = 1, then f is constant. If f(1) > 1, then f is increasing. If 0 < f(1) < 1, then f is decreasing. Solution (i) Fix x R. Then since lim y x f(y) = lim y 0 f(x + y), lim y x f(y) = lim y 0 f(x + y) = lim y 0 f(x)f(y) = f(x) lim f(y) = f(x). y 0 This shows that f is continuous at x. Since x was arbitrary, f is continuous. (ii) First suppose that there exists x 0 > 0 such that f(x 0 ) 0. Let A = {x 0 : f(x) 0}. Since A, s = inf A is finite. We know s 0. We first claim that s > 0. Indeed, if it were not so, then 0 = inf A. Then for any n N, 1/n is not a lower bound for A, and there exists x n [0, 1/n) such that f(x n ) 0. This yields a sequence (x n ). Since lim n x n = 0 and since f is continuous at 0, 1 = f(0) = lim n f(x n ). But since f(x n ) (, 0] for each n, lim n f(x n ) (, 0], which means 1 = f(0) 0, a contradiction. Therefore s > 0. This means that s/2 < s. But since s is a lower bound for A, s/2 / A, and f(s/2) > 0. Then f(s) = f( s 2 + s 2 ) = f(s/2)2 > 0, a contradiction. This gives that f(x) > 0 for all x > 0. Next, note that for any x R, 1 = f(0) = f(x + ( x)) = f(x)f( x). This means that neither f(x) nor f( x) can be zero, and we can divide both sides by f(x) to obtain f( x) = f(x) 1 ). If x > 0, then this fact together with the previous paragraph yields that f( x) = f(x) 1 > 0. (iii) We first note that f(0 x) = f(0) = 1 and f(x) 0 = 1, so the statement holds if r = 0. Next, we prove by induction on r that f(rx) = f(x) r for all r N. The base

7 HOMEWORK ASSIGNMENT 6 7 case r = 1 is true since f(1 x) = f(x) = f(x) 1. Next, assume f(rx) = f(x) r for some r N. Then f((r + 1)x) = f(rx + x) = f(rx)f(x) = f(x) r f(x) = f(x) r+1. This finishes the induction and shows that f(rx) = f(x) r for all r N. Next, suppose r is a negative integer. Then r N. Applying the previous paragraph together with the fact from (ii) that f( x) = f(x) 1 for any x, we deduce that f(rx) = f( ( r)x) = f(( r)x) 1 = [f(x) r ] 1 = f(x) r. Next, fix q Z, q 0. Then by the previous paragraph, ( f(x) = f q ( 1 ( 1 ) q, q x)) = f q x ( 1 ) and taking q th roots on both sides gives f x = f(x) 1 q q. Let r be a rational number. Then we may write r = p for some p, q Z, q 0. Then using the previous paragraph q and the beginning of this paragraph, ( p ) ( f(rx) = f q x = f p ( 1 ( 1 ) p q x)) = f q x 1 = [f(x) q ] p = f(x) p q = f(x) r. Next, let r be any real number. Fix a sequence (q n ) of rational numbers such that lim n q n = r. Then lim n q n x = rx. Since f is continuous, lim n f(q n x) = f(rx). But since q n is rational for each n, f(q n ) = f(x) qn. By continuity of the function g(y) = f(x) y, and since lim n q n = r, we deduce that lim n f(x) qn = f(x) r. Therefore f(rx) = lim n f(q n x) = lim n f(x) qn = f(x) r. (iv) Let b = f(1). We know b > 0. For any x R, by (iii), f(x) = f(x 1) = f(1) x = b x. Thus f is just the exponential function b x. If b = 1, this is constant. If b > 1, this is increasing. If 0 < b < 1, this is increasing.