Lecture 15: Consequences of Continuity. Theorem Suppose a; b 2 R, a<b, and f :[a; b]! R. If f is continuous and s 2 R is

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1 Lecture 15: Cosequeces of Cotiuity 15.1 Itermediate Value Theorem The followig result is kow as the Itermediate Value Theorem. Theorem Suppose a; b 2 R, a<b, ad f :[a; b]! R. If f is cotiuous ad s 2 R is such that either f(a) s f(b) orf(b) s f(a), the there exists c 2 [a; b] such that f(c) =s. Proof Suppose f(a) <f(b) ad f(a) <s<f(b). Let c = supfx : x 2 [a; b];f(x) sg: Suppose f(c) <s. The c<bad, sice f is cotiuous at c, there exists a >0such that f(x) <sfor all x 2 (c; c + ). But the f c + <s; 2 cotradictig the deitio of c. Similarly, if f(c) > s, the c > a ad there exists > 0 such that f(x) >sfor all x 2 (c, ;c), agai cotradictig the deitio of c. Hece we must have f(c) =s. Example Suppose a 2 R, a>0, ad cosider f(x) =x, a where 2 Z, >1. The f(0) =,a <0 ad f(1 + a) =(1+a), a X =1+a + a i, a i i=2 X =1+(, 1)a + a i > 0; i i=2 where! = i i!(, i)! : Hece, by the Itermediate Value Theorem, there exists a real umber >0 such that = a. We call the th root of a, ad write = p a; or = a 1 : Moreover, if a 2 R, a<0, 2 Z + is odd, ad is the th root of,a, the (,) =(,1) () =(,1)(,a) =a: 15-1

2 Lecture 15: Cosequeces of Cotiuity 15-2 That is,, is the th root of a. Deitio If = p q 2 Q with q 2 Z+, the we dee for all real x 0. x =( qp x) p Exercise Explai why the equatio x 5 +4x 2, 16 = 0 has a solutio i the iterval (0; 2). Exercise Give a example of a closed iterval [a; b], a; b 2 R ad a fuctio f :[a; b]! R which do ot satisfy the coclusio of the Itermediate Value Theorem. Exercise Show that if I R is a iterval ad f : I! R is cotiuous, the f(i) isaiterval. Exercise Suppose f :(a; b)! R is cotiuous ad strictly mootoic. Let (c; d) =f((a; b)). Show that f,1 :(c; d)! (a; b) is strictly mootoic ad cotiuous. Exercise Let 2 Z +. Show that the fuctio f(x) = p x is cotiuous o (0; +1). Exercise Use the method of bisectio to give aother proof of the Itermediate Value Theorem Extreme Value Theorem Theorem compact. Suppose D R is compact ad f : D! R is cotiuous. The f(d) is Proof Give a sequece fy g 2I i f(d), choose a sequece fx g 2I such that f(x )= y. Sice D is compact, fx g 2I has a coverget subsequece fx k g 1 with k=1 lim k!1 x k = x 2 D: Let y = f(x). The y 2 f(d) ad, sice f is cotiuous, Hece f(d) is compact. y = lim k!1 f(x k ) = lim k!1 y k : Exercise Proof the previous theorem usig the ope cover deitio of a compact set. The followig theorem is kow as the Extreme Value Theorem. Suppose D R is compact ad f : D! R is cotiuous. The there exists Theorem a 2 D such that f(a) f(x) for all x 2 D ad there exists b 2 D such that f(b) f(x) for all x 2 D.

3 Lecture 15: Cosequeces of Cotiuity 15-3 Proof Let s = sup f(d) ad t = if f(d). The s 2 f(d), so there exists a 2 D such that f(a) =s, ad t 2 f(d), so there exists b 2 D such that f(b) =t. As a cosequece of the Extreme Value Theorem, a cotiuous fuctio o a closed bouded iterval attais both a maximum ad a miimum value. Exercise Fid a example of a closed bouded iterval [a; b] ad a fuctio f :[a; b]! R such that f attais either a maximum or a miimum value o [a; b]. Exercise Fid a example of a bouded iterval I ad a fuctio f : I! R which iscotiuous o I such that f attais either a maximum or a miimum value o I. Exercise Suppose K R is compact ad a=2 K. Show that there exists b 2 K such that jb, aj jx, aj for all x 2 K. Suppose D R is compact, f : D! R is oe-to-oe, ad E = f(d). Propositio The f,1 : E! D is cotiuous. Proof Let V R be a ope set. We eed to show that f(v \ D) =U \ E for some ope set U R. Let C = D \ (R V ). The C is a closed subset of D, ad so is compact. Hece f(c) is a compact subset of E. Thus f(c) is closed, ad so U = R f(c) is ope. Moreover, U \ E = E f(c) =f(v \ D). Thus f,1 is cotiuous. Exercise Suppose f :[0; 1] [ (2; 3]! [0; 2] by f(x) = x; if 0 x 1, x, 1; if 2 <x 3. Show that f is cotiuous, oe-to-oe, ad oto, but that f,1 is ot cotiuous Uiform cotiuity Deitio Suppose D R ad f : D! R. Wesay f is uiformly cotiuous o D if for every >0 there exists >0 such that for ay x; y 2 D, wheever jx, yj <. jf(x), f(y)j < Exercise Suppose D R ad f : D! R is Lipschitz. Show that f is uiformly cotiuous o D. Clearly, iff is uiformly cotiuous o D the f is cotiuous o D. cotiuous fuctio eed ot be uiformly cotiuous. However, a

4 Lecture 15: Cosequeces of Cotiuity 15-4 Example Dee f :(0; +1) by f(x) = 1. Give ay >0, choose 2 x Z+ such that 1 <. Let (+1) x = 1 ad y = 1. The +1 jx, yj = 1, 1 +1 = 1 ( +1) <: However, jf(x), f(y)j = j, ( +1)j =1: Hece, for example, there does ot exist a >0 such that wheever jx, yj <. cotiuous o (0; +1). Example jf(x), f(y)j < 1 2 Thus f is ot uiformly cotiuous o (0; +1), although f is Dee f : R! R by f(x) =2x. Let >0 be give. If = 2, the jf(x), f(y)j =2jx, yj < wheever jx, yj <. Hece f is uiformly cotiuous o R. Exercise Show that f(x) =x 2 is ot uiformly cotiuous o (,1; +1). Propositio uiformly cotiuous o D. Suppose D R is compact ad f : D! R is cotiuous. The f is Proof Let >0begive. For every x 2 D, choose x such that wheever y 2 D ad jx, yj < x. Let jf(x), f(y)j < 2 J x =(x, x 2 ;x+ x 2 ): The fj x : x 2 Dg is a ope cover of D. Sice D is compact, there must exist x 1 ;x 2 ;:::;x, 2 Z +, such that J x1 ;J x2 ;:::;J x is a ope cover of D. Let be the smallest of x1 2 ; x2 2 ;:::; x 2 : Now let x; y 2 D with jx, yj <. The for some iteger k, 1 k, x 2 J xk, that is, jx, x k j < x k 2 :

5 Lecture 15: Cosequeces of Cotiuity 15-5 Moreover, Thus jy, x k jjy, xj + jx, x k j <+ x k 2 x k : jf(x), f(y)j jf(x), f(x k )j + jf(x k ), f(y)j < = : Exercise Suppose D R ad f : D! R is uiformly cotiuous. Show that if fx g 2I is a Cauchy sequece i D, the ff(x )g 2I is a Cauchy sequece i f(d). Exercise Suppose f :(0; 1)! R is uiformly cotiuous. Show that f(0+) exists. Exercise Suppose f : R! R is cotiuous ad lim x!,1 f(x) = 0 ad lim x!+1 f(x) = 0. Show that f is uiformly cotiuous.