Part 2.2 Continuous functions and their properties v1 2018

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1 Part 2.2 Continuous functions and their properties v 208 Intermediate Values Recall R is complete. This means that ever non-empt subset of R which is bounded above has a least upper bound. That is: (A R : A φ and M : a A,a M) = luba eists. And the definition of luba is that, if λ = luba then λ is an upper bound: a A,a λ, λ is the least of all upper bounds; if µ is an upper bound for A then λ µ. Alternativel For all δ > 0, λ δ is not an upper bound for A which means a A : λ a > λ δ. We also recall an earlier lemma, previousl given for limits but here stated for continuous functions: Lemma 2.2. If g : A R is continuous at a R then. if g(a) > 0 there eists δ > 0 such that a < δ implies g() > 0, 2. if g(a) < 0 there eists δ > 0 such that a < δ implies g() < 0. Theorem (Bolzano 87) Intermediate Value Theorem Suppose that f is a function continuous on a closed and bounded interval [a,b]. For all γ between f(a) and f(b) there eist c : a c b for which f(c) = γ. Important Do get the order of the quantifiers correct, for all first and there eists second, i.e. γ between f(a) and f(b) c : a c b and f(c) = γ.

2 On a graph ou would be starting with a point γ on the -ais and finding a point c on the -ais which maps to it. f(b) γ f f(a) a c b Proof of I V Th m We first translate and reflect the function f. There are two cases; If f(a) f(b) then f(a) γ f(b). Define g() = f() γ, then g(a) 0 and g(b) 0. If f(a) > f(b) then f(b) γ < f(a). This time define g() = γ f(), then again g(a) 0 and g(b) 0. Summing up, g() = and in both cases g(a) 0 g(b). { f() γ if f(a) f(b) γ f() if f(a) > f(b). If either g(a) = 0 or g(b) = 0 the proof is finished, simpl choose c = a or b respectivel. Thus we ma assume that we have strict inequalities in g(a) < 0 < g(b) and it suffices to find c (a,b) : g(c) = 0. Consider the set S = { [a,b] : g() < 0}. Then S φ since a S, while S [a,b] and so S is bounded above b b. Therefore, b the Completeness Aiom of R there eists c R : c = lubs. There are three possibilities g(c) < 0, g(c) > 0 and g(c) = 0. 2

3 Case : Assume g(c) < 0. Continuit of g :BLemma2.2.8thereeistsδ > 0suchthatif c < δ then g() < 0. If δ is sufficientl small then { : c < δ} [a,b]. Consider the point 0 = c + δ/2. Then 0 { : c < δ} and so g( 0 ) < 0, i.e 0 S, b the definition of S. But 0 S means 0 c since c is an upper bound for S. On the other hand 0 = c+δ/2 > c. Hence 0 c and 0 > c, a contradiction, which means our assumption is false and we do not have g(c) < 0. Case 2 Assume g(c) > 0. Continuit of g :BLemma2.2.8thereeistsδ > 0suchthatif c < δ then g() > 0, that is / S. Definition of c : The fact that c = lubs means that c is the least of all upper bounds for S in which case c δ, with the δ just found, is not an upper bound for S. In turn this means there eists S and c > c δ. But S means g( ) < 0 while c > c δ, written as c < δ implies g() > 0. Hence g( ) < 0 and g() > 0, a contradiction, which means our assumption is false and we do not have g(c) < 0. Case 3 Because ever other possibilit leads to a contradiction we must therefore have g(c) = 0. There is a good chance ou will have used this result, for eample b finding roots of a polnomial b looking for a sign change. 3

4 Eample Let p() = Show that there is a zero of this polnomial between 0 and 4. Is there a zero between 0 and 2.5? Solution p(0) = 6andp(4) = 6sop(0) < 0 < p(4),i.e. 0isanintermediate value between p(0) and p(4). Since p is a polnomial it is continuous so we can appl the Intermediate Value Theorem with γ = 0 to deduce that there eists 0 < c < 4 for which p(c) = 0. Since p(2.5) = there is no sign change between 0 and 2.5 so we cannot appl the Intermediate Value Theorem with γ = 0 to show there is a zero in [0,2.5]. This is a weakness of this method to find roots for it is not hard to see that = is a root of p() in [0,2.5]. In fact, from the graph ou can see two roots between 0 and Eample Show that for all real a,b > 0 there is a solution to asin = bcos in [0,π/2]. Solution in Tutorial Let f() = asin bcos. We see that f(0) = b and f(π/2) = a so f(0) < 0 < f(π/2). Since f is continuous on [0,π/2] the Intermediate Value Theorem implies there eists c (0, π/2) such that f(c) = 0, i.e. asinc = bcosc. Eample (A special case of the) Fied Point Theorem. If f : [0,] [0,] is continuous then there eists c [0,] such that f(c) = c. Solution Define g() = f (), a function continuous on [0,]. B definition 0 f () for all 0. In particular f (0) 0 and so g(0) = f (0)

5 Similarl, f () so g() = f () = 0. Thatis, g() 0 g(0). SoapplI.V.Thmtog on[0,]tofindc : g(c) = 0, i.e. f (c) = c. This result should not be a surprise. Being continuous on a closed interval the function f is tied down at f(0) and f(). Since these values are between 0 and the graph between them has to cross the line =. See Figure. The same result should hold with = replaced b an continuous function between (0,0) and (,). For eample see Figure 2 where = 3. f(0) f = f(0) f = 3 f() f() c c Figure : = Figure 2: = 3 Eample If f : R [,8] is continuous then there eists c R such that f(c) = c 3. Solution in Tutorial If there is a solution of f(c) = c 3 then, since f(c) 8 we have c 3 8, i.e. c 2. This could be seen on the graph: 5

6 8 = 3 =f() 2 So we need onl appl the Intermediate Value Theorem on the interval [,2]. Let g() = f() 3. Then g() = f() 3 = 0 since f() for all [,2]. Also g(2) = f(2) = 0 since f() 8 for all [,2]. Thus g() 0 g(2), i.e. 0 is an intermediate value. Appl the Intermediate Value Theorem to g on [,2] with γ = 0 to show there eists c [,2] such that g(c) = 0, that is, f(c) = c 3. 6