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Class notes Nov. 13 The Intermediate Value Theorem We start by stating the Intermediate Value Theorem (Theorem 4.5.1. in Abbott) and give its proof. Theorem 1. Let f :[a, b] R be a continuous function and let L be a real number satisfying f (a) L f (b) or f (a) L f (b) . Then there exists a point a c b such that f (c) L . Proof: Let us assume for convenience that f (a) L f (b) . Define the function g ( x) f ( x) L . Then g is continuous on [a, b] , and g (a) f (a) L 0, g (b) f (b) L 0 . Then the conclusion of the theorem is equivalent to showing that g must be zero somewhere inside the interval [a, b] . Define the following sets: A1 {x [a, b]: g ( x) 0}, A2 {x [a, b]: g ( x) 0} . Both sets are bounded, so by the Axiom of Completeness they both have a supremum and an infimum. Let a1 inf A1 , b1 sup A1 , a2 inf A2 , b2 sup A2 . Notice that a A1 , b A2 , so inf A1 a,sup A2 b . Since b1 sup A1 there exists a sequence ( xn ) A1 : lim xn b1 . But since the function g is n continuous, we have lim g ( xn ) g (b1 ) , and since g ( xn ) 0, for all n 1, g (b1 ) 0 , n therefore b1 b . This allows us to construct a sequence converging to b1 from above: yn b1 : lim yn b1 . Since b1 sup A1 we have g ( yn ) 0, for all n 1 and n lim g ( yn ) g (b1 ) 0 . So we found a point a b1 sup A1 b such that both g (b1 ) 0 n and g (b1 ) 0 must be true. Then g (b1 ) 0 and the proof is complete. QED Following you will find some hints to some of the homework problems due Nov15. Exercise 4.4.4. We are given f :[a, b] R, f ( x) 0 on [a, b] . Now, for the function to be bounded on [a, b] , we need to show that there exists M 0 : have that f ( x) 0 so 1 0 . The statement f ( x) 1 f 1 M . We already f ( x) 1 M f ( x) is equivalent to 1 0 , so what you need to show is that the function has a positive lower bound M on the interval [a, b] . Can you say anything about inf{ f ( x) : x [a, b]} ? For Exercise 4.4.8 use the approach in Exercise 4.4.7 below. f ( x) Exercise 4.4.7. We have g uniformly continuous on (a, b] and [b,c),a<b<c . We write the uniform continuity for the two intervals. Let 0 . We know that there exists 1 0 such that for any x, y (a, b] :| x y | 1 ,| g ( x) g ( y ) | (1). You will see 2 later why we chose in the above. 2 Also, since g is uniformly continuous on [b,c) , there exists 2 0 such that for any x, y [b, c) :| x y | 2 ,| g ( x) g ( y ) | (2). 2 Let min{1 , 2 } . From (1) and (2) it follows that if x, y (a, c) :| x y | and either x, y (a, b] or x, y [b, c) then | g ( x) g ( y ) | . However, we should ask ourselves 2 what happens if x b y :| x y | ? Notice that in this case | x b | and | y b | and | g ( x) g (b) | ,| g (b) g ( y ) | . Then we write 2 2 | g ( x) g ( y ) || g ( x) g (b) | | g (b) g ( y ) | 2 2 and we are done. Remark. If in the above we knew that the function is uniformly continuous on two overlapping intervals, such as (a, b],[d , c), a d b c then there would be no need of discussing the case x b y :| x y | separately. Think about this when doing exercise 4.4.8. Exercise 4.4.11. (Topological Characterization of Continuity) Let g : R R . Define the preimage of a set A: g 1 ( A) {x R : g ( x) A} . Show that g is continuous if and only if for any open set O R, g 1 (O) is open. Proof: We need to prove two implications here. Let’s assume first that g is continuous, and let O R be open. We will prove that g 1 (O) is open by proving that all points in g 1 (O) are interior points. Let a g 1 (O) . Then g (a) O and since O is open, there exists a neighborhood of g ( a ) contained in O, that is: there exists 0 such that ( g (a) , g (a) ) O . Since the function is continuous at a there exists 0 such that for any x :| x a | ,| g ( x) g (a) | , or equivalently g ( x) ( g (a) , g (a) ) O . But if g ( x) O it follows that x g 1 (O) for all x :| x a | , so (a , a ) g 1 (O) , proving that g 1 (O) is open. Conversely, let us assume g 1 (O) is open for any open set O. Let a R and let 0 . Define the open set O ( g (a) , g (a) ) . According to our assumption g 1 (O) must be open as well, and we also know that a g 1 (O) . Since g 1 (O) is open we can find a -neighborhood of a contained in g 1 (O) , that is there exists 0 such that (a , a ) g 1 (O) . This last statement implies that g ( x) O for all x :| x a | , or equivalently | g ( x) g (a) | for all x :| x a | , proving that g is continuous at a. QED