5.5 Deeper Properties of Continuous Functions

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1 5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS Deeper Properties of Continuous Functions Intermediate Value Theorem and Consequences When one studies a function, one is usually interested in the values the function can have. In other words, one is interested in the range of the function. More precisely, if f is a function with domain D, one tries to answer questions of the type: 1. What kind of a set is the range R of f? 2. Under which conditions is the range of f bounded? 3. Under which conditions is it closed? 4. Does f have a minimum? a maximum? You should recognize some of these questions. You studied them in Calculus I. We answer some of these questions in this section. We begin by stating and proving a very important theorem: the intermediate value theorem. Intuitively, this theorem says that the idea of a continuous function is that its graph can be drawn without lifting the pencil. In other words, its graph has no holes or breaks. Thus, it takes on all values between any two it achieves. More precisely: Theorem (Intermediate Value Theorem) Suppose that f is continuous on the closed interval I = [a, b], let A = f (a), and B = f (b) and suppose that A B. If C is a number between A and B, then there exists a number c in (a, b) such that f (c) = C. If in addition f is strictly monotone on I then c is unique. Proof. The theorem has two parts: existence of c and uniqueness of c. We prove each part separately. Existence of c The reader should draw a picture corresponding to the situation of the theorem and represent on the picture the various quantities involved in the proof. Assume all the hypotheses of the theorem are satisfied. If A B, then either A > B or A < B. We do the proof in the case A > B. The case A < B is identical, and is left to the reader. We look at g (x) = f (x) C. Since f is continuous on [a, b] and C is a constant, g is also continuous on [a, b]. Furthermore, by our assumption, g (a) > 0. Since g is continuous, by theorem , g will remain strictly positive to the right of a. For the same reason, g is strictly negative at b, and therefore to the left of b. Thus, the set S = {x I such that f (x) C > 0} is not empty. Furthermore, it is a subset of I and therefore is bounded. Hence S has a supremum. Let c = sup S. The claim is that c is the number in the theorem, in other words, f (c) = C. We prove this fact by contradiction. Assume that f (c) C. Then, either f (c) > C or f (c) < C.

2 196 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION case 1: f (c) > C. By continuity of f, f (x) will remain strictly greater than C for a while. In other words, there is a number x 1 > c such that f (x 1 ) > C. Thus, x 1 S and x 1 > c. This contradicts the fact that c is an upper bound for S. case2: f (c) < C. Again, by continuity of f, f (x) < C to the left of c. Thus, there exists x 1 < c such that f (x 1 ) < C. Thus, x 1 < c is an upper bound for S, which contradicts the fact that c = sup S. Therefore, f (c) = C Uniqueness of c We do a proof by contradiction. Assume c is not unique; that is there exists at least two such numbers, call them c 1 and c 2. Without loss of generality, we may assume that c 1 < c 2. f is either strictly increasing or strictly decreasing. case 1: f is strictly increasing. In this case, f (c 1 ) < f (c 2 ). Therefore, we have: C = f (c 1 ) < f (c 2 ) = C which is a contradiction. case 2: f is strictly decreasing. In this case, f (c 1 ) > f (c 2 ). Therefore, we have: C = f (c 1 ) > f (c 2 ) = C which is also a contradiction. This theorem is very important for both theoretical and practical reasons. On the theoretical side, as we will see shortly, many important results depend on this theorem. On the practical side, this theorem can be used to prove that certain equations have solutions. It can also be used to find approximations of these solutions. In fact, these approximations can be computed with as much accuracy as one needs. We illustrate this by an example. Example Show that x x 3 207x 2 70x = 0 has a solution between 2 and 3. Find an approximation of this solution correct to 2 decimal places. Let f (x) = x x 3 207x 2 70x Then, f is continuous for all real numbers, so it is continuous on [2, 3]. In addition, f (2) = 528 and f (3) = 322. So, 0 is between f (2) and f (3). By the intermediate value theorem, there exists a number c between 2 and 3 such that f (c) = 0. To find an approximation, we compute f (2.1), f (2.2),..., f (2.9) until we find two consecutive values for which f changes sign. In this case, f (2.6) = and f (2.7) = So, repeating the same argument as above, we find that c is between 2.6 and 2.7. We repeat the same procedure with f (2.61), f (2.62),..., f (2.69). We find that f (2.64) = and f (2.65) = By the same argument as above, c is between 2.64 and So, we know that the 2.6 part is correct. We need

3 5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS 197 to repeat the procedure one more time since we want the solution correct to 2 decimal places. We compute f (2.641), f (2.642),..., f (2.649). We find that f (2.645) = and f (2.646) = Therefore, c is between and Thus, 2.64 is part of the exact answer. Corollary Let f be continuous on an interval I (any interval). Then, the range of f is also an interval. Proof. See homework. This theorem tells us that if the domain of f consists of one piece, so will its range. Some of the problems below as the reader to sketch some graphs relating a function and its range. The reader should do them before reading further, they will help in the understanding of these concepts. Note in particular that the theorem does not say that the range and the domain are the same type of intervals. One could be closed, the other could be open. One could be bounded, the other not. In the previous theorem, I was any interval. When we study f on a closed interval, we can reach even more powerful conclusions, as the next theorems show. Corollary Let f be continuous on a closed interval [a, b]. Then, the range of f is bounded. Proof. See homework. Theorem Let f be defined and continuous on [a, b]. Then, f attains the supremum and the infimum of its range. Proof. We show that f attains the supremum of its range. Let R denote the range of f. By the previous theorem, we know R is bounded, thus bounded above. Since it is not empty (f (a) R), it has a supremum. Let C = sup R. We need to show there exists c [a, b] such that f (c) = C. Define g (x) = C f (x) for x [a, b]. Then, either g (x) = 0 for some x in [a, b] or g (x) > 0 for all x in [a, b]. case 1: g (x) = 0 for some x in [a, b]. Then, we are done. case 2: g (x) > 0 for all x in [a, b]. We show this cannot happen. Suppose it did. Then, g would be continuous and non zero on [a, b] (why?). Therefore, 1 the function would also be continuous on [a, b] (why?). Thus, by the g (x) 1 previous theorem, would be bounded on [a, b]. Because C = sup R, g given ɛ > 0, one can find x [a, b] such that 0 < g (x) < ɛ (why?), thus 1 g (x) > 1 ɛ. It follows that 1 cannot be bounded, which is a contradiction. g Corollary Let f be defined and continuous on [a, b], then the range of f is a closed interval. Proof. See exercises at the end of the section

4 198 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION Corollary Let f be defined and continuous on [a, b]. Then, there exists x 1 and x 2 in [a, b] such that for all x [a, b], we have: f (x 1 ) f (x) f (x 2 ) Proof. See exercises at the end of the section We have seen some interesting properties of continuous functions. First, we saw that if I is an interval and f is a continuous function, then f (I) is also an interval. Thus, a continuous function preserves intervals. We also saw that the image of a closed interval was a closed interval. More precisely, f ([a, b]) = [m, M] where m = inf (f ([a, b])), M = sup (f ([a, b])). We might ask ourselves what other properties of sets are preserved by continuous functions. These are important questions that have been asked by mathematicians. If you want to know some of these answers, keep taking math courses! Topological Interpretation of Continuity If you had taken a topology class before real analysis, you would have seen a different definition of continuity. What we list below as a theorem is taken as the topological definition of continuity. We show the two definitions, though very different, are equivalent. In proving the theorem, we will use our definition of continuity. Theorem Let E be a subset of R and f : E R a function. Then: f continuous f 1 (V ) open in E for every open subset V of R Proof. We outline the proof and leave the details as homework. Suppose f is continuous on E. Let V R be open, show that f 1 (V ) is open. Pick a f 1 (V ), show that a is an interior point that is there exists δ > 0 such that (a δ, a + δ) f 1 (V ). But f (a) V which is open, so there exists ɛ > 0 such that (f (a) ɛ, f (a) + ɛ) V. Since f is continuous, for each ɛ, there exists a δ. See how the δ you get from continuity can be used to show (a δ, a + δ) f 1 (V ). Suppose f 1 (V ) open in E for every open subset V of R, show f is continuous on E. We need to show f is continuous at every point of E. The steps are similar to those of the other direction Exercises 1. Prove that x4 + 2x x x6 + 2x x 7 = 0 has a solution between 1 and

5 5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS Same question as the previous problem for g (x) = 5 x x x 3 between 1 and Prove corollary Prove corollary In the proof of theorem there are three statements you were asked to justify. They are the statements with (why?). Justify these statements. 6. Prove corollary Prove corollary Sketch the graph of a continuous function on an open interval whose range is also an open interval. 9. Sketch the graph of a continuous function on an open interval whose range is a closed interval. 10. Sketch the graph of a function on an interval of the form [a, ) whose range is a closed interval. 11. Illustrate with an example the fact that if f is not continuous at a point, then its range may not be an interval, even if its domain is a closed interval. 12. Finish proving theorem

6 200 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION Hints for the Exercises 1. Prove that x4 + 2x x6 + 2x = 0 has a solution between 1 and x 1 x 7 7. Find an approximation of this solution correct to 2 decimal places. Hint: As in the example in the notes, you will have to use the intermediate value theorem. However, you will have to be a little more creative as the expression on the left of the equal sign is not defined at 1 and 7. Answer: The solution correct to 2 decimal places is: Same question as the previous problem for g (x) = 5 x x x 3 between 1 and 3. Hint: Same as previous problem. 3. Prove corollary Hint: Use the definition of an interval and the intermediate value theorem. 4. Prove corollary Hint: We do the proof in several steps, outlined here. (a) Justify why f is bounded on some neighborhood of a. (b) Define S = {c [a, b] such that f is bounded on [a, c)}. Prove that sup S exists using the completeness axiom. (c) Let d = sup S. Prove by contradiction that d = b. (d) Explain why part c proves the result. 5. In the proof of theorem there are three statements you were asked to justify. They are the statements with (why?). Justify these statements. Hint: none 6. Prove corollary Hint: Use the theorems of this section. 7. Prove corollary Hint: Use the theorems of this section. 8. Sketch the graph of a continuous function on an open interval whose range is also an open interval. 9. Sketch the graph of a continuous function on an open interval whose range is a closed interval. 10. Sketch the graph of a function on an interval of the form [a, ) whose range is a closed interval.

7 5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS Illustrate with an example the fact that if f is not continuous at a point, then its range may not be an interval, even if its domain is a closed interval. 12. Finish proving theorem Hint: See the hints in the notes.

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