Continuity. Chapter 4

Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of D R with the property that for each x D there is exactly one y R with (x, y) in this subset. We write f(x) = y. Our high school notion of continuity might be that the function is continuous if it can be drawn without lifting our pencil - we will see if we can come up with definitions and theorems to corroborate this notion. Definition 4.. Suppose f : D R. (i) A function f is continuous at c D if and only if, for all ǫ > 0, there exists a δ > 0 such that for all x D, x c < δ implies f(x) f(c) < ǫ. (ii) A function f is called continuous if and only if it is continuous at each point in its domain D. With our topological notions this is equivalent to Definition 4.3. A function f is continuous at c D if and only if, for any neighborhood V of f(c), there exists a neighborhood U of c such that, if x U D, then f(x) V. Example 4.4. Suppose D = N, and suppose f(n) = n. Is f continuous? Proof. Yes. Indeed, let ǫ > 0 and n 0 D. Set δ = 1/. Then for all x N, x n 0 < δ implies f(x) f(n 0 ) = 0 < ǫ. So much for our no-pencil-lifting ideas. It turns out a function is always continuous at an isolated point of the domain (see Problem 4.6). Example 4.5. Let D = R and f(x) = x. Prove f is continuous at x = 1. 35

36 4. Continuity Proof. We need to show x 1 is arbitrarily small when x is close enough to one. Note x 1 = x + 1 x 1. Thus we need to control the size of x + 1 to make the difference small. Let us suppose arbitrarily that x 1 < 1. By the triangle inequality this implies x 1 < 1 or x <. Then x + 1 x + 1 < 3. Set δ = min{1, ǫ/3}. Then x 1 < δ implies x 1 = x + 1 x 1 < 3 x 1 < ǫ. We need to know when a function is not continuous at a point in its domain. Lemma 4.6. The function f is not continuous at c D if there exists an ǫ 0 > 0 such that for all δ > 0 there exists x 0 D with x 0 c < δ and f(x 0 ) f(c) ǫ 0. Proof. In logic notation continuity at a point is The negation of this is ( ǫ > 0)( δ > 0)( x D)( x c < δ f(x) f(c) < ǫ). ( ǫ 0 > 0)( δ > 0)( x 0 D)( x 0 c < δ and f(x 0 ) f(c) ǫ 0 ). Example 4.7. (Dirichlet function) Consider the function on the real line 1 x Q, f(x) = 0 x Q c. Show the Dirichlet function is not continuous anywhere. Proof. Set ǫ 0 = 1/, δ > 0, and let c R. By Theorem.17 (the density of rationals) and its corollaries there exists r Q and ξ Q c such that and δ + c < r < c + δ δ + c < ξ < c + δ. That is, r c < δ and ξ c < δ. Either f(c) = 0 or f(c) = 1, but f(r) = 1 and f(ξ) = 0. Thus there exists x 0 D (either r or ξ), such that x 0 c < δ and f(x 0 ) f(c) = 1 ǫ 0. Example 4.8. Consider the function on the real line given by x x Q f(x) = c, 1 x x Q. Show the function is continuous at x = 1/ and discontinuous everywhere else.

4.1. Continuity and Limits 37 Proof. To show the function is continuous at x = 1/ we need to show f(x) f( 1 ) is small when x is close enough to 1/. We compute f(x) f( 1 x ) = 1 x Q c, 1 x x Q. Thus f(x) f( 1 ) = x 1 for all x R. Let ǫ > 0. Set δ = ǫ, and the result follows. The proof the function is discontinuous everywhere else follows the previous example. In Example 4.0 we show the existence of a function which is continuous on the irrationals and discontinuous on the rationals. 4.1. Continuity and Limits In this section we establish a connection between continuity and limits. We first define the quantity lim f(x) = L. To make sense of the limit we need for the function f to be defined near c. This puts some restrictions on the domain. Example 4.9. Suppose D = N. What should mean? lim f(n) = L n 5 Definition 4.10. Let f : D R with x 0 a cluster point of D (Definition 3.4). Then f has a limit L at x 0 if and only if, for all ǫ > 0, there exists δ > 0 such that for all x D, 0 < x x 0 < δ implies f(x) L < ǫ. We denote the limit lim f(x) = L. We recall a cluster point of D may not be in D. Hence, we require 0 < x x 0 sin x (so x 6= x 0 ). Thus, for example, lim makes sense. x 0 x The following theorem may seem familiar. Indeed, some elementary texts take (iii) as the definition of continuity. Theorem 4.11. Let f : D R and let c be a cluster point of D with c D. The following are equivalent. (i) f is continuous at c. (ii) All sequences {x n } n N D, x n c as as n implies {f(x n )} n N converges to f(c). (iii) lim f(x) = f(c). Remark 4.1. Combining (i) and (ii), we see (ii) says lim n f(x n) = f( lim n x n). That is, the limit passes through continuous functions.

38 4. Continuity Remark 4.13. To show (ii) implies (i) we will prove the contrapositive. Some of the negations here are involved. For reference we have (i) ( ǫ > 0)( δ > 0)( x D)( x c < δ f(x) f(c) < ǫ), (i) ( ǫ 0 > 0)( δ > 0)( x 0 D)( x 0 c < δ and f(x 0 ) f(c) ǫ 0 ). Moreover, not (ii) is ( {x n } n N D) x n c and ( ǫ 0 > 0)( N N)( n 0 N)( f(x n0 ) f(c) ǫ 0 ), and (iii) ( ǫ > 0)( δ > 0)( x D)(0 < x c < δ f(x) f(c) < ǫ). Proof. We show (i) implies (ii). Let {x n } n N be a sequence as stated. We have to show the sequence {f(x n )} n N converges. This requires us to find a N N so that f(x n ) f(c) < ǫ. To find such a N we return to the definition of continuity. Let ǫ > 0. There exists a δ > 0 such that x D, x c < δ implies f(x) f(c) < ǫ. Since the sequence {x n } n N converges to c, there exist N such that x n c < δ for n N. However, for such n, f(x n ) f(c) < ǫ as required. Next we show not (i) implies not (ii). Set δ = 1/n, n N in the negation of (i). Then ǫ 0 > 0 exists such that for each n N, there exists x n D such that x n c < 1/n and f(x n ) f(c) ǫ 0. This is not (ii). To see (i) and (iii) are equivalent we first notice the only difference between them is (iii) has 0 < x c < δ while (i) has x c < δ. Thus (i) implies (iii) is trivial. To get the converse note that c D. Thus, if x = c, f(x) f(c) = 0 < ǫ, and there is no harm in dropping the condition 0 < x c. That is, (iii) implies (i). Example 4.14. Show the function f(x) = x is continuous on D = [0, ]. Proof. Homework Problem 3.1 shows that, for any sequence x n x, x n x as n. Thus by Theorem 4.11 f is continuous on its domain. Theorem 4.15. (Algebra of Continuous Functions) Suppose f, g : D R are continuous functions at c D and with c a cluster point of D. Then at c the functions (1) αf, for α R, () f ± g, (3) f g, (4) f/g, if g 6= 0 on D, are continuous. Proof. Let {x n } D be any sequence converging to c. By the algebra of limits for sequences, Theorem 3.1, the functions in the list satisfy (ii) in Theorem 4.11. Theorem 4.11 now provides the result. Remark 4.16. Note that, if c D is not a cluster point of D, then f : D R is continuous at c. Thus, no loss of generality occurs in requiring c to be a cluster point of the domain. See Problem 4.6.

4.1. Continuity and Limits 39 Theorem 4.17. (Composite Functions) Let f : A B and g : B R. Suppose f is continuous at c A with c a cluster point of A, and g continuous at f(c). Then the composite function g(f(x)) = g f(x) is continuous at c. Proof. Let {x n } A be any sequence converging to c. By assumption and Theorem 4.11, {f(x n )} n N converges to f(c) B. In particular, z n = f(x n ) B converges to f(c) B. Since g is continuous at f(c), the analysis in the first paragraph of the proof of Theorem 4.11 shows {g(z n )} n N converges to g(f(c)). That is, {g(f(x n ))} n N converges to g(f(c)) for all sequences x n A with x n c as n. Theorem 4.11 provides the result. Example 4.18. Prove cos x is continuous on R. Proof. Homework Problem 4.10 shows sin x is continuous on the real numbers. Note that cos x = sin(x + π ). One easily checks that f = x + π/ is a continuous function. Theorem 4.17 gives the result. Example 4.19. Prove tan x is continuous on [0, π/). Proof. Since sin x and cos x are continuous on the stated interval and cos x is non zero there, Theorem 4.15 applies and provides the result. Example 4.0. Consider the function defined on (0, 1) given by 0 x Q f(x) = c, x Q, x = m/n. 1 n Show the function is continuous on the irrationals and discontinuous on the rationals. Proof. Let r (0, 1) be rational and {ξ n } n N be a sequence of irrational numbers in (0, 1) converging to r. If f were continuous at r, then according to Remark 4.1, 0 = lim n f(ξ n) = f( lim n ξ n) = f(r) 6= 0. This is a contradiction and so f is not continuous on the rational numbers in (0, 1). To show f is continuous on the irrational numbers in (0, 1), let ξ (0, 1) be irrational. We must show f(x n ) f(ξ) as n for all sequences in (0, 1) converging to ξ. Since f(ξ) = 0 and f(x n ) = 0 if x n is irrational, we may suppose the x n are rational for n N. Thus, we suppose x n = p n /q n in reduced form and n N. Since f(x n ) = 1/q n for each n N, we need to show 1/q n 0 as n. Suppose to the contrary. Then an ǫ 0 > 0 exists so that for each N N there exists a n 0 > N so that 1/q n0 ǫ 0. This implies the existence of a subsequence (let N = k) {q nk } k N such that q nk 1/ǫ 0 for all k N. Since the q nk are bounded for k N, the set E = x nk = p n k q nk k N (0, 1) is finite. It follows any convergent sequence in E must eventually be constant (cf. Problem 3.13) and hence rational - this contradicts ξ being irrational.

40 4. Continuity 4.. More on limits The next theorem is quite similar to Theorem 4.11, and the proof is almost identical. Theorem 4.1. (Sequential Characterization of Limits) Let f : D R, and let c be a cluster point of D (not necessarily in D). The following two statements are equivalent. (i) lim f(x) = L, (ii) All sequences {x n } n N D, such that x n 6= c for n N, x n c as n, implies {f(x n )} n N converges to L. Proof. To show (i) implies (ii), let ǫ > 0. A δ exists so that f(x) L < ǫ for all x D and 0 < x c < δ. Choose N so that 0 < x n c < δ for all n N. Then f(x n ) L < ǫ. This is (ii). For the converse we prove the contrapositive - not (i) implies not (ii). For each n N, set δ = 1/n. Not (i) implies there exists x n D such that 0 < x n c < 1/n and f(x n ) L ǫ 0 for some fixed ǫ 0 > 0. This is not (ii). We can get much mileage from the last theorem. Example 4.. Show that the function f(x) = sin(1/x) on D = (0, ) has no limit at x = 0. Proof. We just need to find a sequence {x n } n N D with x n as n and with {f(x n } n N having no limit or different limits for different {x n } n N with the mentioned properties. Consider a n = π(4n + 1), b n = 1 nπ. Then f(a n ) = sin(π(4n + 1)/) = 1 and f(b n ) = sin nπ = 0 for all n N. Thus (ii) is not true in Theorem 4.1, and the limit does not exist. Theorem 4.3. (Algebra of Limits) Let f, g : D R and c be a cluster point of dd. Suppose lim f(x) = L 1, and lim g(x) = L. Then (a) For all α R, lim αf(x) = αl 1. (b) lim (f(x) + g(x)) = L 1 + L. (c) lim f(x)g(x) = L 1 L. (d) If g(x) 6= 0 on D and L 6= 0, then lim f(x) g(x) = L 1 L. Proof. Let {x n } n N D be any sequence converging to c. Then by the algebra of limits for sequences, Theorem 3.1, the above are true for sequences. Theorem 4.1 now provides the result.

4.3. Properties of Continuous Functions 41 4.3. Properties of Continuous Functions We are now in a position to establish some global properties of continuous functions. The first result gives conditions for a function to have a minimum and maximum. Here are two functions which do not have a maximum. Example 4.4. f(x) = x on (0, 1). Example 4.5. f(x) = x 0 x < 1, 0 x = 1. Note the first function is continuous - so we expect some restriction on the domain for a maximum to exist. The second function is not continuous. Theorem 4.6. (Extreme-Value Theorem) Suppose I is a closed bounded interval and f : I R is continuous. Then x m, x M exists in I such that f(x m ) = min f, f(x M ) = max f. I I Remark 4.7. We note that f bounded means While f not bounded means ( M)( x I)( f(x) M). ( M)( x I)( f(x) > M). Proof. We first show the function is bounded. Suppose f is not bounded on I. For each n N let M = n in the negation. Then there exists a sequence {x n } n N I such that f(x n ) > n. Since the sequence is bounded the Bolzano-Weierstrass theorem applies and a convergent subsequence of {x n } n N exists - say {x nk } k N converging to x 0 as k. Exercise 3.10 shows that x 0 I. Since f is continuous (Remark 4.1) lim f(x n k ) = f( lim x n k ) = f(x 0 ). k k However, f(x nk ) n k k and {f(x nk )} k N does not converge - a contradiction and f is bounded. Consider the set E = {f(x) x I}. Note E is nonempty and bounded. By the Completeness Axiom E has a supremum. Set M = sup E. We need to find x M I such that f(x M ) = M. By the properties of supremum (Corollary 3.6) there exists a sequence {z n } such that M 1/n < f(z n ) M. The Bolzano- Weierstrass theorem applies again and a subsequence {z nk } k N exists converging to x M I (this defines x M ). By continuity and the squeeze theorem, Theorem 3.4, we have M = lim f(z n k ) = f( lim z n k ) = f(x M ). k k A similar argument shows the existence of x m I such that f(x m ) = inf E. We need a technical lemma before proceeding.

4 4. Continuity Lemma 4.8. (Sign Preserving Property) Let f : D R be continuous at x 0 D. Suppose f(x 0 ) > 0. Then δ > 0 exists so that f(x) > f(x 0 )/ for x D and x x 0 < δ. Proof. Set ǫ = f(x 0 )/ in the definition of continuity. Then a δ > 0 exists so that f(x 0) < f(x) f(x 0 ) < f(x 0), for all x D and x x 0 < δ. That is, for all x D T ( δ + x 0, δ + x 0 ). f(x 0 ) < f(x) The next theorem is consistent with the notion continuous functions can be drawn without lifting a pencil. Definition 4.9. A function, f : [a, b] R, is said to have the intermediate-value property if and only if, given any y 0 between f(a) and f(b), there exists a x 0 [a, b] with f(x 0 ) = y 0. Theorem 4.30. (Bolzano Intermediate-Value Theorem) Let f : [a, b] R be continuous. Then f has the intermediate-value property. Before proceeding to the proof, let us drop the hypothesis and see what happens. Example 4.31. Consider f(x) = x on D = [0, 1] [, 3]. If y 0 = 1.5, there is no x 0 D such that f(x 0 ) = y 0. Thus an interval is required. Example 4.3. Here f : [0, 1] R with 0 0 x < 1, f(x) = 1 x = 1. If y 0 = 1/, there is no x 0 [0, 1] such that f(x 0 ) = y 0. Thus continuity is required. Proof. Suppose f(a) < y 0 < f(b). Set E = {x a x b, f(x) y 0 }. The set E is nonempty any bounded since it is a subset of [a, b]. By the completeness axiom, E has a supremum, say x 0. Moreover, x 0 E (Exercise 3.18). We will show f(x 0 ) = y 0. We first show f(x 0 ) y 0. By the properties of supremums, Corollary 3.6, there exist a sequence {x n } n N E converging to x 0 as n. By continuity we have lim f(x n) = f( lim x n) = f(x 0 ). n n Moreover, since f(x n ) y 0 for all n N, it follows f(x 0 ) y 0. Now suppose f(x 0 ) < y 0. We need a contradiction. Set g(x) = y 0 f(x) on [a, b]. By our assumption, g(x 0 ) > 0. By the sign-preserving property, Lemma 4.8, g is positive on [a, b] T ( δ + x 0, δ + x 0 ) for some δ > 0. Since a < x 0 < b, there must exist x > x 0 with g(x) > 0. That is f(x) < y 0 for some x > x 0. This contradicts x 0 = sup E. Thus by trichotomy, f(x 0 ) = y 0.

4.4. Inverses 43 Corollary 4.33. The continuous image of a closed interval is again a closed interval. Proof.. Let f : [a, b] R be continuous on [a, b]. By the Extreme-Value Theorem, 4.6, x m and x M in [a, b] exist so that f(x m ) f(x) f(x M ) for all x [a, b]. By the Intermediate-Value Theorem, 4.30, given any y [f(x m ), f(x M )], there exists x [a, b] such that f(x) = y. That is, f([a, b]) = [f(x m ), f(x M )]. 4.4. Inverses Its natural to ask when a continuous injective function has a continuous inverse. You might think it is automatic. It is not. Example 4.34. Consider the function on D = [0, 1] [, 3) given by x 0 x 1, f(x) = 4 x x < 3. The function is continuous and injective, and therefore has an inverse. However, its inverse is discontinuous. 1.5 f(x) 1 0.5 0 0 0.5 1 1.5.5 3 x 3.5 f -1 (x) 1.5 1 0.5 0 0 0.5 1 1.5 x Figure 1. Top: graph of f(x). Bottom: graph of f 1 (x)

44 4. Continuity Theorem 4.35. Suppose f : I R with I a closed bounded interval, f continuous on I, and f injective. Then f 1 is continuous. Proof. The inverse exists since f is injective. To show the inverse is continuous, we use the sequential characterization of continuity, Theorem 4.11. Let {y n } n N be any sequence in the range of f (that is, in the domain of f 1 ) converging to y 0 F (I). The injectivity of f implies the existence of x n I such that f(x n ) = y n for all n N. Suppose we can show {x n } n N converges to some x 0 I. Continuity of f implies f(x n ) f(x 0 ) as n. That is, y 0 = f(x 0 ). In particular, this implies f 1 (y n ) f 1 (y 0 ) as n. Theorem 4.11 then implies f 1 is continuous on f 1 (I). To show the sequence {x n } n N converges we apply Homework Exercise 3.7 - we show all convergent subsequences have the same limit. Let {x nk } k N be any convergent subsequence to say z 0 I (we know z 0 I since I is closed and bounded). By continuity f(x nk ) f(z 0 ) as n. But f(x nk ) = y nk. Since {y n } n N is a convergent sequence, {y nk } k N converges to y 0. That is, f(z 0 ) = y 0. Set x 0 = f 1 (y 0 ). Then all convergent subsequences of {x n } n N converge to x 0. Therefore, {x n } n N converges to x 0. Corollary 4.36. The functions sin x, cos x, and tan x defined on [ π, π ], [0, π], [ π 4, π 4 ] respectively, have continuous inverses. Proof. Homework Problem 4.10, Examples 4.18, and 4.19 show sin x, cos x, and tan x are continuous on the respective domains. They are seen to be injective there as well. 4.5. Uniform Continuity We will have occasion to require a stronger form of continuity. Definition 4.37. A function f : D R is uniformly continuous on D if and only if, for all ǫ > 0, there exists a δ > 0 such that for all x, a D, x a < δ implies f(x) f(a) < ǫ. Note the difference with continuity - the δ works for all x, a D and only depends on ǫ. This means there is a minimum δ. See the figure below. Lemma 4.38. A function f : D R is not uniformly continuous on D, if there exists a ǫ 0 > 0 and two sequences {x n } n N, {y n } n N D such that x n y n < 1/n and f(x n ) f(y n ) ǫ 0 for all n N. Proof. In logic notation uniform continuity is ( ǫ > 0)( δ > 0)( x 0, x D)( x x 0 < δ f(x) f(x 0 ) < ǫ). By the way continuity on D (but not necessarily uniform) is ( ǫ > 0)( x 0 D)( δ > 0)( x D)( x x 0 < δ f(x) f(x 0 ) < ǫ). Note the difference. The negation of uniform continuity is ( ǫ 0 > 0)( δ > 0)( x 0, x D)( x x 0 < δ and f(x) f(x 0 ) ǫ 0 ). Let δ = 1/n and the result follows.

4.5. Uniform Continuity 45 9 8 7 6 y axis 5 4 3 1 0 0 0.1 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x axis Figure. Graph of f = 1/x. Note the dependence of the size of the δ neighborhood (on the x-axis) on the location of the ǫ neighborhood (on the y-axis). The δ neighborhood continually shrinks as the continuity point moves towards the origin. Example 4.39. Show the function f(x) = 1/x on (0, ) is not uniformly continuous. Proof. Pick x n = 1/n and y n = 1/n. Then x n y n < 1/n and f(x n ) f(y n ) = n n n 1. Lemma 4.38 now applies. Example 4.40. Show the function f(x) = 1/x on [1, ) is uniformly continuous. Proof. We first calculate f(x) f(y) = x y xy x y on [1, ). Thus we choose δ = ǫ in the definition of uniformly continuity and the result follows. Theorem 4.41. Suppose I is a closed, bounded interval and f : I R is continuous on I. Then f is uniformly continuous on I. Proof. Suppose f is not uniformly continuous on I. Then two sequences {x n } n N, {y n } n N I exist and a ǫ 0 > 0 exists such that x n y n < 1/n and f(x n ) f(y n ) ǫ 0 for all n N. Since {x n } n N is bounded, the Bolzano-Weierstrass

46 4. Continuity theorem applies and convergent subsequence exist such that x nk x I as k. Consider the corresponding subsequence {y nk } k N (same indices). We claim {y nk } k N converges to x. Indeed, lim y n k = lim x nk +(y nk x nk ) = x+0 = x. k k Since the absolute function is continuous (Exercise 4.) and f is continuous, we have lim f(x n k ) f(y nk ) = f(x) f(x) ǫ 0 > 0. This is a contradiction. k

4.5. Uniform Continuity 47 It might be fun to find conditions on a function f so that it has properties physicists and engineers expect. Here is one possibility. Theorem 4.4. Suppose I is a closed, bounded interval and f : I R is continuous on I. Then f is uniformly continuous on I. f has a minimum and maximum. f enjoys the intermediate-value property. If f injective, f 1 is continuous. Here is a summary of our theorems so far. The Completeness Axiom Monotone-Convergence Theorem Nested-Cell Theorem Archimedean Principle Density of Q Bolzano-Weierstrass Theorem Uniformly Continuous Extreme-Value Thm Cauchy Theorem

48 4. Continuity 4.6. Homework Exercise 4.1. Prove that f(x) = x is continuous on R. Exercise 4.. Suppose f : R R is continuous and f(x) M for some M. Let {x n } n N be any sequence converging to x. Prove, by just applying appropriate theorems and the previous problem, that lim n f(x n) = f(x) M. Exercise 4.3. Prove that f(x) = 1/x is continuous on R, x 6= 0, by using the definition of continuity and by applying theorems. Exercise 4.4. Prove the following function is continuous at x = 0 x sin 1 f(x) = x x 6= 0, 0 x = 0. Exercise 4.5. Give an example of a function defined on the real numbers continuous only at the origin. Exercise 4.6. Suppose f : D R, c D and c is not a cluster point of D. Show f is continuous at c. Hint: examine Example 4.4 and recall Problem 3.0. Exercise 4.7. Give an example of a function f : [0, 1] R which is discontinuous exactly at one point and a function g : [0, 1] R discontinuous only on Q such that f g is continuous nowhere. Exercise 4.8. Give an example of a continuous function defined on (0, 1) and a Cauchy sequence {x n } n N (0, 1) such that {f(x n )} n N is not Cauchy. Exercise 4.9. Let f be a continuous function defined on the real numbers, and suppose f(x) = c for all x Q and for some c R. Prove that f(x) = c for all x R. Exercise 4.10. Use the inequality sin(x) x for x R to show that the sine function is continuous at x = 0. Use this fact with the identity 1 1 sin(x) sin(u) = sin (x u) cos (x + u) to prove that the sine function is continuous at any point of R. Exercise 4.11. Let c be a cluster point of a set D. Suppose f + g and fg have limits at c. Does it follow f and g have limits at c? Exercise 4.1. Let 1 x Q, f(x) = 1 x Q c. Show that f does not have a limit at any point. Exercise 4.13. Suppose f : D R and c is a cluster point of D. If a < f(x) < b for all x D and f has a limit at c, show a lim f(x) b. Exercise 4.14. Prove the following Squeeze Theorem. Let f, g, h : D R with f(x) g(x) h(x) and c be a cluster point of D. If both f and h have limits at c and converge to L, then g has a limit at c. Moreover, lim g(x) = L.

4.6. Homework 49 Exercise 4.15. Make a definition for the meaning of lim f(x) = L lim + f(x) = L. Prove that lim f(x) = L if and only if lim f(x) = L and lim + f(x) = L. Exercise 4.16. A function f : R R is called additive if f(x + y) = f(x) + f(y) for all x, y R. (a) Show f(n) = nf(1) for all n N. (b) Show f n m x = n mf(x) for all n, m N and x R. (c) Show f(x y) = f(x) f(y) for all x, y R. Then show f( x) = f(x) (f is odd) and f(0) = 0. (d) Show f(r) = rf(1) for all r Q. (e) If f is continuous at one point, f is continuous everywhere. (f) If f is continuous on R, f(x) = xf(1) for all x R. Exercise 4.17. Give an example to show the continuous image of an open interval may not be an open interval. In contrast recall Corollary 4.33 which shows in continuous image of closed interval is a closed interval. Exercise 4.18. Show that f(x) = x 4 3x + 1 has two roots in [0, ]. Exercise 4.19. Find an interval of length one that contains a root of the equation xe x = 1. You may assume e x is a continuous function on R. Exercise 4.0. Show that an odd degree polynomial has at least one real root. Exercise 4.1. Let f : [0, 1] R be continuous with f(0) = f(1). Show c [0, 1/] exists so that f(c) = f c + 1. Exercise 4.. Suppose ϕ(x) is continuous on R, lim ϕ(x) = 7 and lim ϕ(x) = x x 6. Show for each 6 < y < 7, there exists x R with ϕ(x) = y. Exercise 4.3. Show the equation cos x = x has a solution in (0, π/). sin x = x has a solution in (0, π/)? Does Exercise 4.4. Let f : [a, b] R be continuous and injective. Show f is strictly monotone: for any x, y [a, b], x < y implies f(x) < f(y) or for any x, y [a, b], x < y implies f(x) > f(y). Exercise 4.5. Use the Intermediate-Value Theorem to show every positive number p > 1 has a unique n th root by considering the function f(x) = x n on the interval [0, p] (cf. Theorem.1). Exercise 4.6. (Brouwer Fixed Point Theorem) Let f : [a, b] [a, b] be continuous. Prove that f has a fixed point. That is, there exists at least one x [a, b] such that f(x) = x. Exercise 4.7. Let f be a continuous function from R to R. (a) Assume f is not bounded and odd (f( x) = f(x) for all x R). Show f is onto.

50 4. Continuity (b) Give an example of an unbounded f which is not onto. Show your answer is correct. Exercise 4.8. Show f(x) = x is not uniformly continuous on R. Exercise 4.9. Show f(x) = sin(1/x) is not uniformly continuous on (0, π). Exercise 4.30. Show f(x) = mx + b is uniformly continuous on R. Exercise 4.31. Show f(x) = 1/x is not uniformly continuous on (0, ). But is uniformly on [1, ). Exercise 4.3. Suppose f is continuous on [a, b] and uniformly continuous on D, where D is either [b, c] or [b, ). Show that f is uniformly continuous on [a, b] S D. Exercise 4.33. Show f(x) = x is uniformly continuous on [0, ). Use Exercise 4.3 with D = [1, ) or show it directly. Exercise 4.34. Let f and g be uniformly continuous on D. Show that f + g is uniformly continuous on D. Show by example that fg may not be uniformly continuous on D. Exercise 4.35. Suppose D is bounded and f is uniformly continuous on D. Show that f is bounded on D.