Chapter 7 Continuous Functions In this chapter, we define continuous functions and study their properties. 7.1. Continuity Continuous functions are functions that take nearby values at nearby points. Definition 7.1. Let f : A ! R, where A ⊂ R, and suppose that c 2 A. Then f is continuous at c if for every > 0 there exists a δ > 0 such that jx − cj < δ and x 2 A implies that jf(x) − f(c)j < . A function f : A ! R is continuous if it is continuous at every point of A, and it is continuous on B ⊂ A if it is continuous at every point in B. The definition of continuity at a point may be stated in terms of neighborhoods as follows. Definition 7.2. A function f : A ! R, where A ⊂ R, is continuous at c 2 A if for every neighborhood V of f(c) there is a neighborhood U of c such that x 2 A \ U implies that f(x) 2 V: The -δ definition corresponds to the case when V is an -neighborhood of f(c) and U is a δ-neighborhood of c. Note that c must belong to the domain A of f in order to define the continuity of f at c. If c is an isolated point of A, then the continuity condition holds auto- matically since, for sufficiently small δ > 0, the only point x 2 A with jx − cj < δ is x = c, and then 0 = jf(x) − f(c)j < . Thus, a function is continuous at every isolated point of its domain, and isolated points are not of much interest. If c 2 A is an accumulation point of A, then the continuity of f at c is equivalent to the condition that lim f(x) = f(c); x!c meaning that the limit of f as x ! c exists and is equal to the value of f at c. 121 122 7. Continuous Functions Example 7.3. If f :(a; b) ! R is defined on an open interval, then f is continuous on (a; b) if and only if lim f(x) = f(c) for every a < c < b x!c since every point of (a; b) is an accumulation point. Example 7.4. If f :[a; b] ! R is defined on a closed, bounded interval, then f is continuous on [a; b] if and only if lim f(x) = f(c) for every a < c < b; x!c lim f(x) = f(a); lim f(x) = f(b): x!a+ x!b− Example 7.5. Suppose that 1 1 1 A = 0; 1; ; ;:::; ;::: 2 3 n and f : A ! R is defined by 1 f(0) = y ; f = y 0 n n for some values y0; yn 2 R. Then 1=n is an isolated point of A for every n 2 N, so f is continuous at 1=n for every choice of yn. The remaining point 0 2 A is an accumulation point of A, and the condition for f to be continuous at 0 is that lim yn = y0: n!1 As for limits, we can give an equivalent sequential definition of continuity, which follows immediately from Theorem 6.6. Theorem 7.6. If f : A ! R and c 2 A is an accumulation point of A, then f is continuous at c if and only if lim f(xn) = f(c) n!1 for every sequence (xn) in A such that xn ! c as n ! 1. In particular, f is discontinuous at c 2 A if there is sequence (xn) in the domain A of f such that xn ! c but f(xn) 6! f(c). Let's consider some examples of continuous and discontinuous functions to illustrate the definition. p Example 7.7. The function f : [0; 1) ! R defined by f(x) = x is continuous on [0; 1). To prove that f is continuous at c > 0, we note that for 0 ≤ x < 1, p p x − c 1 jf(x) − f(c)j = x − c = p p ≤ p jx − cj; x + c c p so given > 0, we can choose δ = c > 0 in the definition of continuity. To prove that f is continuous at 0, we note that if 0 ≤ x < δ where δ = 2 > 0, then p jf(x) − f(0)j = x < . 7.1. Continuity 123 Example 7.8. The function sin : R ! R is continuous on R. To prove this, we use the trigonometric identity for the difference of sines and the inequality j sin xj ≤ jxj: x + c x − c j sin x − sin cj = 2 cos sin 2 2 x − c ≤ 2 sin 2 ≤ jx − cj: It follows that we can take δ = in the definition of continuity for every c 2 R. Example 7.9. The sign function sgn : R ! R, defined by 8 1 if x > 0, <> sgn x = 0 if x = 0, :>−1 if x < 0, is not continuous at 0 since limx!0 sgn x does not exist (see Example 6.8). The left and right limits of sgn at 0, lim f(x) = −1; lim f(x) = 1; x!0− x!0+ do exist, but they are unequal. We say that f has a jump discontinuity at 0. Example 7.10. The function f : R ! R defined by ( 1=x if x 6= 0, f(x) = 0 if x = 0, is not continuous at 0 since limx!0 f(x) does not exist (see Example 6.9). The left and right limits of f at 0 do not exist either, and we say that f has an essential discontinuity at 0. Example 7.11. The function f : R ! R defined by ( sin(1=x) if x 6= 0, f(x) = 0 if x = 0 is continuous at c 6= 0 (see Example 7.21 below) but discontinuous at 0 because limx!0 f(x) does not exist (see Example 6.10). Example 7.12. The function f : R ! R defined by ( x sin(1=x) if x 6= 0, f(x) = 0 if x = 0 is continuous at every point of R. (See Figure 1.) The continuity at c 6= 0 is proved in Example 7.22 below. To prove continuity at 0, note that for x 6= 0, jf(x) − f(0)j = jx sin(1=x)j ≤ jxj; so f(x) ! f(0) as x ! 0. If we had defined f(0) to be any value other than 0, then f would not be continuous at 0. In that case, f would have a removable discontinuity at 0. 124 7. Continuous Functions 1 0.4 0.8 0.3 0.2 0.6 0.1 0.4 0 0.2 −0.1 0 −0.2 −0.2 −0.3 −0.4 −0.4 −3 −2 −1 0 1 2 3 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 Figure 1. A plot of the function y = x sin(1=x) and a detail near the origin with the lines y = ±x shown in red. Example 7.13. The Dirichlet function f : R ! R defined by ( 1 if x 2 , f(x) = Q 0 if x2 = Q is discontinuous at every c 2 R. If c2 = Q, choose a sequence (xn) of rational numbers such that xn ! c (possible since Q is dense in R). Then xn ! c and f(xn) ! 1 but f(c) = 0. If c 2 Q, choose a sequence (xn) of irrational numbers such that xn ! c; for example if c = p=q, we can take p p 2 x = + ; n q n p since xn 2 Q would imply that 2 2 Q. Then xn ! c and f(xn) ! 0 but f(c) = 1. Alternatively, by taking a rational sequence (xn) and an irrational sequence (~xn) that converge to c, we can see that limx!c f(x) does not exist for any c 2 R. Example 7.14. The Thomae function f : R ! R is defined by ( 1=q if x = p=q 2 where p and q > 0 are relatively prime, f(x) = Q 0 if x2 = Q or x = 0: Figure 2 shows the graph of f on [0; 1]. The Thomae function is continuous at 0 and every irrational number and discontinuous at every nonzero rational number. To prove this claim, first suppose that x = p=q 2 Q n f0g is rational and nonzero. Then f(x) = 1=q > 0, but for every δ > 0, the interval (x − δ; x + δ) contains irrational points y such that f(y) = 0 and jf(x) − f(y)j = 1=q. The definition of continuity therefore fails if 0 < ≤ 1=q, and f is discontinuous at x. Second, suppose that x2 = Q is irrational. Given > 0, choose n 2 N such that 1=n < . There are finitely many rational numbers r = p=q in the interval (x − 1; x + 1) with p, q relatively prime and 1 ≤ q ≤ n; we list them as fr1; r2; : : : ; rmg. Choose δ = minfjx − rkj : k = 1; 2; : : : ; ng 7.2. Properties of continuous functions 125 Figure 2. A plot of the Thomae function in Example 7.14 on [0; 1] to be the distance of x to the closest such rational number. Then δ > 0 since x2 = Q. Furthermore, if jx − yj < δ, then either y is irrational and f(y) = 0, or y = p=q in lowest terms with q > n and f(y) = 1=q < 1=n < . In either case, jf(x) − f(y)j = jf(y)j < , which proves that f is continuous at x2 = Q.