Functions. Chapter Continuous Functions

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1 Chapter 3 Functions 3.1 Continuous Functions A function f is determined by the domain of f: dom(f) R, the set on which f is defined, and the rule specifying the value f(x) of f at each x dom(f). If f is defined by a formula without mentioning its domain, the domain is understood to be the natural domain: the largest subset of R on which the formula is well defined. For example: the domain of f(x) = 1 x is R \ {0}; the domain of g(x) = x is {x R : x 0}. We may use x f(x) to denote the function f(x). For example, x 1 x is another way writing the function f(x) = 1 x. Definition We say that f is continuous at x 0 in dom(f) if, for every sequence (x n ) in dom(f) (x n dom(f) for all n) such that x n x 0, we have f(x n ) f(x 0 ). f is said to be continuous if it is continuous at every x 0 dom(f). Theorem f is continuous at x 0 dom(f) if and only if ε > 0, δ > 0, x dom(f) with x x 0 < δ, f(x) f(x 0 ) < ε. Proof. First suppose that f satisfies the above ε δ property. Suppose that (x n ) is a sequence in dom(f) and x n x 0. We will show that f(x n ) f(x 0 ). Fix ε > 0. There is δ > 0 such that if x dom(f) and x x 0 < δ then f(x) f(x 0 ) < ε. Since x n x 0, there is N N such that if n > N, then x n x 0 < δ. Since x n dom(f) for all n, combining the above two sentences, we see that if n > N, then f(x) f(x 0 ) < ε. Thus, f(x n ) f(x 0 ). Now suppose that f satisfies the property in the definition. Suppose that f does not satisfy the ε δ property. Then ε > 0, δ > 0, x dom(f) with x x 0 < δ, f(x) f(x 0 ) ε. The ε now is fixed. We have the freedom to choose different δ. The above formula implies that for every n N, there is x n dom(f) such that x n x 0 < 1 n and f(x n) f(x 0 ) ε. Then (x n ) is a sequence in dom(f). Since x 0 1 n < x n < x n for all n, we have x n x 0. Since f(x n ) f(x 0 ) ε for all n, we do not have f(x n ) f(x 0 ), which is a contradiction. 50

2 Examples. 1. Using the definition and limit theorems, it is straightforward to check that f(x) = C, f(x) = x, f(x) = x, f(x) = 1 x (x 0), f(x) = x, and f(x) = x (x 0) are continuous.. The familiar functions: sin x, cos x, tan x, cot x, e x, and log x for x > 0 are all continuous in their domains. Rigorous proof requires definitions of these functions. We now simply accept that they are continuous. Given two functions f and g. The sum f + g is defined on dom(f) dom(g), and for every x dom(f) dom(g), the value (f + g)(x) is defined to be f(x) + g(x). The functions f g and fg are defined similarly. The quotient f/g is defined on dom(f) {x dom(g) : g(x) 0}, and (f/g)(x) = f(x)/g(x). Theorem Let f and g be continuous at x 0 R, and k R. Then kf, f, f + g, f g, and fg are continuous at x 0. If g(x 0 ) 0, then f/g is continuous at x 0. Proof. These results follow from the limit theorems for sequences. For example, we show that f/g is continuous at x 0 if g(x 0 ) 0. In this case x 0 dom(f/g). Suppose that (x n ) is a sequence in dom(f/g) and x n x 0. Since f and g are continuous at x 0, we have f(x n ) f(x 0 ) and g(x n ) g(x 0 ). Since g(x n ) and g(x 0 ) are not 0, we have f(x n )/g(x n ) f(x 0 )/g(x 0 ). So f/g is continuous at x 0. That kf, f + g, f g, and fg are continuous at x 0 follow from a similar argument. Corollary Let f and g be continuous, and k R. Then kf, f, f + g, f g, fg, and f/g are all continuous. Proof. For example, from the previous theorem, f/g is continuous at every x 0 dom(f/g), so f/g is continuous. Examples. 1. A polynomial function P (x) = a 0 + a 1 x + a x + a n x n is continuous. Proof. We know that f(x) = a 0 and f(x) = x are continuous. Applying the product rule repeatedly, we see that f(x) = a k x k is continuous for any 0 k n. Since P (x) is the sum of these functions, we see that P (x) is continuous.. Let P (x) and Q(x) be two polynomial functions. Then R(x) = P (x) Q(x) domain (all x such that Q(x) 0). is continuous in its 3. If f(x) and g(x) are continuous at x 0, then max{f, g} and min{f, g} are continuous at x 0. The reason is that max{f(x), g(x)} = f(x)+g(x) + f(x) g(x) and min{f(x), g(x)} = f(x)+g(x) f(x) g(x). 51

3 Let f and g be two functions. The composite g f is defined by: dom(g f) = {x dom(f) : f(x) dom(g)}, and for every x dom(g f), g f(x) = g(f(x)). Theorem If f is continuous at x 0 and g is continuous at f(x 0 ), then g f is continuous at x 0. Proof. Suppose that (x n ) is a sequence in dom(g f) and x n x 0. Then (x n ) is a sequence in dom(f) and (f(x n )) is a a sequence in dom(g). Since f is continuous at x 0, f(x n ) f(x 0 ). Since g is continuous at f(x 0 ), g(f(x n )) g(x 0 ). Corollary If f and g are both continuous, then g f is continuous. Examples. 1. For x > 0 and y R, the power function x y is defined as e y log x. For a fixed y R, x x y is the composition of x log x, x yx, and x e x, and so is continuous. For a fixed x > 0, y x y is the composition of y y log x and y e y, and so is continuous.. e sin x and sin(e x ) are continuous, sin x is continuous at those x such that sin x f(x) = x sin( 1 x ) is defined on R \ {0}, and is continuous on this domain. Now we extend f to R by letting f(0) = 0. We claim that f is also continuous at 0. The is proved by definition. First, we see that f(x) x for all x R. Suppose that x n 0. Then f(x n ) x n for all n, from a corollary of the squeeze theorem, we get f(x n ) 0 = f(0). Thus, f is continuous at f(x) = sin( 1 x ) is defined on R \ {0}, and is continuous on its domain. We will show that it is not possible to extend f to R such that f is continuous at 0. Suppose such extension 1 1 exists. Let x n = nπ+ π and y n = nπ π, n N. Then x n 0 and y n 0. We have f(x n ) = sin(nπ + π ) = 1 and f(y n) = sin(nπ π ) = 1 for all n N. Since f is continuous at 0, 1 = lim f(x n ) = f(0) = lim f(y n ) = 1, which is a contradiction. Homework. 17.1, (a,b,c), Properties of Continuous Functions Theorem [Extreme Value Theorem] Let f be a continuous function on [a, b]. Then f is bounded, and f assumes its maximum and minimum values on [a, b]. In other words, let S = f([a, b]) := {f(x) : x [a, b]}. Then max S and min S exist. This means that there exist x 0, y 0 [a, b] such that f(x 0 ) f(x) f(y 0 ) for all x [a, b]. 5

4 Proof. First we show that S = f([a, b]) is bounded above. If this is not true, then for any n N, there is s n S such that s n > n. Then there is x n [a, b] such that f(x n ) > n. So f(x n ). Apply the Bolzano-Weierstrass theorem to the bounded sequence (x n ). We see that (x n ) has a convergent subsequence (x nk ). Let x 0 be its limit. Since a x nk b for all k, by the comparison theorem, x 0 [a, b]. Since f is continuous on [a, b], f(x nk ) f(x 0 ) R. However, since (f(x nk )) is a subsequence of (f(x n )), we have f(x nk ), which is a contradiction. Thus, f is bounded above. So sup S R. For any n N, sup S 1 n is not an upper bound of S. So there is s n S such that s n > sup S 1 n. Since s n S, sup S s n > sup S 1 n, which implies that s n sup S. For every n, s n = f(x n ) for some x n [a, b]. Apply the Bolzano-Weierstrass theorem to (x n ). We see that (x n ) has a subsequence (x nk ), which converges to some y 0 [a, b]. Since f is continuous at y 0, s nk = f(x nk ) f(y 0 ). Since (s nk ) is a subsequence of (s n ), and s n sup S, we have s nk sup S. Thus, f(y 0 ) = sup S, which means sup S S, and so that max S exists. The proof that min S exists is similar. Theorem 3... [Intermediate Value Theorem] Suppose f is continuous on [a, b] and f(a) f(b). Then for any y that lies between f(a) and f(b) (i.e., f(a) < y < f(b) or f(b) < y < f(a)) there exists x (a, b) such that f(x) = y. The theorem is intuitively clear from a geometric observation. We now give a rigorous proof. Proof. We assume that f(a) < y < f(b). The other case is similar. Let S = {x [a, b] : f(x) < y}. Since a S, S. Since S [a, b], it is bounded above. Thus, x 0 = sup S exists and a x 0 b. For each n N, x 0 1 n is not an upper bound of S. So there is s n S such that x 0 s n > x 0 1 n. So s n x 0. Since f is continuous at x 0, f(s n ) f(x 0 ). Since f(s n ) < y for all n, we get f(x 0 ) y. Since f(b) > y, b S. Since x 0 = sup S, x n S for any n N. Let t n = min{b, x n } for n N. Then t n [a, b] \ S and x 0 < t n < x n. Thus, t n x 0. That t n [a, b] \ S implies that f(t n ) y. That t n x 0 implies that f(t n ) f(x 0 ). From f(t n ) y for each n we get f(x 0 ) y. So f(x 0 ) = y. We now give some applications of the Intermediate Value Theorem. Corollary Suppose f is continuous on an interval I. Then f(i) = {f(x) : x I} is either an interval or a single point. Proof. Let J = f(i), M = sup J and m = inf J. Here M could be +, and m could be. If M = m, then f(i) = J contains only one point. Now suppose that m < M. Then J [m, M]. Suppose y (m, M). Then y is neither an upper bound nor a lower bound of J. So there are y 1, y S such that y 1 < y < y. There are a < b I such that f(a) = y 1 and f(b) = y, or f(a) = y and f(b) = y 1. Since f is continuous on [a, b], from the Intermediate Value Theorem, there is x (a, b) such that f(x) = y. Thus, y J. So we find that (m, M) J [m, M], which implies that f(i) = J is an interval. 53

5 Corollary 3... Suppose that f is continuous on [0, 1], and f(x) [0, 1] for every x [0, 1]. Then f has a fixed point, which means that there exists x [0, 1] such that f(x) = x. Proof. Let g(x) = f(x) x. Then g is continuous on [0, 1]. Note that g(0) = f(0) 0 and g(1) = f(1) 1 0. Suppose that f has no fixed point. Then g(x) 0 for all x [0, 1]. Thus, g(0) > 0 > g(1). From the Intermediate Value Theorem, there is x (0, 1) such that g(x) = 0, which is a contradiction. Theorem [Existence of Power Root] For every n N and x R with x 0, there is a unique y R with y 0 such that y n = x. Proof. We first prove the existence. If x = 0, we may take y = 0. Now suppose x > 0. Let f(y) = y n. Then f is continuous, and we want y > 0 such that f(y) = x. Observe that f(0) = 0 < x, and f(1 + x) = (1 + x) n 1 + nx > x. Applying the Intermediate Value Theorem, we see that there exists y (0, 1 + x) with f(y) = x. Now we prove the uniqueness. Suppose y 1, y 0 satisfy y1 n = yn = x. Then we must have y 1 = y because if, say y 1 > y, then y1 n > yn, which is a contradiction. Homework. 18., 18.5, Uniform Continuity Recall that from a theorem, f defined on S is continuous, if and only if x 0 S, ε > 0, δ > 0, x S with x x 0 < δ, f(x) f(x 0 ) < ε. (3.1) The point is that the δ depends on both ε and x 0. We say that f is uniformly continuous on S if we can find δ without knowing x 0, i.e., ε > 0, δ > 0, x 0 S, x S with x x 0 < δ, f(x) f(x 0 ) < ε. (3.) 54