5. Some theorems on continuous functions

Transcription

1 5. Some theorems on continuous functions The results of section 3 were largely concerned with continuity of functions at a single point (usually called x 0 ). In this section, we present some consequences of a function being continuous throughout an interval. Both the results and their proofs are of a rather different nature. The intermediate value theorem Pictures make it seem obvious that a reasonable function must cross the line. In other words, if η is a number between f(a) and f(b), we expect there to be a number ξ in (a, b) such that f(ξ) = η. But what exactly is reasonable? As the second picture shows, the statement can fail if f is discontinuous. Our theorem shows that continuity is what is needed. (It is worth remarking that not every continuous function can be represented by drawing a smooth line.) Prepare for a use of the notion of least upper bound. 5.1 (The intermediate value theorem) Suppose that f is continuous on [a, b], and that either f(a) < η < f(b) or f(a) > η > f(b). Then there is a number ξ in (a, b) such that f(ξ) = η. Proof. Suppose that f(a) < η < f(b) (the second case can then be derived by considering f). Define E = {x [a, b] : f(x) η}. Then a E and b is not in E. Since E [a, b], it is certainly bounded: let ξ = sup(e). We show that f(ξ) = η. Suppose that f(ξ) < η. Then ξ < b, so f is (at least) right-continuous at ξ. Hence there exists δ > 0 such that (i) ξ +δ < b and (ii) f(ξ +δ) η. But this says that ξ +δ E, contradicting the fact that ξ is an upper bound of E. Now suppose that f(ξ) > η. Then ξ > a, so f is (at least) left-continuous at ξ. Hence there exists δ > 0 such that (i) ξ δ > a and (ii) f(x) > η for all x in [ξ δ, ξ]. But this implies that no members of E are in [ξ δ, ξ], so all members of E are in [a, ξ δ]. This contradicts the fact that ξ is the least upper bound of E. 1

2 So we conclude that f(ξ) = η. 5.2 COROLLARY. If f is continuous on [a, b] and f(a) η f(b), then there exists ξ [a, b] such that f(ξ) = η. Proof. If η is f(a) or f(b), take ξ to be a or b. Otherwise, f(a) < η < f(b), and 5.1 applies. 5.3 COROLLARY. Suppose that f and g are continuous on a, b], and that f(a) < g(a) and f(b) > g(b). Then there exists ξ (a, b) such that f(ξ) = g(ξ). Proof. Apply the theorem to f g. The theorem is very useful for establishing the existence and rough location of solutions of equations. Some typical examples: Example 1. Show that there exists x between 0 and π/2 such that tan x = x Example 2. Show that there are at least two positive values of x such that 2 x = 6x, and in each case state an integer n such that x is between n and n + 1. Example 3. Show that there are at least two solutions of the equation 2 cos x = 1/x between 0 and π/2. We now give some general results derived from the intermediate value theorem If p is a polynomial of odd degree, then there is at least one solution of the equation p(x) = 0. 2

3 Proof. Assume the leading coefficient is positive. By 1.18, p(x) tends to as x and to as x. So p(x) certainly has both positive and negative values. It is continuous, by 3.5. By the IVT, it must take the value 0 somewhere For n = 2, 3,..., every positive number has a unique positive nth root. Proof. Let the positive number be y 0. Let f(x) = x n. This is continuous, and f(0) = 0 < y 0. We only need to show that there is a number b such that f(b) y 0 : the statement then follows, by the IVT Suppose that f is continuous on the interval I = [a, b], and maps I into I, so that for each x in I, the image f(x) is also in I. Then there exists a number x 0 in I such that f(x 0 ) = x 0 (that is, a fixed point for f). Proof Suppose that f is continuous on [a, b], and x 1, x 2,..., x n are points of [a, b]. Suppose that n i=1 f(x i) = S. Then there exists x in [a, b] such that f(x) = S/n. Proof. The next result is an essential step in establishing the properties of cos x and sin x from the power series definitions There exists x 0 in (0, 2) such that cos x 0 = 0. Note. Of course, this x 0 is the number we usually call π/2. In the formal development, one can take this as a way of defining π. Proof. The function cos is continuous and cos 0 = 1. Also, cos 2 = 3

4 Boundedness Recall that every bounded sequence has a convergent subsequence (Theorem 1.20). 5.9 THEOREM. If a function f is continuous on [a, b], then it is bounded on [a, b] and attains its bounds there. In other words, there are points of [a, b] where f attains its supremum and infimum. Proof. (i) Suppose that f is unbounded on [a, b]. Then for each n 1, there is a point x n of [a, b] such that f(x n > n. By the theorem quoted, (x n ) has a subsequence (x nk ) that converges, say to x 0, as k. Then x 0 is in [a, b], so f is continuous at x 0 (or at least right-continuous if x 0 = a, etc.). By 3.9, the sequence [f(x nk )] tends to f(x 0 ) as k, so is certainly bounded. But this contradicts the fact that f(x nk ) > n k for each k. (ii) Now let M = sup{f(x) : a x b}. Alternative proof of (ii). Suppose that f(x) < M for all x [a, b] (so that the value M is not attained). Let g(x) = 1/(M f(x)). Then g is defined and continuous on [a, b], so by (i), there exists K such that g(x) K for all x [a, b]. But this implies that f(x) M 1/K for x [a, b], contradicting the fact that M is the least upper bound. Easy examples show that both statements in 5.9 fail for open intervals. The function 1/x is continuous on (0, 1), but unbounded there. The function f(x) = x has infimum 0 on (0, 1), but does not attain the value 0 at any point of the interval COROLLARY. If f is continuous on [a, b] and f(x) > 0 for all x [a, b], then there exists δ > 0 such that f(x) δ for all x [a, b]. Proof. Let m = inf{f(x) : a x b}. Then m = f(x 1 ) for some x 1 [a, b], so m > 0 and the statement holds with δ = m. Uniform continuity Our third theorem relates to a more subtle property. We shall need it when we come to define definite integrals. Suppose that f is continuous on an interval I (which could be the whole of R). If ε > 0 is given, each point x of I will have its own corresponding δ. There might not be any δ 4

5 that works for all x I simultaneously. Uniform continuity is (roughly) the statement that there is such a δ. The precise definition is: f is uniformly continuous on I if, given ε > 0, there exists δ > 0 such that for any two points x 1, x 2 I with x 1 x 2 δ, we have f(x 1 ) f(x 2 ) ε. Both x 1 and x 2 are variable here. Note that uniformly continuous on its own, without the on I, is meaningless. Clearly: (1) uniform continuity on I implies continuity there (the definition implies continuity at x 1 for each x 1 I); (2) if f is Lipschitz on I, then it is uniformly continuous there: for if f(x 1 ) f(x 2 ) M x 1 x 2 for all x 1, x 2 I, then the above definition is satisfied with δ = ε/m. Example 4. f(x) = x 2. We show that f is not uniformly continuous on R (though of course it is continuous). In fact, the definition fails with ε = 1. For: try any δ > 0 and take x 1 = x (where x > 0) and x 2 = x + δ. Then (x + δ) 2 x 2 > 2δx, which is greater than 1 when x > 1/2δ. (Roughly, the point is that the gradient of x 2 is unbounded.) However, f is Lipschitz, and hence uniformly continuous, on any bounded interval I = [ M, M], since if x, y I, then f(y) f(x) = (y + x)(y x) 2M y x. It s easy to show that if f and g are uniformly continuous on I, then so is f + g. However, the same is not true for fg, as the above example shows (take f(x) = g(x) = x) PROPOSITION. If f is continuous on a closed interval [a, b], then it is uniformly continuous there. Proof. Suppose not. Then there exists ε > 0 such that for each n 1, there exist x n and y n in [a, b] such that x n y n 1 n while f(x n) f(y n ) > ε. (1) By 1.20, (x n ) has a subsequence (x nk ) that converges, say to x 0. Then (y nk ) also converges to x 0, since x n y n 0 as n. Since f is continuous on [a, b], both f(x nk ) and f(y nk ) tend to f(x 0 ) as k. Hence f(x nk ) f(y nk ) 0. But this contradicts (1). 5