# Theorem

EXERCISES 4.3 1. Prove Theorem 4.3.3. 2. Show that the following functions are not uniformly continuous on the given domain. 1 *a. f(x) = *. Dom f = (0.00) b. 8(x) Dom g = c. h(x) = sin Dom h = (0,00) x2 (0.00) 3. Prove that each of the following functions is uniformly continuous on the indicated set. *a. f(x) x€ [0, …) b. 8(x) = r?, XEN . 1 + x 1 c. h(x) XER d. k(x) = cosx, ER r? +1′ r sinx e. e(x) x€ (0,00) *l. f(x) = x€ (0, 1) x + 1 4. Show that each of the following functions is a Lipschitz function. 1 *a. f(x) Dom f = (a,00), a > 0 b. 8(x) – X x Dom 8 = (0,00) +1′ ch(x) = sin Dom h = = (a,00), a > 0 d. p(x) a polynomial, Dom p = = (-a, a), a > 0 5. *a. Show that f(x) = Vx satisfies a Lipschitz condition on (a,00), a > 0. b. Prove that Vx is uniformly continuous on (0.00). c. Show that f does not satisfy a Lipschitz condition on (0,00). 6. Suppose E C R and f, 8 are Lipschitz functions on E. a. Prove that f + g is a Lipschitz function on E. b. If in addition f and g are bounded on E, or the set E is compact, prove that fg is a Lipschitz function on E. 7. Suppose E C R and f, 8 are uniformly continuous real-valued functions on E. a. Prove that f + g is uniformly continuous on E. *b. If, in addition, f and g are bounded, prove that fg is uniformly continuous on E. c. Is part (b) still true if only one of the two functions is bounded? 8. Suppose E CR and f: E-R is uniformly continuous. If {x,} is a Cauchy sequence in E, prove that {f(x.)} is a Cauchy sequence. 9. Let f:(a,b) → R be uniformly continuous on (a, b). Use the previous exercise to show that f can be defined at a and b such that f is continuous on (a, b). 10. Suppose that E is a bounded subset of R and f: E-R is uniformly continuous on E. Prove that f is bounded on E. 4.3 Uniform Continuity In the previous section we discussed continuity of a function at a point and on a set. By Definition 4.2.1, a function f:E → R is continuous on E if for each p E E, given any € > 0, there exists a 8 > O such that \(x) – f(p)] < e for all x € En Nop). In gen- eral, for a given e > 0, the choice of 8 that works depends not only on e and the func- tion f, but also on the point p. This was illustrated in Example 4.1.2(g) for the function f(x) = 1/x.x € (0,00). Functions for which a choice of 8 independent of p is possible are given a special name. 4.3.1 DEFINITION Lei ECR and f: E™R. The function is uniformly continuous on E if given e > 0, there exists a 8 >0 such that \f(x) – f(y)] 0 be given. Take S = €/2C. If x, y E E with lx – ył < 8, then by the above \(x) – f(y)] = 2Clx – yl < 2C8 = e. Therefore, f is uniformly continuous on E. In this example, the choice of 8 depends both on e and the set E. In the exercises you will be asked to show that this result is false if the set E is an unbounded interval. (b) Let f(x) = sin x. As in Example 4.2.2(f), Vly) – f(x)= ly – x1 for all x, y E R. Consequently, f is uniformly continuous on R. 4.3 Uniform Continuity 145 146 Chapter 4 Limits and Continuity Proof. Exercise 1. O (c) In this example we show that the function f(x) = 1/x, x € (0.00) is not uniformly continuous on (0,00). Suppose, on the contrary, that f is uniformly continuous on (0,00). Then, as in Example 4.1.2(g), if we take e = 1, there exists a 8 >0 such that Vix) = f()) = 1*-} 0. Suppose x, y E (a,). Then THEOREM If KCR is compact and f: K+R is continuous on K, then f is uni- formly continuous on K. Proof. Let e > 0 be given. Since f is continuous, for each pe K, there exists a Op > O such that \F(x) – f() < (3) for all x E KA N28,(). The collection {N$, ()}pek is an open cover of K. Since K is compact, a finite num- ber of these will cover K. Thus there exist a finite number of points Pos. such that ., Pn in K \f(x) – fb)! = Fazla yl. ry Hence, given e > 0, if we choose & such that 0 < s < a?e, then as a consequence of the above inequality, \f(x) – f(y)] < e for all x, y E (a,00) with bx – y < 8. cÚN(). Let Lipschitz Functions Both of the functions in Example 4.3.2(a) and (b), and the function f(x) = 1/x with Dom f = [a, 00), a > 0, are examples of an extensive class of functions. If E CR, a function f: EnR satisfies a Lipschitz condition on E if there exists a positive con- stant M such that \f(x) – f(y)] = Mpx – y for all x, y E E. Functions satisfying the above inequality are usually referred to as Lip- schitz functions. As we will see in the next chapter, functions for which the derivative is bounded are Lipschitz functions. As a consequence of the following theorem, every Lipschitz function is uniformly continuous. However, not every uniformly continuous function is a Lipschitz function. For example, the function f(x) = Vr is uniformly continuous on (0,00), but f does not satisfy a Lipschitz condition on [0, 0) (see Ex- ercise 5). 8 = min{de, : i = 1,…,n}. Then 8 > 0. Suppose x, y E K with [x – yl < 8. Since x € K, x € N8,(Pi) for some i. Furthermore, since fx – yl

Purchase answer to see full attachment