# Instructors Manual - Introduction to Real Analysis by Robert G. Bartle

By Robert G. Bartle

This introductory textbook information the rigorous remedy of the elemental thought of features of 1 actual variable. Topological strategies reminiscent of open set and closed set were accrued jointly to supply a extra unified dialogue.

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Extra info for Instructors Manual - Introduction to Real Analysis

Example text

Since f (b) = f (a) is excluded, we must have f (b) > f (a). 13. Assume that h is continuous on [0, 1] and let c1 < c2 be the two points in [0, 1] where h attains its supremum. If 0 < c1 , choose a1 , a2 such that 0 < a1 < c1 < a2 < c2 . Let k satisfy sup{h(a1 ), h(a2 )} < k < h(c1 ) = h(c2 ); then there exist 42 Bartle and Sherbert three numbers bj such that a1 < b1 < c1 < b2 < a2 < b3 < c2 where k = h(bj ), a contradiction. Now consider the points where h attains its inﬁmum. 14. Let x > 0 and consider the case m, p, n, q ∈ N.

14. 12. 15. Suppose that |f (x)| ≤ K for x ∈ I. For x, y ∈ I, apply the Mean Value Theorem to get |f (x) − f (y)| = |(x − y)f (c)| ≤ K|x − y|. 16. (a) Given ε > 0 there exists nε ∈ N such that if x ≥ nε , then |f (x) − b| < ε. Hence if x ≥ nε and h > 0, there exists yx ∈ (x, x + h) such that f (x + h) − f (x) − b = |f (yx ) − b| < ε. h Since ε > 0 is arbitrary, then lim (f (x + h) − f (x))/h = b. x→∞ (b) Assume that b = 0 and let ε < |b|/2. Let nε be as in part (a). Since lim f x→∞ exists, we may also assume that if x, y ≥ nε , then |f (x) − f (y)| < ε.

Moreover |f (nε )/x| < ε if x is suﬃciently large; since f is bounded on [nε , ∞), then |nε f (yε )/x| < ε if x is suﬃciently large. Therefore, lim f (x)/x = b. x→∞ 17. Apply the Mean Value Theorem to the function g − f on [0, x]. 18. 1, and let x < c < y be such that 0 < |x − y| < δ. Since f (x) − f (y) = f (x) − f (c) + f (c) − f (y), a simple calculation shows that f (x) − f (y) x − c f (x) − f (c) c − y f (y) − f (c) = · + · . x−y x−y x−c x−y y−c 47 Chapter 6 — Differentiation Since both (x − c)/(x − y) and (c − y)/(x − y) are positive and have sum 1, it follows that f (x) − f (y) −f x−y x−c ≤ x−y x−c < + x−y (c) f (x) − f (c) − f (c) + x−c c−y ε = ε.