Math 409-502Harold P. [email protected] continuityRecall that a function f is continuous on an interval if for every point a in the interval and forevery positive number e there exists a positive number δ such that |f (x) − f (a)| < e for all xsatisfying the inequality |x − a| < δ.Here the choice of δ may depend on both the point a and the number e.A function f is uniformly continuous on an interval if for every positive number e there exists apositive number δ such that |f (x) − f (a)| < e for all x and a satisfying the inequality |x −a| < δ.Here the choice of δ may depend only on the number e.Math 409-502 November 5, 2004 — slide #2Examples for uniform continuity1. On the interval (0, 1), the function x2is uniformly continuous.Proof: Fix e > 0. For all points x and a in the interval (0, 1), |x2− a2| = |(x − a)(x + a)|≤ 2|x − a|. Therefore the choice δ = e/2 works in the definition of uniform continuity.2. On the interval (0, 1), the function 1/x is not uniformly continuous.Proof: Take e = 1. Suppose there were a positive δ such that¯¯¯¯1x−1a¯¯¯¯< 1 for all x and asatisfying the inequality |x − a| < δ. Leaving a arbitrary, set x = a +12δ. Then¯¯¯¯1x−1a¯¯¯¯=12δa(a +12δ), which tends to ∞ as a → 0+. This contradiction shows that the function is notuniformly continuous on the interval (0, 1).Math 409-502 November 5, 2004 — slide #3Uniform continuity on compact intervalsTheorem. A continuous function on a compact interval is automatically uniformly continuouson the interval.Proof (different from the proof in the book). Fix e > 0. Suppose there were no positive δ thatworks uniformly on the interval.Bisect the interval. For at least one half, there is no δ that works uniformly on that half. Bisectagain, and iterate.We get nested compact intervals on each of which there is no δ that works for the given fixed e.The intervals converge to some point a in the original interval.The function is continuous at a, so there is some positive δ for which the values of the functionare all within e of each other on the interval (a −δ, a + δ).The contradiction shows that the function must be uniformly continuous after all.Math 409-502 November 5, 2004 — slide #4Homework1. Read section 13.5, pages 190–192.2. The interval [0, ∞) is not compact. Show nonetheless that the function√x is uniformlycontinuous on this unbounded interval.3. The interval (0, 1) is not compact. Determine (with proof) whether sin(1/x) is uniformlycontinuous on this open interval.Math 409-502 November 5, 2004 — slide