Continuity and Uniform Continuity521May 12, 20101. Throughout S will denote a subset of the real numbers R and f : S → Rwill be a real valued function defined on S. The set S may be bounded likeS = (0, 5) = {x ∈ R : 0 < x < 5}or infinite likeS = (0, ∞) = {x ∈ R : 0 < x}.It may even be all of R. The value f(x) of the function f at the point x ∈ Swill be defined by a formula (or formulas).Definition 2. The function f is said to be continuous on S iff∀x0∈ S ∀ε > 0 ∃δ > 0 ∀x ∈ S|x − x0| < δ =⇒ |f(x) −f(x0)| < ε.Hence f is not continuous1on S iff∃x0∈ S ∃ε > 0 ∀δ > 0 ∃x ∈ S|x − x0| < δ and |f(x) − f (x0)| ≥ ε.Definition 3. The function f is said to be uniformly continuous on S iff∀ε > 0 ∃δ > 0 ∀x0∈ S ∀x ∈ S|x − x0| < δ =⇒ |f(x) − f(x0)| < ε.Hence f is not uniformly continuous on S iff∃ε > 0 ∀δ > 0 ∃x0∈ S ∃x ∈ S|x − x0| < δ and |f(x) − f (x0)| ≥ ε.1For an example of a function which is not continuous see Example 22 below.14. The only difference between the two definitions is the order of the quan-tifiers. When you prove f is continuous your proof will have the formChoose x0∈ S. Choose ε > 0. Let δ = δ(x0, ε). Choose x ∈ S.Assume |x − x0| < δ. ··· Therefore |f(x) − f(x0)| < ε.The expression for δ(x0, ε) can involve both x0and ε but must be independentof x. The order of the quanifiers in the definition signals this; in the proof xhas not yet been chosen at the point where δ is defined so the definition ofδ must not involve x. (The ··· represent the proof that |f(x) − f(x0)| < εfollows from the earlier steps in the proof.) When you prove f is uniformlycontinuous your proof will have the formChoose ε > 0. Let δ = δ(ε). Choose x0∈ S. Choose x ∈ S.Assume |x − x0| < δ. ··· Therefore |f(x) − f(x0)| < ε.so the expression for δ can only involve ε and must not involve either x orx0.It is obvious that a uniformly continuous function is continuous: if we canfind a δ which works for all x0, we can find one (the same one) which worksfor any particular x0. We will see below that there are continuous functionswhich are not uniformly continuous.Example 5. Let S = R and f(x) = 3x + 7. Then f is uniformly continuouson S.Proof. Choose ε > 0. Let δ = ε/3. Choose x0∈ R. Choose x ∈ R. Assume|x − x0| < δ. Then|f(x) − f(x0)| = |(3x + 7) − (3x0+ 7)| = 3|x − x0| < 3δ = ε.Example 6. Let S = {x ∈ R : 0 < x < 4} and f(x) = x2. Then f isuniformly continuous on S.Proof. Choose ε > 0. Let δ = ε/8. Choose x0∈ S. Choose x ∈ S. Thus0 < x0< 4 and 0 < x < 4 so 0 < x + x0< 8. Assume|x − x0| < δ. Then|f(x) − f(x0)| = |x2− x20| = (x + x0)|x − x0| < (4 + 4)δ = ε.27. In both of the preceeding proofs the function f satisfied an inequality ofform|f(x1) − f(x2)| ≤ M|x1− x2| (1)for x1, x2∈ S. In Example 5 we had|(3x1+ 7) − (3x2+ 7)| ≤ 3|x1− x2|and in Example 6 we had|x21− x22| ≤ 8|x1− x2|for 0 < x1, x2< 4. An inequality of form (1) is called a Lipschitz inequalityand the constant M is called the corresponding Lipschitz constant.Theorem 8. If f satisfies (1) for x1, x2∈ S, then f is uniformly continuouson S.Proof. Choose ε > 0. Let δ = ε/M. Choose x0∈ S. Choose x ∈ S. Assumethat |x − x0| < δ. Then|f(x) − f(x0)| ≤ M|x − x0| < Mδ = ε.9. The Lipschitz constant depend might depend on the interval. For example,|x21− x22| = (x1+ x2)|x1− x2| ≤ 2a|x1− x2|for 0 < x1, x2< a but the function f(x) = x2does not satisfy a Lipschitzinequality on the whole interval (0, ∞) since|x21− x22| = (x1+ x2)|x1− x2| > M|x1− x2|if x1= M and x2= x1+ 1. In fact,Example 10. The function f(x) = x2is continuous but not uniformly con-tinuous on the interval S = (0, ∞).Proof. We show f is continuous on S, i.e.∀x0∈ S ∀ε > 0 ∃δ > 0 ∀x ∈ S|x − x0| < δ =⇒ |x2− x20| < ε.3Choose x0. Let a = x0+ 1 and δ = min(1, ε/2a). (Note that δ depends onx0since a does.) Choose x ∈ S. Assume |x −x0| < δ. Then |x − x0| < 1 sox < x0+ 1 = a so x, x0< a so|x2− x20| = (x + x0)|x − x0| ≤ 2a|x − x0| < 2aδ ≤ 2aε2a= εas required.We show that f is not uniformly continuous on S, i.e.∃ε > 0 ∀δ > 0 ∃x0∈ S ∃x ∈ S|x − x0| < δ and |x2− x20| ≥ ε.Let ε = 1. Choose δ > 0. Let x0= 1/δ and x = x0+ δ/2. Then |x − x0| =δ/2 < δ but|x2− x20| =1δ+δ22−1δ2= 1 +δ24> 1 = εas required. (Note that x0is large when δ is small.)11. According to the Mean Value Theorem from calculus for a differentiablefunction f we havef(x1) − f(x2) = f0(c)(x2− x1).for some c between x1and x2. (The slope (f(x1) − f(x2))/(x1− x2) of thesecant line joining the two points (x1, f(x1)) and (x2, f(x2)) on the graph isthe same as the slope f0(c) of the tangent point at the intermediate point(c, f(c)).) If x1and x2lie in some interval S and |f0(c)| ≤ M for all c ∈ Swe conclude that the Lipschitz inequality (1) holds on S. We don’t want touse the Mean Value Theorem without first proving it, but we certainly canuse it to guess an appropriate value of M and then prove the inequality byother means.12. For example, consider the function f(x) = x−1defined on the intervalS = (a, ∞) where a > 0. For x1, x2∈ S the Mean Value Theorem says thatx−11− x−12= −c−2(x1− x2) where c is between x1and x2. If x1, x2∈ S thenc ∈ S (as c is between x1and x2) and hence c > a so c−2< a−2. We canprove the inequality|x−11− x−12| ≤ a−2|x2− x2|4for x1, x2≥ a as follows. First a2≤ x1x2since a ≤ x1and a ≤ x2. Then|x−11− x−12| =|x1− x2|x1x2≤|x1− x2|a2(2)where we have used the fact that α−1< β−1if 0 < α < β. It follows thatthat the function f(x) is uniformly continuous on any interval (a, ∞) wherea > 0. Notice however that the Lipschitz constant M = a−2depends onthe interval. In fact, the function f(x) = x−1does not satisfy a Lipshitzinequality on the interval (0, ∞).13. We can discover a Lipscitz inequality for the square root function f(x) =√x in much the same way. Consider the function f(x) =√x defined on theinterval S …