Kenneth Kuttler

112 CHAPTER 6. CONTINUOUS FUNCTIONS

6.7 Uniform ContinuityThere is a theorem about the integral of a continuous function which requires the notion ofuniform continuity. Uniform continuity is discussed in this section. Consider the functionf (x) = 1

x for x ∈ (0,1) . This is a continuous function because, by Theorem 6.0.8, it iscontinuous at every point of (0,1) . However, for a given ε > 0, the δ needed in the ε,δdefinition of continuity becomes very small as x gets close to 0. The notion of uniformcontinuity involves being able to choose a single δ which works on the whole domain of f .Here is the definition.

Definition 6.7.1 Let f be a function. Then f is uniformly continuous if for everyε > 0, there exists a δ depending only on ε such that if |x− y|< δ then | f (x)− f (y)|< ε.

It is an amazing fact that under certain conditions continuity implies uniform continuity.

Theorem 6.7.2 Let f : K→ Fq be continuous where K is a sequentially compact setin Fp. Then f is uniformly continuous on K.

Proof: If this is not true, there exists ε > 0 such that for every δ > 0 there exists a pairof points, xδ and yδ such that even though ∥xδ − yδ∥ < δ , ∥ f (xδ )− f (yδ )∥ ≥ ε. Takinga succession of values for δ equal to 1,1/2,1/3, · · · , and letting the exceptional pair ofpoints for δ = 1/n be denoted by xn and yn,

∥xn− yn∥<1n,∥ f (xn)− f (yn)∥ ≥ ε.

Now since K is sequentially compact, there exists a subsequence,{

xnk

}such that xnk →

z ∈ K. Now nk ≥ k and so∥∥xnk − ynk

∥∥< 1k . Consequently, ynk → z also. xnk is like a person

walking toward a certain point and ynk is like a dog on a leash which is constantly gettingshorter. Obviously ynk must also move toward the point also. Indeed,∥∥ynk − z

∥∥≤ ∥∥ynk − xnk

∥∥+∥∥xnk − z∥∥≤ 1

k+∥∥xnk − z

∥∥and the right side converges to 0 as k→ ∞.

By continuity of f and Theorem 6.1.1,

0 = ∥ f (z)− f (z)∥= limk→∞

∥∥ f(xnk

)− f

(ynk

)∥∥≥ ε,

an obvious contradiction. Therefore, the theorem must be true.The following corollary follows from this theorem and Theorems 4.8.14, 4.8.13 which

give closed and bounded sets are sequentially compact.

Corollary 6.7.3 Suppose K is any closed and bounded set in Fp. Then if f is continuouson K, it follows that f is uniformly continuous on K.

6.8 Exercises1. A function f : Fp→ Fq is Holder continuous if there exists a constant, K such that

∥ f (x)− f (y)∥ ≤ K ∥x− y∥α

for all x,y ∈ D. Show every Holder continuous function is uniformly continuous.When α = 1, this is called a Lipschitz function or Lipschitz continuous function.

112 CHAPTER 6. CONTINUOUS FUNCTIONS6.7 Uniform ContinuityThere is a theorem about the integral of a continuous function which requires the notion ofuniform continuity. Uniform continuity is discussed in this section. Consider the functionf (x) = + for x € (0,1). This is a continuous function because, by Theorem 6.0.8, it iscontinuous at every point of (0,1). However, for a given € > 0, the 6 needed in the €,6definition of continuity becomes very small as x gets close to 0. The notion of uniformcontinuity involves being able to choose a single 6 which works on the whole domain of f.Here is the definition.Definition 6.7.1 Le: f be a function. Then f is uniformly continuous if for every€ > 0, there exists a 6 depending only on € such that if |\x — y| < 6 then |f (x) —f(y)| <e.It is an amazing fact that under certain conditions continuity implies uniform continuity.Theorem 6.7.2 Ler f :K — F4 be continuous where K is a sequentially compact setin F?. Then f is uniformly continuous on K.Proof: If this is not true, there exists € > 0 such that for every 6 > 0 there exists a pairof points, xg and yg such that even though ||xs —ys|| < 6, ||f (xs) —f (vs)|| => €. Takinga succession of values for 6 equal to 1,1/2,1/3,---, and letting the exceptional pair ofpoints for 6 = 1/n be denoted by x, and yp,1IIxn —Yall <> IF On) — FOn)Il 2 &Now since K is sequentially compact, there exists a subsequence, {Xn, } such that xp,z€K. Now n > k and so ||Xng —Yng | < i: Consequently, yn, — z also. Xn, is like a personwalking toward a certain point and y,, is like a dog on a leash which is constantly gettingshorter. Obviously y,,, must also move toward the point also. Indeed,1[me — 2] S [dm ne || + [Pom — el] S + [Pm al]kand the right side converges to 0 as k — ©».By continuity of f and Theorem 6.1.1,0= [IF @)—F (= him [If Gn) — F On) II Eean obvious contradiction. Therefore, the theorem must be true. JJThe following corollary follows from this theorem and Theorems 4.8.14, 4.8.13 whichgive closed and bounded sets are sequentially compact.Corollary 6.7.3 Suppose K is any closed and bounded set in F?. Then if f is continuouson K, it follows that f is uniformly continuous on K.6.8 Exercises1. A function f : F? + F? is Holder continuous if there exists a constant, K such thatIf (2) — f(y) || SK |lx—yl]*for all x,y € D. Show every Holder continuous function is uniformly continuous.When @ = 1, this is called a Lipschitz function or Lipschitz continuous function.