Uniform continuity

In document Basic Analysis: Introduction to Real Analysis (Page 125-131)

Continuous Functions

3.4 Uniform continuity

Note: 1.5–2 lectures (Continuous extension can be optional)


Uniform continuity

We made a fuss of saying that theδ in the definition of continuity depended on the pointc. There

are situations when it is advantageous to have aδ independent of any point. Let us give a name to

this concept.

Definition 3.4.1. LetSR, and let f: SRbe a function. Suppose for anyε>0 there exists

a δ >0 such that whenever x,c∈S and |x−c|<δ, then |f(x)−f(c)|<ε. Then we say f is

uniformly continuous.

It is not hard to see that a uniformly continuous function must be continuous. The only difference in the definitions is that in uniform continuity, for a givenε >0 we pick aδ >0 that works for

allcS. That is,δ can no longer depend onc, it only depends onε. The domain of definition of

the function makes a difference now. A function that is not uniformly continuous on a larger set, may be uniformly continuous when restricted to a smaller set. Note thatxandcare not treated any differently in the definition.

Example 3.4.2: f: [0,1]→R, defined by f(x):=x2is uniformly continuous.

Proof: Note that 0≤x,c≤1. Then


=|x+c||x−c| ≤(|x|+|c|)|x−c| ≤(1+1)|x−c|.

Therefore, givenε>0, letδ :=ε/2. If|x−c|<δ, thenx2−c2<ε.

On the other hand,g: R→R, defined byg(x):=x2is not uniformly continuous.

Proof: Suppose it is uniformly continuous, then for allε>0, there would exist aδ >0 such

that if|x−c|<δ, thenx2−c2

<ε. Takex>0 and letc:=x+δ/2. Write


=|x+c||x−c|= (2x+δ/2)δ/2≥δx.

Therefore,x<ε/δ for allx>0, which is a contradiction.

Example 3.4.3: The function f: (0,1)R, defined by f(x):=1/xis not uniformly continuous.

Proof: Givenε >0, thenε>|1/x−1/y|holds if and only if ε >|1/x−1/y|= |y−x| |xy| = |y−x| xy , or |x−y|<xyε.

Therefore, to satisfy the definition of uniform continuity we would have to haveδ ≤xyε for allx,y

The examples show that if f is defined on an interval that is either not closed or not bounded, then f can be continuous, but not uniformly continuous. For a closed and bounded interval[a,b],

we can, however, make the following statement.

Theorem 3.4.4. Let f: [a,b]Rbe a continuous function. Then f is uniformly continuous.

Proof. We prove the statement by contrapositive. Suppose f is not uniformly continuous. We will prove that there is somec[a,b]where f is not continuous. Let us negate the definition of

uniformly continuous. There exists anε >0 such that for everyδ >0, there exist pointsx,yin

[a,b]with|xy|<δ and|f(x)−f(y)| ≥ε.

So for theε>0 above, we find sequences{xn}and{yn}such that|xn−yn|<1/nand such that |f(xn)−f(yn)| ≥ε. By , there exists a convergent subsequence{xnk}. Let

c:=limxnk. Asa≤xnk≤b, thena≤c≤b. Write

|ynk−c|=|ynk−xnk+xnk−c| ≤ |ynk−xnk|+|xnk−c|<1/nk+|xnk−c|.


k−c|both go to zero whenkgoes to infinity,{ynk}converges and the limit isc. We

now show that f is not continuous atc. We estimate

|f(xnk)− f(c)|=|f(xnk)−f(ynk) + f(ynk)−f(c)| ≥ |f(xnk)−f(ynk)| − |f(ynk)−f(c)| ≥ε− |f(ynk)−f(c)|.

Or in other words

|f(xnk)−f(c)|+|f(ynk)−f(c)| ≥ε.

At least one of the sequences{f(xnk)}or{f(ynk)}cannot converge to f(c), otherwise the left hand

side of the inequality would go to zero while the right-hand side is positive. Thus f cannot be continuous atc.


Continuous extension

Before we get to continuous extension, we show the following useful lemma. It says that uniformly continuous functions behave nicely with respect to Cauchy sequences. The new issue here is that for a Cauchy sequence we no longer know where the limit ends up; it may not end up in the domain of the function.

Lemma 3.4.5. Let f: SRbe a uniformly continuous function. Let{xn}be a Cauchy sequence in S. Then{f(xn)}is Cauchy.

Proof. Let ε >0 be given. There is aδ >0 such that |f(x)−f(y)|<ε whenever |x−y|<δ.

Find an MN such that for alln,k≥M we have|xn−xk|<δ. Then for alln,k≥M we have


An application of the above lemma is the following result. It says that a function on an open interval is uniformly continuous if and only if it can be extended to a continuous function on the closed interval.

Proposition 3.4.6. A function f: (a,b)→Ris uniformly continuous if and only if the limits


→af(x) and Lb:=xlim→bf(x) exist and the function ef: [a,b]→Rdefined by

e f(x):=      f(x) if x(a,b), La if x=a, Lb if x=b, is continuous.

Proof. One direction is not difficult. If ef is continuous, then it is uniformly continuous by

. As f is the restriction of feto(a,b), then f is also uniformly continuous (easy exercise).

Now suppose f is uniformly continuous. We must first show that the limitsLaandLbexist. Let us concentrate onLa. Take a sequence{xn}in(a,b)such that limxn=a. The sequence{xn}is Cauchy, so by the sequence{f(xn)}is Cauchy and thus convergent. We have some numberL1:=lim f(xn). Take another sequence{yn}in(a,b)such that limyn=a. By the same reasoning we getL2:=lim f(yn). If we show thatL1=L2, then the limitLa=limx→af(x)exists. Letε >0 be given. Findδ >0 such that|x−y|<δ implies|f(x)−f(y)|<ε/3. FindM∈Nsuch

that for nM we have |a−xn|<δ/2,|a−yn|<δ/2, |f(xn)−L1|<ε/3, and |f(yn)−L2|<ε/3.

Then forn≥Mwe have

|xn−yn|=|xn−a+a−yn| ≤ |xn−a|+|a−yn|<δ/2+δ/2=δ.


|L1−L2|=|L1−f(xn) +f(xn)−f(yn) + f(yn)−L2| ≤ |L1−f(xn)|+|f(xn)−f(yn)|+|f(yn)−L2| ≤ε/3+ε/3+ε/3=ε.

Therefore,L1=L2. ThusLaexists. To show thatLbexists is left as an exercise.

Now that we know that the limits La and Lb exist, we are done. If limx→af(x)exists, then limx→a ef(x)exists (See ). Similarly withLb. Hence feis continuous ataandb.

And since f is continuous atc(a,b), then ef is continuous atc∈(a,b).

A common application of this proposition (together with ) is the following. Suppose f: (−1,0)∪(0,1)→Ris uniformly continuous, then limx→0f(x)exists and the function has what is called anremovable singularity, that is, we can extend the function to a continuous function on(1,1).


Lipschitz continuous functions

Definition 3.4.7. A function f: SRisLipschitz continuous , if there exists aKR, such that |f(x)f(y)| ≤K|xy| for allxandyinS.

A large class of functions is Lipschitz continuous. Be careful, just as for uniformly continuous functions, the domain of definition of the function is important. See the examples below and the exercises. First we justify the use of the wordcontinuous.

Proposition 3.4.8. A Lipschitz continuous function is uniformly continuous.

Proof. Let f: SRbe a function and letK be a constant such that|f(x)−f(y)| ≤K|x−y|for

allx,yinS. Letε>0 be given. Takeδ :=ε/K. For anyxandyinSsuch that|x−y|<δ we have

|f(x)− f(y)| ≤K|x−y|<Kδ =Kε

K =ε. Therefore, f is uniformly continuous.

We interpret Lipschitz continuity geometrically. If f is a Lipschitz continuous function with some constantK. We rewrite the inequality to say that forx6=ywe have

f(x)f(y) x−y ≤ K. The quantity f(x)−f(y)

x−y is the slope of the line between the points x,f(x)

and y,f(y)

, that is, a secant line. Therefore, f is Lipschitz continuous if and only if every line that intersects the graph of

f in at least two distinct points has slope less than or equal toK. See .

x y

slope= f(xx)−yf(y)

Figure 3.7:The slope of a secant line. A function is Lipschitz if|slope|=

f(x)−f(y) x−y

≤Kfor allxandy.

Example 3.4.9: The functions sin(x)and cos(x)are Lipschitz continuous. We have seen ( ) the following two inequalities.

|sin(x)−sin(y)| ≤ |x−y| and |cos(x)−cos(y)| ≤ |x−y|.

Hence sin and cos are Lipschitz continuous withK=1.

Example 3.4.10: The function f: [1,∞)→R defined by f(x):=√x is Lipschitz continuous.

Proof: √x −√y = xy √x +√y = √|xx−y| +√y.

Asx≥1 andy≥1, we see that √x+1√y ≤ 12. Therefore √x −√y= x−y √x +√y ≤ 1 2|x−y|.

On the other hand f: [0,∞)Rdefined by f(x):=√xis not Lipschitz continuous. Let us see why: Suppose we have

x −√y


for someK. Lety=0 to obtain√x≤Kx. IfK>0, then forx>0 we then get1/Kx. This cannot possibly be true for allx>0. Thus no suchK>0 exists and f is not Lipschitz continuous.

The last example is a function that is uniformly continuous but not Lipschitz continuous. To see that √x is uniformly continuous on [0,∞) note that it is uniformly continuous on [0,1] by . It is also Lipschitz (and therefore uniformly continuous) on[1,∞). It is not hard (exercise) to show that this means that√xis uniformly continuous on[0,∞).



Exercise3.4.1: Let f:S→Rbe uniformly continuous. Let A⊂S. Then the restriction f|A is uniformly


Exercise3.4.2: Let f:(a,b)→Rbe a uniformly continuous function. Finish the proof of

by showing that the limitlim


Exercise3.4.3: Show that f:(c,∞)Rfor some c>0and defined by f(x):=1/xis Lipschitz continuous. Exercise3.4.4: Show that f:(0,∞)Rdefined by f(x):=1/xis not Lipschitz continuous.

Exercise3.4.5: Let A,B be intervals. Let f:ARand g:B→Rbe uniformly continuous functions such

that f(x) =g(x)for x∈A∩B. Define the function h:A∪B→Rby h(x):= f(x)if x∈A and h(x):=g(x)if


a) Prove that if A∩B6=/0, then h is uniformly continuous.

b) Find an example where A∩B=/0and h is not even continuous.

Exercise3.4.6(Challenging): Let f:R→Rbe a polynomial of degree d≥2. Show that f is not Lipschitz


Exercise 3.4.7: Let f: (0,1)R be a bounded continuous function. Show that the function g(x):=

x(1−x)f(x)is uniformly continuous.

Exercise3.4.8: Show that f:(0,∞)→Rdefined by f(x):=sin(1/x)is not uniformly continuous.

Exercise3.4.9(Challenging): Let f:Q→Rbe a uniformly continuous function. Show that there exists a


a) Find a continuous f:(0,1)Rand a sequence{xn}in(0,1)that is Cauchy, but such that{f(xn)}is

not Cauchy.

b) Prove that if f:R→Ris continuous, and{xn}is Cauchy, then{f(xn)}is Cauchy.

Exercise3.4.11: Prove:

a) If f: S→Rand g: S→Rare uniformly continuous, then h:S→Rgiven by h(x):= f(x) +g(x)is

uniformly continuous.

b) If f:S→Ris uniformly continuous and a∈R, then h: S→Rgiven by h(x):=a f(x) is uniformly


Exercise3.4.12: Prove:

a) If f:S→Rand g:S→Rare Lipschitz, then h:S→Rgiven by h(x):=f(x) +g(x)is Lipschitz.

b) If f:S→Ris Lipschitz and a∈R, then h:S→Rgiven by h(x):=a f(x)is Lipschitz. Exercise3.4.13:

a) If f:[0,1]Ris given by f(x):=xmfor an integer m≥0, show f is Lipschitz and find the best (the

smallest) Lipschitz constant K (depending on m of course). Hint:(xy)(xm−1+xm−2y+xm−3y2+···+

xym−2+ym−1) =xmym.

b) Using the previous exercise, show that if f:[0,1]→Ris a polynomial, that is, f(x):=amxm+am−1xm−1+

···+a0, then f is Lipschitz.

Exercise 3.4.14: Suppose for f: [0,1] R we have |f(x)−f(y)| ≤K|x−y| for all x,y in [0,1], and

f(0) = f(1) =0. Prove that|f(x)| ≤K/2for all x[0,1]. Further show by example thatK/2is the best

possible, that is, there exists such a continuous function for which|f(x)|=K/2for some x[0,1].

Exercise3.4.15: Suppose f:R→Ris continuous and periodic with period P>0. That is, f(x+P) =f(x)

for all x∈R. Show that f is uniformly continuous.

Exercise3.4.16: Suppose f:S→Rand g:[0,∞)→[0,∞) are functions, g is continuous at0, g(0) =0,

and whenever x and y are in S we have|f(x)f(y)| ≤g(|xy|). Prove that f is uniformly continuous.

Exercise 3.4.17: Suppose f:[a,b]Ris a function such that for every c∈[a,b]there is a Kc>0and

anεc>0for which|f(x)−f(y)| ≤Kc|x−y|for all x and y in(c−εc,c+εc)∩[a,b]. In other words, f is

“locally Lipschitz.”

a) Prove that there exists a single K>0such that|f(x)f(y)| ≤K|xy|for all x,y in[a,b].

b) Find a counterexample to the above if the interval is open, that is, find an f:(a,b)Rthat is locally

In document Basic Analysis: Introduction to Real Analysis (Page 125-131)