Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

3.5: Uniform Continuity

( \newcommand{\kernel}{\mathrm{null}\,}\)

We discuss here a stronger notion of continuity.

Definition 3.5.1: Uniformly Continuous

Let D be a nonempty subset of R. A function f:DR is called uniformly continuous on D if for any ε>0, there exists δ>0 such that if u,vD and |uv|<δ, then

|f(u)f(v)|<ε.

Example 3.5.1

Any constant function f:DR, is uniformly continuous on its domain.

Solution

Indeed, given ε>0, |f(u)f(v)|=0<ε for all u,vD regardless of the choice of δ.

The following result is straightforward from the definition.

Theorem 3.5.1

If f:DR is uniformly continuous on D, then f is continuous at every point x0D.

Example 3.5.2

Let f:RR be given by f(x)=7x2. We will show that f is uniformly continuous on R.

Solution

Let ε>0 and choose δ=ε/7. Then, if u,vR and |uv|<δ, we have

|f(u)f(v)|=|7u2(7v2)|=|7(uv)|=7|uv|<7δ=ε

Example 3.5.3

Let f:[3,2]R be given by f(x)=x2. This function is uniformly continuous on [3,2].

Solution

Let ε>0. First observe that for u,v[3,2] we have |u+v||u|+|v|6. Now set δ=ε/6. Then, for u,v[3,2] satisfying |uv|<δ, we have

|f(u)f(v)|=|u2v2|=|uv||u+v|6|uv|<6δ=ε.

Example 3.5.4

Let f:RR be given by f(x)=x2x2+1. We will show that f is uniformly continuous on R.

Solution

Let ε>0. We observe first that

u2u2+1v2v2+1=|u2(v2+1)v2(u2+1)(u2+1)(v2+1)|=|uv||u+v|(u2+1)(v2+1)|uv|(|u|+|v|)(u2+1)(v2+1)|uv|((u2+1)+(v2+1))(u2+1)(v2+1)|uv|(1v2+1+1u2+1)2|uv|,

(where we used that |x|x2+1 for all xR, which can be easily seen by considering separately the cases |x|<1 and |x|1).

Now set δ=ε/2. In view of the previous calculation, given u,vR satisfying |uv|<δ we have

|f(u)f(v)|=|u2u2+1v2v2+1|2|uv|<2δ=ε.

Definition 3.5.2: Hölder Continuity

Let D be a nonempty subset of R. A function f:DR is said to be Hölder continuous if there are constants 0 and α>0 such that

|f(u)f(v)||uv|α for every u,vD.

The number α is called Hölder exponent of the function. If α=1, then the function f is called Lipschitz continuous.

Theorem 3.5.2

If a function f:DR is Hölder continuous, then it is uniformly continuous.

Proof

Since f is Hölder conitnuous, there are constants 0 and α>0 such that

|f(u)f(v)||uv|α for every u,vD.

If =0, then f is constant and, thus, uniformly continuous. Suppose next that >0. For any ε>0, let δ=(ε)1/α. Then, whenever u,vD, with |uv|<δ we have

|f(u)f(v)||uv|α<δα=ε.

The proof is now complete.

Example 3.5.5

  1. Let D=[a,), where a>0.(2) Let D=[0,).

Solution

  1. Then the function f(x)=x is Lipschitz continuous on D and, hence, uniformly continuous on this set. Indeed, for any u,vD, we have

|f(u)f(v)|=|uv|=|uv|u+v12a|uv|,

which shows f is Lipschitz with =1/(2a).

Annotation 2020-08-30 161036.png

Figure 3.4: The square root function.

  1. Then f is not Lipschitz continuous on D, but it is Hölder continuous on D and, hence, f is also uniformly continuous on this set.

Indeed, suppose by contradiction that f is Lipschitz continuous on D. Then there exists a constant >0 such htat

|f(u)f(v)|=|uv||uv| for every u,vD.

Thus, for every nN, we have

|1n0||1n0|.

This implies

n or n2 for every nN.

This is a contradiction. Therefore, f is not Lipschitz continuous on D.

Let us show that f is Hölder continuous on D. We are going to prove that

|f(u)f(v)||uv|1/2 for every u,vD.

The inequality in (3.9) holds obviously for u=v=0. For u>0 or v>0, we have

|f(u)f(v)|=|uv|=|uvu+v|=|uv||uv|u+v|u|+|v|u+v|uv|=|uv|.

Note that one can justify that inequality

|u|+|v|u+v1

by squaring both sides since they are both positive. Thus, (3.9) is satisfied.

While every uniformly continuous function on a set D is also continuous at each point of D, the converse is not true in general. The following example illustrates this point.

Example 3.5.6

Let f:(0,1)R be given by

f(x)=1x.

Annotation 2020-08-30 162555.png

Figure 3.5: Continuous but not uniformly continuous on (0,).

Solution

We already know that this function is continuous at every ˉx(0,1). We will show that f is not uniformly continuous on (0,1). Let ε=2 and δ>0. Set δ0=min{δ/2,1/4}, x=δ0, and y=2δ0. Then x,y(0,1) and |xy|=δ0<δ, but

|f(x)f(y)|=|1x1y|=|yxxy|=|δ02δ20|=|12δ0|2=ε.

This shows f is not uniformly continuous on (0,1).

The following theorem offers a sequential characterization of uniform continuity analogous to that in Theorem 3.3.3.

Theorem 3.5.3

Let D be a nonempty subset of R and f:DR. Then f is uniformly continuous on D if and only if the following condition holds

(C) for every two sequences {un}, {vn} in D such that limn(unvn)=0, it follows that limn(f(un)f(vn))=0.

Proof

Suppose first that f is uniformly continuous and let {un}, {vn} be sequences in D such that limn(unvn)=0. Let ε>0. Choose δ>0 such that |f(u)f(v)|<ε whenever u,vD and |uv|<δ. Let NN be such that |unvn|<δ for nN. For such n, we have |f(un)f(vn)|<ε. This shows limn(f(un)f(vn))=0.

To prove the converse, assume condition (C) holds and suppose, by way of contradiction, that f is not uniformly continuous. Then there exists ε0>0 such that for any δ>0, there exists u,vD with

|uv|<δ and |f(u)f(v)|ε0.

Thus, for every nN, there exist un,vnD with

|unvn|1/n and |f(un)f(vn)|ε0.

It follows that for such sequences, limn(unvn)=0, but {f(un)f(vn)} does not converge to zero, which contradicts the assumption.

Example 3.5.7

Using this theorem, we can give an easier proof that the function in Example 3.5.6 is not uniformly continuous.

Solution

Consider the two sequences un=1/(n+1) and vn=1/n for all n2. Then clearly, limn(unvn)=0, but

limn(f(un)f(vn))=limn(11/(n+1)11/n)=limn(n+1n)=10.

The following theorem shows one important case in which continuity implies uniform continuity.

Theorem 3.5.4

Let f:DR be a continuous function. Suppose D is compact. Then f is uniformly continuous on D.

Proof

Suppose by contradition that f is not uniformly continuous on D. Then there exists ε0>0 such that for any δ>0, there exists u,vD with

|uv|<δ and |f(u)f(v)|ε0.

Thus, for every nN, there exists un,vnD with

|unvn|1/n and |f(un)f(vn)|ε0.

Since D is compact, there exist u0D and a subsequence {unk} of {un} such that

unku0 as k.

Then

|unkvnk|1nk.

for all k and, hence, we also have

vnku0 as k.

By the continuity of f,

f(unk)f(u0) and f(vnk)f(u0).

Therefore, f converges to zero, which is a contradiction. The proof is now complete.

We now prove a result that characterizes uniform continuity on open bounded intervals. We first make the observation that if f:DR is uniformly continuous on D and AD, then f is uniformly continuous on A. More precisely, the restriction fA:AR is uniformly continuous on A (see Section 1.2 for the notation). This follows by noting that if |f(u)f(v)|<ε whenever u,vD with |uv|<δ, then we also have |f(u)f(v)|<ε when we restrict u,v to be in A.

Theorem 3.5.5

Let a,bR and a<b. A function f:(a,b)R is uniformly continuous if and only if f can be extended to a continuous function ˜f:[a,b]R (that is, there is a continuous function ˜f:[a,b]R such that f=˜f(a,b)).

Proof

Suppose first that there exists a continuous function ˜f:[a,b]R such that f=˜f(a,b). By Theorem 3.5.4, the function ˜f is uniformly continuous on [a,b]. Therefore, it follows from our early observation that f is uniformly continuous on (a,b).

For the converse, suppose f:(a,b)R is uniformly continuous. We will show first that limxa+f(x) exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the εδ condition of Theorem 3.2.2 holds.

Let ε>0. Choose δ0>0 so that |f(u)f(v)|<ε whenever u,v(a,b) and |uv|<δ0. Set δ=δ0/2. Then, if u,v(a,b), |ua|<δ, and |va|<δ we have

|uv||ua|+|av|<δ+δ=δ0

and, hence, |f(u)f(v)|<ε. We can now invoke Theorem 3.2.2 to conclude limxa+f(x) exists. In a similar way we can show that limxbf(x) exists. Now define, ˜f:[a,b]R by

˜f(x)={f(x), if x(a,b);limxa+f(x), if x=a;limxbf(x), if x=b.

By its definition ˜f(a,b)=f and, so, tildef is continuous at every x(a,b). Moreover, limxa+˜f(x)=limxa+f(x)=˜f(a) and limxb˜f(x)=limxbf(x)=˜f(b), so ˜f is also continuous at a and b by Theorem 3.3.2. Thus ˜f is the desired continuous extension of f.

Exercise 3.5.1

Prove that each of the following functions is uniformly continuous on the given domain:

  1. f(x)=ax+b,a,bR, on R.
  2. f(x)=1/x on [a,), where a>0.
Answer

Add texts here. Do not delete this text first.

Exercise 3.5.2

Prove that each of the following functions is not uniformly continuous on the given domain:

  1. f(x)=x2 on R.
  2. f(x)=sin1x on (0,1).
  3. f(x)=ln(x) on (0,).
Answer

Add texts here. Do not delete this text first.

Exercise 3.5.3

Determine which of the following functions are uniformly continuous on the given domains.

  1. f(x)=xsin(1x) on (0,1).
  2. f(x)=xx+1 on [0,).
  3. f(x)=1|x1| on (0,1).
  4. f(x)=1|x2| on (0,1).
Answer

Add texts here. Do not delete this text first.

Exercise 3.5.4

Let DR and kR. Prove that if f,g:DR are uniformly continuous on D, then f+g and kf are uniformly continuous on D.

Answer

Add texts here. Do not delete this text first.

Exercise 3.5.5

Give an example of a subset D of R and uniformly continuous functions f,g:DR such that fg is not uniformly continuous on D.

Answer

Add texts here. Do not delete this text first.

Exercise 3.5.6

Let D be a nonempty subset of R and let f:DR. Suppose that f is uniformly continuous on D. Prove that if {xn} is a cauchy sequence with xnD for every nN, then {f(xn)} is also a Caucy sequence.

Answer

Add texts here. Do not delete this text first.

Exercise 3.5.7

Let a,bR and let f:(a,b)R.

  1. Prove that if f is uniformly continuous, then f is bounded.
  2. Prove that if f is continuous, bounded, and monotone, then it is uniformly continuous.
Answer

Add texts here. Do not delete this text first.

Exercise 3.5.8

Let f be a continuous function on [a,). Suppose

limxf(x)=c.

  1. Prove that f is bounded on [a,).
  2. Prove that f is uniformly continuous on [a,).
  3. Suppose further that c>f(a). Prove that there exists x0[a,) such that

f(x0)=inf{f(x):x[a,)}.

Answer

Add texts here. Do not delete this text first.


This page titled 3.5: Uniform Continuity is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Lafferriere, Lafferriere, and Nguyen (PDXOpen: Open Educational Resources) .

Support Center

How can we help?