3.5: Uniform Continuity
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We discuss here a stronger notion of continuity.
Let D be a nonempty subset of R. A function f:D→R is called uniformly continuous on D if for any ε>0, there exists δ>0 such that if u,v∈D and |u−v|<δ, then
|f(u)−f(v)|<ε.
Any constant function f:D→R, is uniformly continuous on its domain.
Solution
Indeed, given ε>0, |f(u)−f(v)|=0<ε for all u,v∈D regardless of the choice of δ.
The following result is straightforward from the definition.
If f:D→R is uniformly continuous on D, then f is continuous at every point x0∈D.
Let f:R→R be given by f(x)=7x−2. We will show that f is uniformly continuous on R.
Solution
Let ε>0 and choose δ=ε/7. Then, if u,v∈R and |u−v|<δ, we have
|f(u)−f(v)|=|7u−2−(7v−2)|=|7(u−v)|=7|u−v|<7δ=ε
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 |u−v|<δ, we have
|f(u)−f(v)|=|u2−v2|=|u−v||u+v|≤6|u−v|<6δ=ε.
Let f:R→R 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+1−v2v2+1∣=|u2(v2+1)−v2(u2+1)(u2+1)(v2+1)|=|u−v||u+v|(u2+1)(v2+1)≤|u−v|(|u|+|v|)(u2+1)(v2+1)≤|u−v|((u2+1)+(v2+1))(u2+1)(v2+1)≤|u−v|(1v2+1+1u2+1)≤2|u−v|,
(where we used that |x|≤x2+1 for all x∈R, 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,v∈R satisfying |u−v|<δ we have
|f(u)−f(v)|=|u2u2+1−v2v2+1|≤2|u−v|<2δ=ε.
Let D be a nonempty subset of R. A function f:D→R is said to be Hölder continuous if there are constants ℓ≥0 and α>0 such that
|f(u)−f(v)|≤ℓ|u−v|α for every u,v∈D.
The number α is called Hölder exponent of the function. If α=1, then the function f is called Lipschitz continuous.
If a function f:D→R is Hölder continuous, then it is uniformly continuous.
- Proof
-
Since f is Hölder conitnuous, there are constants ℓ≥0 and α>0 such that
|f(u)−f(v)|≤ℓ|u−v|α for every u,v∈D.
If ℓ=0, then f is constant and, thus, uniformly continuous. Suppose next that ℓ>0. For any ε>0, let δ=(εℓ)1/α. Then, whenever u,v∈D, with |u−v|<δ we have
|f(u)−f(v)|≤ℓ|u−v|α<ℓδα=ε.
The proof is now complete. ◻
- Let D=[a,∞), where a>0.(2) Let D=[0,∞).
Solution
- Then the function f(x)=√x is Lipschitz continuous on D and, hence, uniformly continuous on this set. Indeed, for any u,v∈D, we have
|f(u)−f(v)|=|√u−√v|=|u−v|√u+√v≤12√a|u−v|,
which shows f is Lipschitz with ℓ=1/(2√a).
Figure 3.4: The square root function.
- 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 ℓ>0 such htat
|f(u)−f(v)|=|√u−√v|≤ℓ|u−v| for every u,v∈D.
Thus, for every n∈N, we have
|1√n−0|≤ℓ|1n−0|.
This implies
√n≤ℓ or n≤ℓ2 for every n∈N.
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)|≤|u−v|1/2 for every u,v∈D.
The inequality in (3.9) holds obviously for u=v=0. For u>0 or v>0, we have
|f(u)−f(v)|=|√u−√v|=|u−v√u+√v|=√|u−v|√|u−v|√u+√v≤√|u|+|v|√u+√v√|u−v|=√|u−v|.
Note that one can justify that inequality
√|u|+|v|√u+√v≤1
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.
Let f:(0,1)→R be given by
f(x)=1x.
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 |x−y|=δ0<δ, but
|f(x)−f(y)|=|1x−1y|=|y−xxy|=|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.
Let D be a nonempty subset of R and f:D→R. 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→∞(un−vn)=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→∞(un−vn)=0. Let ε>0. Choose δ>0 such that |f(u)−f(v)|<ε whenever u,v∈D and |u−v|<δ. Let N∈N be such that |un−vn|<δ for n≥N. 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,v∈D with
|u−v|<δ and |f(u)−f(v)|≥ε0.
Thus, for every n∈N, there exist un,vn∈D with
|un−vn|≤1/n and |f(un)−f(vn)|≥ε0.
It follows that for such sequences, limn→∞(un−vn)=0, but {f(un)−f(vn)} does not converge to zero, which contradicts the assumption. ◻
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 n≥2. Then clearly, limn→∞(un−vn)=0, but
limn→∞(f(un)−f(vn))=limn→∞(11/(n+1)−11/n)=limn→∞(n+1−n)=1≠0.
The following theorem shows one important case in which continuity implies uniform continuity.
Let f:D→R 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,v∈D with
|u−v|<δ and |f(u)−f(v)|≥ε0.
Thus, for every n∈N, there exists un,vn∈D with
|un−vn|≤1/n and |f(un)−f(vn)|≥ε0.
Since D is compact, there exist u0∈D and a subsequence {unk} of {un} such that
unk→u0 as k→∞.
Then
|unk−vnk|≤1nk.
for all k and, hence, we also have
vnk→u0 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:D→R is uniformly continuous on D and A⊂D, then f is uniformly continuous on A. More precisely, the restriction f∣A:A→R 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,v∈D with |u−v|<δ, then we also have |f(u)−f(v)|<ε when we restrict u,v to be in A.
Let a,b∈R 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 limx→a+f(x) exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the ε−δ 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 |u−v|<δ0. Set δ=δ0/2. Then, if u,v∈(a,b), |u−a|<δ, and |v−a|<δ we have
|u−v|≤|u−a|+|a−v|<δ+δ=δ0
and, hence, |f(u)−f(v)|<ε. We can now invoke Theorem 3.2.2 to conclude limx→a+f(x) exists. In a similar way we can show that limx→b−f(x) exists. Now define, ˜f:[a,b]→R by
˜f(x)={f(x), if x∈(a,b);limx→a+f(x), if x=a;limx→b−f(x), if x=b.
By its definition ˜f∣(a,b)=f and, so, tildef is continuous at every x∈(a,b). Moreover, limx→a+˜f(x)=limx→a+f(x)=˜f(a) and limx→b−˜f(x)=limx→b−f(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:
- f(x)=ax+b,a,b∈R, on R.
- f(x)=1/x on [a,∞), where a>0.
- Answer
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Exercise 3.5.2
Prove that each of the following functions is not uniformly continuous on the given domain:
- f(x)=x2 on R.
- f(x)=sin1x on (0,1).
- f(x)=ln(x) on (0,∞).
- Answer
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Exercise 3.5.3
Determine which of the following functions are uniformly continuous on the given domains.
- f(x)=xsin(1x) on (0,1).
- f(x)=xx+1 on [0,∞).
- f(x)=1|x−1| on (0,1).
- f(x)=1|x−2| on (0,1).
- Answer
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Exercise 3.5.4
Let D⊂R and k∈R. Prove that if f,g:D→R are uniformly continuous on D, then f+g and kf are uniformly continuous on D.
- Answer
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Exercise 3.5.5
Give an example of a subset D of R and uniformly continuous functions f,g:D→R such that fg is not uniformly continuous on D.
- Answer
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Exercise 3.5.6
Let D be a nonempty subset of R and let f:D→R. Suppose that f is uniformly continuous on D. Prove that if {xn} is a cauchy sequence with xn∈D for every n∈N, then {f(xn)} is also a Caucy sequence.
- Answer
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Exercise 3.5.7
Let a,b∈R and let f:(a,b)→R.
- Prove that if f is uniformly continuous, then f is bounded.
- Prove that if f is continuous, bounded, and monotone, then it is uniformly continuous.
- Answer
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Exercise 3.5.8
Let f be a continuous function on [a,∞). Suppose
limx→∞f(x)=c.
- Prove that f is bounded on [a,∞).
- Prove that f is uniformly continuous on [a,∞).
- Suppose further that c>f(a). Prove that there exists x0∈[a,∞) such that
f(x0)=inf{f(x):x∈[a,∞)}.
- Answer
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