Uniform Continuity is a special kind of continuity. In this page, we will learn about uniformly continuous functions with examples and related theorems.
Notation:
I= An interval.
ℝ= The set of all real numbers.
MOTIVATION:
Let f: I → ℝ be a continuous function and c ∈ I. As f is continuous at c, for every ε>0 there exists a δ>0 such that
|f(x)-f(c)| < ε whenever 0<|x-c|<δ.
If we discuss the continuity at some other point then for the same ε chosen above, we may get different δ.Thus we can say δ depends on ε and the point c chosen. For this reason, lets denote δ:= δ(ε, c). Consider all such possible δ’s and take infimum of them, denoted by δ’. That is,
δ’:= inf{δ(ε, c): c ∈ I}.
Then δ’ ≥0.
If δ0>0, then it works uniformly over the entire interval in the following sense: For every ε>0 and any two points x1, x2 in I satisfying |x1– x2|<δ, we have |f(x1)- f(x2)|<ε. If this happens, then we say f is uniformly continuous on I.
Remark: If there is no positive δ’ exists, then the function cannot be uniformly continuous, although it’s a continuous function.
Definition of Uniform Continuity
A function f: I → ℝ is called an uniformly continuous function, if for every two points x1, x2 in I satisfying |x1– x2|<δ, we have |f(x1)- f(x2)|<ε.
An Example
Consider the function f: ℝ → ℝ be defined by f(x) = x2.
We will check its uniform continuity on two intervals [a, b] and [a, ∞) where a>0.
Let ε>0.
On [a,b] with a>0: |f(x1)- f(x2)| = |(x1-x2)(x1+x2)| < 2b|x1– x2|. Thus choose δ=$\frac{\epsilon}{2b}$. As δ does not depend on points, so we conclude that f is uniformly continuous.
On [a, ∞) with a>0: On [a,b], we have seen that whenever b becomes larger and larger, we cannot fix a single δ which will work on the whole domain. As a result, f is NOT uniformly continuous.
Related Article: Continuous Function
Uniformly Continuous Implies Continuous.
Theorem: If a function f: I → ℝ is uniformly continuous, then it is continuous on I.
Converse Part:
The converse may not be true. For example, f(x) = $\frac{1}{x}$ defined on (0,1]. f is uniformly continuous then it sends a Cauchy sequence in I to a Cauchy sequence in the image. Consider the Cauchy sequence $\{\frac{1}{n}\}$ in (0,1]. But, the sequence $\{f(\frac{1}{n})\}$ = {n} = {1,2,3, …} is not Cauchy. So f is not uniformly continuous.
Theorem: If f: [a,b] → ℝ is continuous, then it is uniformly continuous.
C[0,1]:= The set of all continuous functions on [0,1].
R[0,1]:= The set of all Riemann Integrable functions on [0,1].
B[0,1]:= The set of all bounded functions on [0,1].
NOTE: C[0,1] $\subsetneq$ B[0,1]. Moreover, C[0,1] $\subsetneq$ R[0,1] $\subsetneq$ B[0,1].

This article is written by Dr. Tathagata Mandal, Ph.D in Mathematics from IISER Pune (Algebraic Number Theory), Postdocs at IIT Kanpur & ISI Kolkata. Currently, working as an Assistant Prof. at Adamas University. Thank you for visiting the website.