Let a function f be continuous for every value of x in [a, b]. it means that if x0 ε [a, b], then given ε > 0, there exists δ > 0, such that
│f(x) –f(x0)│< ε whenever│ x – x0│< δ
The number δ will depend upon x0 as well as ε and so we may write it symbolically as δ (ε, x0). For some functions it may happen that given ε > 0 the same δ serves for all x0 ε [a, b] in the condition of condition of continuity. Such functions are called uniformly continuous on [a, b]
Definition for Uniform Continuity
A function f defined on an interval I is said to be uniformly continuous on I if given ε > 0, there exists a δ > 0 such that
│f(x) – f(y) │< wherever │x - y│< δ
It should be noted carefully that uniform continuity is a property associated with an interval and not with a single point. The concept of continuity is local in character where as the concept of uniform is global in character.
Note:- If f uniformly continuous on an interval I, then it is continuous on I.
Note:- A function which is continuous in a closed and bounded interval I = [a, b] is uniformly continuous in [a, b]
Examples on Uniform Continuity
Give an example to show that a function continuous in an open interval may fail to be uniformly continuous in the interval.
Solution
Consider the function f defined on the open interval ] 0, 1[ as follows:
We have, x is continuous in ] 0, 1[ and x ≠ 0 in ] 0, 1[
→ 1/x is continuous in ] 0, 1[
Now, for any δ > 0, we can find m `in` N such that
1/n < δ for every n > m
Let x1 1/2m and x2 = 1/m so that x1, x2 `in` ] 0, 1[
│x1 – x2│= │1/2m – 1/m│= 1/2m < 1/2δ < δ
But │f(x1) – f(x2)│= │2m – m│= m
Which cannot be less than each ε >0
Example2
Show that the function f denoted by f(x) = x3,is uniformly continuous in [–2, 2]
Solution
Let x1, x2 `in` [–2, 2].
We have │f(x2) – f(x1)│=│x23 – x13│
= │(x2 – x1) (x23 + x13 + x1x2│
< │x2 – x1│{│x2│3+│x1│2+│x1││x2│}
< 12│x2 – x1│
... │f(x2) – f(x1)│< ε wherever│x2 – x1│< ε/12
Thus given ε > 0, there exists δ = ε/12 such that
│f(x2) – f(x1)│< ε whenever│x2 – x1│< δ `AA` x1, x2 `in` [–2, 2]
Hence f(x) is uniformly continuous in [–2, 2]
│f(x) –f(x0)│< ε whenever│ x – x0│< δ
The number δ will depend upon x0 as well as ε and so we may write it symbolically as δ (ε, x0). For some functions it may happen that given ε > 0 the same δ serves for all x0 ε [a, b] in the condition of condition of continuity. Such functions are called uniformly continuous on [a, b]
Definition for Uniform Continuity
A function f defined on an interval I is said to be uniformly continuous on I if given ε > 0, there exists a δ > 0 such that
│f(x) – f(y) │< wherever │x - y│< δ
It should be noted carefully that uniform continuity is a property associated with an interval and not with a single point. The concept of continuity is local in character where as the concept of uniform is global in character.
Note:- If f uniformly continuous on an interval I, then it is continuous on I.
Note:- A function which is continuous in a closed and bounded interval I = [a, b] is uniformly continuous in [a, b]
Examples on Uniform Continuity
Give an example to show that a function continuous in an open interval may fail to be uniformly continuous in the interval.
Solution
Consider the function f defined on the open interval ] 0, 1[ as follows:
We have, x is continuous in ] 0, 1[ and x ≠ 0 in ] 0, 1[
→ 1/x is continuous in ] 0, 1[
Now, for any δ > 0, we can find m `in` N such that
1/n < δ for every n > m
Let x1 1/2m and x2 = 1/m so that x1, x2 `in` ] 0, 1[
│x1 – x2│= │1/2m – 1/m│= 1/2m < 1/2δ < δ
But │f(x1) – f(x2)│= │2m – m│= m
Which cannot be less than each ε >0
Example2
Show that the function f denoted by f(x) = x3,is uniformly continuous in [–2, 2]
Solution
Let x1, x2 `in` [–2, 2].
We have │f(x2) – f(x1)│=│x23 – x13│
= │(x2 – x1) (x23 + x13 + x1x2│
< │x2 – x1│{│x2│3+│x1│2+│x1││x2│}
< 12│x2 – x1│
... │f(x2) – f(x1)│< ε wherever│x2 – x1│< ε/12
Thus given ε > 0, there exists δ = ε/12 such that
│f(x2) – f(x1)│< ε whenever│x2 – x1│< δ `AA` x1, x2 `in` [–2, 2]
Hence f(x) is uniformly continuous in [–2, 2]
