Uniformly Continuous Function but not Lipschitz: An Example

If a function satisfies the Lipschitz condition then it is uniformly continuous. But the converse is not true. For example, f(x) = √x on [0, 1].

Definition of Lipschitz Function

If a function f(x) satisfy Lipschitz condition on an interval I ⊂ ℝ then there exists a real number K > 0 such that for all x, y ∈ I we have

|f(x) – f(y)| ≤ K |x – y|.

The constant K is called the Lipschitz constant.

Example of Lipschitz Function

The function f(x) = x² on [0, 1] is an example of Lipschitz functions. This is because for all points x, y in [0, 1] we have that

|f(x) – f(y)| = |x² – y²| = |(x-y)(x+y)| ≤ 2 |x – y|.

Now, we prove that Lipschitz functions are uniformly continuous.

Lipschitz Functions are Uniformly Continuous

Theorem: Let I ⊂ ℝ be an interval and f : I → ℝ be a Lipschitz function. Then f is uniformly continuous on I.

Proof:

As f satisfies Lipschitz condition on I, there is a real number K > 0 such that

|f(x) – f(y)| ≤ K |x – y| …(∗)

holds for all x, y ∈ I.

Now, we will show that f is uniformly continuous on I.

Let ε > 0 be given an arbitrary number. Then we choose δ = ε/K > 0. Now for all x, y ∈ I with |x-y| < δ, from (∗) we have that

|f(x) – f(y)| ≤ K |x – y| < K ⋅ ε/K = ε.

This shows that f is uniformly continuous on I. Hence the theorem follows.

Also Read: Continuous but not Uniformly Continuous: An Example

The Converse is NOT True

The converse of the above theorem is not true. That is, if a function is a uniformly continuous on I, then it may not satisfy Lipschitz condition on I.

Uniformly Continuous but not Lipschitz

Consider the function

f(x) = √x on [0, 1].

As f(x) is defined on a closed bounded interval, it is uniformly continuous. Now, we will show that it does not satisfy Lipschitz condition.

Take x=0, y ∈ (0, 1]. Note that

$|\dfrac{f(x)-f(y)}{x-y}| = \dfrac{1}{\sqrt{y}}$ →∞ as y →0.

So there does not exist any positive K such that $\dfrac{1}{\sqrt{y}} \leq K$. This means that |f(x)-f(y)| = |√x – √y| ≤ K |x – y| does not hold true. In other words, f(x) = √x on [0, 1] is not Lipschitz.

Thus, the function f(x) = √x on [0, 1] is uniformly continuous, but not Lipschitz.

Related Topics:

• Intermediate Value Theorem
• Fixed Point Theorem

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