The function f(x) = x^{2} defined on (0, 1) is uniformly continuous, but it is not on (0, ∞). This will be proved in this article.

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## x^{2} is Uniformly Continuous on (0, 1)

Let us now prove that the function f(x) = x^{2} is uniformly continuous on (0, 1) using the definition of uniform continuity.

### Question: Prove that x^{2} is uniformly continuous on (0, 1).

**Proof:**

We will prove that f(x) = x^{2} is uniformly continuous on (0, 1) using the definition. Let ε>0 be given. Now for x, y ∈ (0, 1) with |x-y| < δ we have that

|f(x) – f(y)| = |x^{2} – y^{2}| = |(x-y) (x+y)|

⇒ |f(x) – f(y)| < 2 |x-y| as 0≤ x, y ≤1.

So choose δ = ε/2.

Then for ε>0 given, we can find a positive δ = ε/2 such that

|f(x) – f(y)| < ε.

This proves that f(x) = x^{2} is uniformly continuous on (0, 1).

**Also Read:** log(x) is uniformly continuous

Continuous but not Uniformly Continuous: An Example

## Prove that x^{2} is NOT Uniformly Continuous on (0, ∞)

For a contradiction, assume that f(x) = x^{2} is uniformly continuous on (0, ∞). So for ε>0, ∃ a positive δ such that

|f(x) – f(y)| < ε whenever |x-y| < δ.

That is, for |x-y| < δ we have that

|x^{2} – y^{2}| < ε

⇒ |(x-y) (x+y)| < ε …(∗)

Take ε = 1.

For any δ>0, we consider y = $\dfrac{1}{\delta}$ and x = δ+y. Then

|(x-y) (x+y)| = $\delta \left( \delta + \dfrac{2}{\delta} \right)$ = δ^{2}+2 > 1 = ε

which contradicts (∗). Hence our assumption was wrong. This means that x^{2} is not uniformly continuous on (0, ∞).

**More Reading:**

- Intermediate Value Theorem
- Fixed Point Theorem
- Uniformly Continuous Function but not Lipschitz: An Example

## FAQ

**Q1: Prove that f(x) = x**

^{2}is uniformly continuous on (0, 1).Answer: As [0, 1] is a closed and bounded interval, f(x) = x^{2} is uniformly continuous on [0, 1], so is on (0, 1).

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.