Uniform Continuity of Functions: Proving Continuity on [0, 1] and Non-Continuity on R

Understanding uniform continuity is crucial in real analysis. In this article, we will explore how to prove the uniform continuity of the function f(x) x^2 on the interval [0, 1] and its non-uniform continuity on the real line R. We will demonstrate these concepts using rigorous mathematical proofs and detailed steps.

Proving Uniform Continuity on [0, 1]

First, let's start by proving that the function f(x) x^2 is uniformly continuous on the interval [0, 1].

Step 1: Basic Definition

By definition, a function f is uniformly continuous on an interval if for every ε > 0, there exists a δ > 0 such that for all x, y in the interval, if , then |f(x) - f(y)| ε.

Step 2: Calculations

Let's calculate |f(x) - f(y)| for our function:

Compute |f(x) - f(y)|:

|f(x) - f(y)| |x^2 - y^2| |x - y||x y|

Given that x, y ∈ [0, 1], we know:

x y ≤ 2

Thus, |f(x) - f(y)| ≤ 2|x - y|

to ensure that |f(x) - f(y)| ε, we can choose δ ε/2.

Then, if |x - y| δ, we have:

|f(x) - f(y)| 2|δ| ε.

This proves that f(x) x^2 is uniformly continuous on [0, 1].

Proving Non-Uniform Continuity on R

Now, we will show that the function f(x) x^2 is not uniformly continuous on the real line R.

Step 1: Counterexample

To show that the function is not uniformly continuous, we need to find a counterexample.

Step 2: Calculations

Consider the points x_n n and y_n n - 1/n for n ∈ N.

Calculate |f(x_n) - f(y_n)|:

We have:

f(x_n) n^2 and f(y_n) (n - 1/n)^2 n^2 - 2n(1/n) 1/n^2 n^2 - 2 1/n^2

So, |f(x_n) - f(y_n)| |n^2 - (n^2 - 2 1/n^2)| |2 - 1/n^2| 2 - 1/n^2 (since n 1)

Calculate |x_n - y_n|:

|x_n - y_n| |n - (n - 1/n)| 1/n

Analyze the limits:

As n → ∞, |x_n - y_n| → 0 and |f(x_n) - f(y_n)| → 2. Hence, we can make |x_n - y_n| arbitrarily small, while |f(x_n) - f(y_n)| remains bounded away from 0 specifically approaching 2.

Further Insights and Theorems

The set [0, 1] is a compact subset of R in the topology generated by the metric d(x, y) |x - y|. Therefore, by the Heine-Cantor theorem, a continuous function from a compact metric space into a metric space is uniformly continuous. However, the set R is not compact in this topology, and we can use this fact to prove that the function f(x) x^2 is not uniformly continuous on R.

Furthermore, the non-uniform continuity on R can be proven by showing that for any ε 0 and any δ 0, there exist points x and y such that |x - y| δ but |f(x) - f(y)| ≥ ε. We can choose a real number x 1 such that x - δ ε. Then, the points x and y x - 0.9δ satisfy |y - x| 0.9δ δ and |f(y) - f(x)| 1.8δ - 0.81δ^2 ε since x 1.

Thus, we have shown that f(x) x^2 is uniformly continuous on [0, 1] but not uniformly continuous on R.