The main purpose of this post is to introduce the concept of uniform continuity. A function satisfies the uniform continuity if the values of the function are close to each other for all elements sufficiently close in its domain. More precisely, a function $$f : D \subseteq \mathbb{R} \rightarrow \mathbb{R}$$ satisfies the uniform continuity if for any positive real number \(\epsilon\) there is a positive real number \(\delta\) such that for all \(x,y \in D\), the assumption \(\vert x – y \vert < \delta\) implies \(\vert f(x) – f(y) \vert < \epsilon.\)

A function is uniformly continuous if it satisfies the uniform continuity. Though each uniformly continuous function is continuous at each point of its domain, some continuous functions may not be uniformly continuous. On the other hand, some continuous functions with certain properties, for example, continuous functions over closed intervals, are uniformly continuous.

#### Hölder condition implies uniform continuity

By definition, a function \(f: D \rightarrow \mathbb{R}\) satisfies \(\alpha\)-Hölder condition if for all \(x,y \in D\), we have $$\vert f(x) – f(y) \vert \leq M \vert x – y \vert ^{\alpha}$$ for some nonnegative real number \(M\).

**Result**. Hölder condition implies uniform continuity. In other words, if a function \(f: D \rightarrow \mathbb{R}\) satisfies \(\alpha\)-Hölder condition, then it is uniformly continuous.

The proof is as follows:

Let \(\epsilon\) be a positive real number. Set \(\delta = (\epsilon/M)^{1/\alpha}\) and observe that if \(\vert x – y \vert < \delta \), then we have $$\vert f(x) – f(y) \vert \leq $$ $$M \vert x – y \vert ^{\alpha} < $$ $$ M \delta^{\alpha} = \epsilon. $$ This means that \(f\) is uniformly continuous and the proof is complete.

**Exercise**. Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be a differentiable function such that its derivative is bounded, i.e. for all \(x \in I\), we have $$\vert f'(x) \vert \leq M$$ for some non-negative real number \(M\). Prove that \(f\) is uniformly continuous.

**Solution**. Let \(a < b\) be arbitrary elements in the interval \(I\). Therefore, \([a,b] \subseteq I\). By the mean value theorem, there is a \(c \in [a,b]\) such that $$f(b) – f(a) = f'(c) (b-a).$$ This implies that $$ \vert f(b) – f(a) \vert \leq M \vert b – a \vert.$$ This means that \(f\) satisfies the Hölder condition, and so, is uniformly continuous.

**Exercise**. Let \(I\) be an open interval and \(f: I \rightarrow \mathbb{R}\) satisfy the \(\alpha\)-Hölder condition where \(\alpha > 1\). Prove that \(f\) is a constant function.

#### Some continuous functions are not uniformly continuous

**Example**. The function \(f(x) = 1/x\) is continuous though not uniformly continuous over the infinite interval \((0,+\infty)\).

**Solution**. By the laws explained in the post on limit of a function, $$\lim_{x \to a} 1/x = 1/a$$ for each \(a \in (0,+\infty)\), and so, \(f\) is continuous over the infinite interval \((0,+\infty)\). However, our claim is that \(f\) is not uniformly continuous over \((0,+\infty)\) and the proof of our claim is as follows:

Set \(\epsilon = 1\), \(\delta > 0\), \(y = \min\{\delta, 1\}\), and \(x = y/2\). Observe that while $$\vert x – y \vert = y/2 \leq \delta / 2 < \delta,$$ we have $$\vert 1/x – 1/y \vert = 1/y \geq 1 = \epsilon.$$

We showed in above that continuous functions are not necessarily uniformly continuous. However, we have the following result though we do not prove it here.

**Result**. Each continuous function over a closed interval \([a,b]\) is uniformly continuous.