📂Analysis

Uniform Continuity of Functions

Definition¹

Let us assume $E \subset \mathbb{R}$ is a non-empty set and define $f : E \to \mathbb{R}$. If for every $\varepsilon > 0$,

$$| x_{1} - x_{2} | < \delta \land x_{1} , x_{2} \in E \implies | f(x_{1}) - f(x_{2}) | < \varepsilon$$

there exists a $\delta>0$ satisfying the above equation, then $f$ is said to be uniformly continuous on $E$.

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• $\land$ represents the logical ‘and’ symbol, known as the logical conjunction.

Explanation

The concept of continuity of a function itself is about a point as in $a \in E$, while uniform continuity considers the entire set $E$. For example, consider the continuous function $f (x) := x^2$.

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Let us assume $E \subset \mathbb{R}$ is a non-empty set and define $f : E \to \mathbb{R}$.

(a) Compact Metric Space: If $f$ is continuous and $E$ is a bounded closed interval, then $f$ is uniformly continuous.

(b) Preservation of Convergence: If $f$ is uniformly continuous and $\left\{ x_{n} \right\}_{n=1}^{\infty}$ is Cauchy, then $\left\{ f(x_{n}) \right\}$ is also Cauchy.

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(1) $E = (0,1)$

If we take $\delta = \dfrac{\varepsilon}{2}$ to be,

then for all $x_{1} , x_{2} \in (0,1)$, when we say $| x_{1} - x_{2} | < \delta$,

$$\begin{align*} | f(x_{1}) - f(x_{2}) | =& | x_{1}^{2} - x_{2}^{2} | \\ =& | x_{1} - x_{2} | | x_{1} + x_{2} | \\ \le & 2 | x_{1} - x_{2} | \\ & < & 2 \delta \\ =& \varepsilon \end{align*}$$

according to the definition, $f$ is uniformly continuous on $E = ( 0 , 1 )$.

(2) $E = \mathbb{R}$

Let’s assume $f$ is uniformly continuous on $E$. Even when $\varepsilon = 1$ is given,

$$| x_{1} - x_{2} | < \delta \land x_{1} , x_{2} \in E \implies | f(x_{1}) - f(x_{2}) | < 1$$

a $\delta$ must exist. However, according to the Archimedean principle, we can choose a $n \in \mathbb{N}$ satisfying $n \delta > 1$. Then, for $x_{1} = n$, $x_{2} = \left( n + \dfrac{ \delta }{2} \right)$,

$$\begin{align*} 1 & < & n \delta \\ <& n \delta + {{ \delta^{2} } \over { 4 }} \\ =& \left| n^2 - \left( n + {{ \delta } \over {2}} \right)^2 \right| \\ =& \left| f( n ) - f \left( n + {{ \delta } \over {2}} \right) \right| \\ =& | f (x_{1} ) - f ( x_{2} ) | \\ <& 1 \end{align*}$$

This leads to a contradiction, given $1 < 1$. Hence, $f$ is not uniformly continuous on $E = \mathbb{R}$.

Considering $g(x) = x$, no matter what domain $E$ we consider, by taking $\delta = \varepsilon$, it satisfies the conditions for uniform continuity. From such examples, it’s intuitive to think that uniformly continuous functions are a type of ‘gentle’ function. It makes sense that $g$ will diverge as $x$ approaches infinity. However, unlike $f(x) = x^2$, it does not grow violently but maintains a certain line. It’s a natural principle that gentle things are easier to handle than violent ones, and it also makes sense that uniformly continuous functions have better conditions than mere continuous functions.

Let’s think about the case in (b) where only continuity is assumed without the assumption of uniform continuity.

$\displaystyle h(x) := {{1} \over {x}}$ is a continuous function, and if we take $\displaystyle x_{n} := {{1} \over {n}}$, then $\left\{ x_{n} \right\}$ is a Cauchy sequence converging to $0$. However, since $\displaystyle h (x_{n} ) = {{1} \over { {{1} \over {n}} }} = n$, it is understood that $\left\{ h ( x_{n} ) \right\}$ is not a Cauchy sequence.