*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# Uniform Continuity

Definition: Let $f : A \to \mathbb{R}$ be a function. We say that $f$ is uniformly continuous on the domain $A$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that if $x, y \in A$ and we have that $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$. |

By the definition of uniform continuity, a function $f$ is uniformly continuous if we are given any $\epsilon > 0$, then we can find a $\delta > 0$ so much so that for any $x, y \in A$, if we have that the distance between $x$ and $y$ is less than $\delta$ then we will have that the distance between $f(x)$ and $f(y)$ will be less than $\epsilon$. We should note that this should work for all $x$ and $y$ that are within $\delta$ from each other. It should be rather obvious, but if a function $f : A \to \mathbb{R}$ is uniformly continuous on $A$, then $f$ must also be continuous on $A$. The following theorem summarizes this nicely.

Theorem 1: If a function $f : A \to \mathbb{R}$ is uniformly continuous on $A$ then $f$ is also continuous on $A$. |

**Proof:**Suppose that $f$ is uniformly continuous. Then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x, y \in A$ satisfy $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$. We want to show that $f$ is continuous on $A$.

- Let $c \in A$. Since $f$ is uniformly continuous, then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ satisfies $\mid x -c \mid < \delta$ then $\mid f(x) - f(c) \mid < \epsilon$. Therefore $f$ is continuous at $c \in A$, and since $c$ is arbitrary, we conclude that $f$ is continuous on all of $A$. $\blacksquare$

We will now look at some examples of functions that are uniformly continuous.

First, consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = c$ where $c \in \mathbb{R}$. Clearly $f$ is a continuous function, and it is not hard to see that $f$ is also uniformly continuous. For $\epsilon > 0$, any $\delta > 0$ will work, since if $x, y \in \mathbb{R}$ then $\mid f(x) - f(y) \mid = \mid c - c \mid = 0 < \epsilon$ regardless of $\delta$. The following graph represents the function $f(x) = \frac{5}{2}$.

Another example of a function that is uniformly continuous is $f : [0, \infty ) \to \mathbb{R}$ defined by $f(x) = x$. To prove this, let $\epsilon > 0$ be given. We want to find $\delta > 0$ such that if $x, y \in [0, \infty)$ satisfy $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$. First look at $\mid f(x) - f(y) \mid = \mid x - y \mid$. If we let $\delta (\epsilon ) = \epsilon$, then if $\mid x - y \mid < \delta$ we have that $\mid x - y \mid = \mid f(x) - f(y) \mid < \delta = \epsilon$. The following graph represents the graph of $f$:

A better explanation to what exactly uniform continuity is can be described with a counter example of a function that is NOT uniformly continuous. Consider the function $f :(0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$, and let $\epsilon > 0$. Suppose that we claim that $\delta > 0$ exists. For some values of $x$ and $y$ that are within $\delta$ from each other, $f(x)$ and $f(y)$ are within $\epsilon$ from each other:

However, now consider the following $x$ and $y$. They are within $\delta$ of each other, but $f(x)$ and $f(y)$ are not within $\epsilon$ of each other. Therefore this $\delta$ does not work.

It shouldn't be too hard to see that for this $\epsilon > 0$ there does not exist any $\delta > 0$ such that if $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$. For example, if we take values of $x$ and $y$ that are arbitrarily close to zero, then we will have to make $\delta$ smaller and smaller since the $f(x)$ and $f(y)$ will be very large, and we need $f(x)$ and $f(y)$ to be within $\epsilon$ of each other. Therefore $f(x) = \frac{1}{x}$ is not uniformly continuous.