The Uniform Continuity Theorem

*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.*

# The Uniform Continuity Theorem

Recall that a function $f : A \to \mathbb{R}$ is said to be Uniformly Continuous on $A$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x, y \in A$ satisfy $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$.

In the special case where $f : I \to \mathbb{R}$ is a continuous function, and $I = [a, b]$ is a closed and bounded interval, then in fact we can guarantee that $f$ is also uniformly continuous. The following theorem known as The Uniform Continuity Theorem summarizes this important result.

Theorem 1 (The Uniform Continuity Theorem): Let $f : I \to \mathbb{R}$ be a continuous function and let $I = [a, b]$ be a closed and bounded interval. Then $f$ is uniformly continuous on $I$. |

**Proof:**Let $f : I \to \mathbb{R}$ be a continuous function from the closed and bounded interval $I = [a, b]$ into the set of real numbers. We will carry through this proof by contradiction. Suppose that $f$ is NOT uniformly continuous on $I$. Then $\exists \epsilon_0 > 0$ such that $\forall \delta > 0$ we have that $\exists x_{\delta}, y_{\delta} \in I$ where $\mid x_{\delta} - y_{\delta} \mid < \delta$ however $\mid f(x_{\delta}) - f(y_{\delta}) \mid ≥ \epsilon_0$.

- Let $\delta_n = \frac{1}{n}$ for all $n \in \mathbb{N}$, and denote $x_{\delta_n} = x_n$ and $y_{\delta_n} = y_n$. We note that $\forall n \in \mathbb{N}$ that $x_n, y_n \in I = [a, b]$. Since $I$ is a bounded interval, it follows that the sequences $(x_n)$ and $(y_n)$ are bounded sequences. Therefore by The Bolzano-Weierstrass Theorem, there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $(x_{n_k})$ converges, that is $\lim_{k \to \infty} x_{n_k} = x$ for some $x \in I$.

- Furthermore we have that $\mid x_{n_k} - y_{n_k} \mid < \frac{1}{n_k}$, or equivalently $x_{n_k} - \frac{1}{n_k} < y_{n_k} < x_{n_k} + \frac{1}{n_k}$. Since $\lim_{k \to \infty} x_{n_k} - \frac{1}{n_k} = x$ and $\lim_{k \to \infty} x_{n_k} + \frac{1}{n_k} = x$, then by The Squeeze Theorem we have that $\lim_{k \to \infty} y_{n_k} = x$.

- Now since $f$ is a continuous function on $I = [a, b]$ then by Sequential Criterion for the Continuity of a Function, we have that $\lim_{k \to \infty} f(x_{n_k}) = f(x) = \lim_{k \to \infty} f(y_{n_k})$. Therefore $\lim_{k \to \infty} f(x_{n_k}) - f(y_{n_k}) = 0$. But for $\epsilon_0$ and $\delta_{n_k} = \frac{1}{n_k}$ we have that $\mid x_{n_k} - y_{n_k} \mid < \frac{1}{n_k}$ however $\mid f(x_{n_k}) - f(y_{n_k}) \mid ≥ \epsilon_0 > 0$, which contradictions the fact that $\lim_{k \to \infty} f(x_{n_k}) - f(y_{n_k}) = 0$ since $\forall \epsilon > 0$ $\exists K_{\epsilon} \in \mathbb{N}$ such that $\forall k ≥ K_{\epsilon}$ we have that $\mid f(x_{n_k}) - f(y_{n_k}) \mid < \epsilon$, and so $\epsilon_0$ would otherwise contradict this.

- Therefore our original assumption that $f$ was not uniformly continuous on $I$ was false, and so $f$ must be uniformly continuous on $I$. $\blacksquare$