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 Continuous Extension Theorem

Recall that from The Uniform Continuity Theorem that if a function $I = [a, b]$ is a closed and bounded interval and $f : I \to \mathbb{R}$ is continuous on $I$, then $f$ must also be uniformly continuous on $I$. The succeeding theorem will help us determine when a function $f$ is uniformly continuous when $I$ is instead a bounded open interval.

Before we look at The Continuous Extension Theorem though, we will need to prove the following lemma.

Lemma 1: If $f : A \to \mathbb{R}$ is a uniformly continuous function and if $(x_n)$ is a Cauchy Sequence from $A$, then $(f(x_n))$ is a Cauchy sequence from $\mathbb{R}$.

• Proof: Let $f : A \to \mathbb{R}$ be a uniformly continuous function and let $(x_n)$ be a Cauchy sequence from $A$. We want to show that $(f(x_n))$ is also a Cauchy sequence. Recall that to show that $(f(x_n))$ is a Cauchy sequence we must show that $\forall \epsilon > 0$ then $\exists N \in \mathbb{N}$ such that $\forall m, n \in \mathbb{N}$, if $m, n ≥ N$ then $\mid f(x_n) - f(x_m) \mid < \epsilon$.

• Since $f$ is uniformly continuous on $A$, then for any $\epsilon > 0$, $\exists \delta_{\epsilon} . 0$ such that for all $x, y \in A$ where $\mid x - y \mid < \delta_{\epsilon}$ we have that $\mid f(x) - f(y) \mid < \epsilon$.

• Now for $\delta_{\epsilon} > 0$, since $(x_n)$ is a Cauchy sequence then $\exists N_{\delta_{\epsilon}} \in \mathbb{N}$ such that $\forall m, n ≥ N_{\delta_{\epsilon}}$ we have that $\mid x_n - x_M \mid < \delta_{\epsilon}$. So this $N_{\delta_{\epsilon}}$ will do for the sequence $(f(x_n))$. So for all $n, m ≥ N_{\delta_{\epsilon}}$ we have that $\mid x_n - x_m \mid < \delta_{\epsilon}$ and from the continuity of $f$ this implies that $\mid f(x_n) - f(x_m) \mid < \epsilon$ and so $(f(x_n))$ is a Cauchy sequence. $\blacksquare$

We are now ready to look at The Continuous Extension Theorem.

Theorem 1 (The Continuous Extension Theorem): If $I = (a,b)$ is an interval, then $f : I \to \mathbb{R}$ is a uniformly continuous function on $I$ if and only if $f$ can be defined at the endpoints $a$ and $b$ such that $f$ is continuous on $[a, b]$.

• Proof: $\Rightarrow$ Suppose that $f$ is uniformly continuous on $I = (a, b)$. Let $(x_n)$ be a sequence in $(a,b)$ that converges to $a$. Then since $(x_n)$ is a convergent sequence, it must also be a Cauchy sequence. By lemma 1, since $(x_n)$ is a Cauchy sequence then $(f(x_n))$ is also a Cauchy sequence, and so $(f(x_n))$ must converge in $\mathbb{R}$, that is $\lim_{n \to \infty} f(x_n) = L$ for some $L \in \mathbb{R}$.

• Now suppose that $(y_n)$ is another sequence in $(a, b)$ that converges to $a$. Then $\lim_{n \to \infty} (x_n - y_n) = a - a = 0$, and so by the uniform continuity of $f$:

(1)

\begin{align} \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) + f(x_n)] \\ \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) ] + \lim_{n \to \infty} f(x_n) \\ \lim_{n \to \infty} f(y_n) = 0 + L = L \end{align}

• So for every sequence $(y_n)$ in $(a, b)$ that converges to $a$, we have that $(f(y_n))$ converges to $L$. Therefore by the Sequential Criterion for Limits, we have that $f$ has the limit $L$ at the point $a$. Therefore, define $f(a) = L$ and so $f$ is continuous at $a$. We use the same argument for the endpoint $b$, and so $f$ is can be extended so that $f$ is continuous on $[a, b]$.

• $\Leftarrow$ Suppose that $f$ is continuous on $[a, b]$. By The Uniform Continuity Theorem, since $[a, b]$ is a closed and bounded interval then $f$ is uniformly continuous. $\blacksquare$