The Continuous Extension 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 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. $\blacksquare$