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 <a class="newpage" href="/cauchy-sequences">Cauchy Sequence</a> 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 $\forall \varepsilon >0$ then $\exists N\in \mathbb {N}$ such that $\forall m,n\in \mathbb {N}$, if $m,n\geq N$ then $\mid f(x_{n})-f(x_{m})\mid <\varepsilon$.

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

• Now for $\delta _{\varepsilon }>0$, since $(x_{n})$ is a Cauchy sequence then $\exists N_{\delta _{\varepsilon }}\in \mathbb {N}$ such that $\forall m,n\geq N_{\delta _{\varepsilon }}$ we have that $\mid x_{n}-x_{M}\mid <\delta _{\varepsilon }$. So this $N_{\delta _{\varepsilon }}$ will do for the sequence $(f(x_{n}))$. So for all $n,m\geq N_{\delta _{\varepsilon }}$ we have that $\mid x_{n}-x_{m}\mid <\delta _{\varepsilon }$ and from the continuity of $f$ this implies that $\mid f(x_{n})-f(x_{m})\mid <\varepsilon$ 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) ${\begin{aligned}\lim _{n\to \infty }f(y_{n})=\lim _{n\to \infty }[f(y_{n})-f(x_{n})+f(x_{n})]\\\lim _{n\to \infty }f(y_{n})=\lim _{n\to \infty }[f(y_{n})-f(x_{n})]+\lim _{n\to \infty }f(x_{n})\\\lim _{n\to \infty }f(y_{n})=0+L=L\end{aligned}}$

• 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. $\blacksquare$

Resources

• The Continuous Extension Theorem, mathonline.wikidot.com