Continuous Extension Theorem
Suppose that \(a < b\) and that \(f: (a,b) \to \mathbb{R}\). Then is uniformly continuous on \((a,b)\) if and only if \(f\) can be continuously extended to \([a,b]\); that is, if and only if there is a continuous function \(g:[a,b] \to \mathbb{R}\) which satisfies $$ \begin{align*} f(x) = g(x), \quad \quad x \in (a,b) \end{align*} $$

Examples

Notes: In other words, we can extend \(f\) to the end points without breaking continuity. So if we take \(f(x)=\sqrt{x}\) where \(f\) is uniformly continuous on \((0,1)\), then it’s easy to see that we can extend \(f\) and define \(g(0) = 0\), \(g(1)=1\) and also let \(g(x)=f(x)\) for all \(x \in (a,b)\). Then \(g(x)\) is uniformly continuous on \([0,1]\). This is not true for \(f(x)=1/x\).

Proof

Let \(\{x_n\}\) be a Cauchy sequence in \((a,b)\) (Since the interval is bounded, such a sequence always exists). Since \(f\) is uniformly continuously on \((a,b)\), then by the lemma, we know that \(f(x_n)\) is also Cauchy. Hence, there exists some \(L \in \mathbb{R}\) such that

$$ \begin{align*} \lim_{n \to \infty} f(x_n) = L \end{align*} $$

Define

$$ \begin{align*} g(x) = \begin{cases} f(x), & x \in (a,b), \\[6pt] L, & x = b \end{cases} \end{align*} $$

We claim that \(g\) is continuous at \(x = b\). By the Sequential Characterization of Continuity, we need to show that for all sequences \(y_n\) such that \(y_n \rightarrow b\), we must have

$$ \begin{align*} g(y_n) \rightarrow g(b) = L \end{align*} $$

By the lemma, we know that \(f(y_n)\) is Cauchy. Then by the triangle inequality we know

$$ \begin{align*} |f(y_n) - L| \leq |f(y_n) - f(x_n)| + |f(x_n) - L| \quad \quad (1) \end{align*} $$

We know we can bound the second term \(|f(x_n) - L|\) since \(f(x_n)\) is Cauchy and hence convergent. Hence, for all \(\epsilon >\), there exists an \(N_1 \in \mathbb{N}\) such that for all \(n > N_1\)

$$ \begin{align*} |f(x_n) - L| < \frac{\epsilon}{2} \end{align*} $$

For the first term, we know that \(f\) is uniformly continuous on \((a,b)\). So there exists a \(\delta > 0\) (that only depends on \((a,b)\) and works for all points) such that for all \(x,y \in (a,b)\)

$$ \begin{align*} |x - y| < \delta \quad \text{ implies } |f(x) - f(y)| < \frac{\epsilon}{2} \quad \quad (2) \end{align*} $$

But now we know that \(x_n \to b\) and we know that \(y_n \to b\). Then as \(n \to \infty\)

$$ \begin{align*} |x_n - y_n| \to 0 \end{align*} $$

That is, there exists some \(N_2 \in \mathbb{N}\) such that whenever \(n \geq N_2\)

$$ \begin{align*} |x_n - y_n| < \delta \end{align*} $$

But then this implies by (2) that

$$ \begin{align*} |f(x_n) - f(y_n)| < \frac{\epsilon}{2} \end{align*} $$

Hence, (1) becomes

$$ \begin{align*} |f(y_n) - L| \leq \epsilon \end{align*} $$

Therefore, \(f(y_n) \to L\). But we defined \(g(y_n) = f(y_n)\) for all \(y_n \in (a,b)\) and also defined \(g(b) = L\). Thus

$$ \begin{align*} \lim\limits_{n \to \infty} g(y_n) = \lim\limits_{n \to \infty} f(y_n) = L = g(b) \end{align*} $$

Thus, \(g\) is continuous at \(b\). A similar argument shows that \(g\) is also continuous at \(a\). This means that \(g\) is continuous on a closed and bounded interval \([a,b]\). Therefore, \(g\) must also be uniformly continuous at \([a,b]\) by Theorem we proved in lecture. \(\ \blacksquare\)

References