0
$\begingroup$

Let $-\infty\leq a<b\leq \infty$. Prove that a function $f$ is uniformly continuous on the interval $(a,b)$ if and only if it has a uniformly continuous extension to all of $\mathbb{R}$.

Definition: A function $f: D \longrightarrow \mathbb{R}$ is uniformly continuous on $D$ if for every $\epsilon > 0$ there exists some $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x-y|<\delta$.

My attempt:

($\longrightarrow$) Suppose $f$ has a uniformly continuous extension to all of $\mathbb{R}$. In other words, there exists a uniformly continuous function $g:\mathbb{R}\longrightarrow \mathbb{R}$ such that $g(x) = f(x)$ for all $x\in(a,b)$. But since $g$ is uniformly continuous on all of $\mathbb{R}$, then of course it is uniformly continuous on $(a,b)$, which means $f$ is uniformly continuous on $(a,b)$.

Does that work? and how would you show the other direction?

$\endgroup$
3
$\begingroup$

You have shown the reverse direction.

($\Leftarrow$) Clearly if $f$ has a uniformly continuous extension to all of $\mathbb{R}$, then it is uniformly continuous on the restriction to $(a,b)$.

($\Rightarrow$) Let $f$ be uniformly continuous on $I=(a,b)$. If $f$ has a uniformly continuous extension to $\overline{I}=[a,b]$, then we can further extend $f$ by defining $f(x) =f(a)$ for $x < a$ and $f(x) = f(b)$ for $x>b$. It is clear that this function is uniformly continuous on $\mathbb{R}$.

To show that $f$ has a uniformly continuous extension to the closure of $I$, choose a sequence $x_i \in I$ such that $x_i \to a$. Let $\epsilon>0$, and pick $\delta>0$ such that if $|x-y|< \delta$, then $|f(x)-f(y)| < \epsilon$. The sequence $x_i$ is Cauchy, hence there exists $N$ such that if $n,m > N$, we have $|x_n-x_m| < \delta$. Then we have $|f(x_n)-f(x_m)| < \epsilon$. Hence $f(x_i)$ is Cauchy and converges to some number $\phi$. To show that the limit does not depend on the sequence, suppose we have another sequence $x_i' \to a$. Since $x_i-x_i' \to 0$, uniform continuity shows that $f(x_i)-f(x_i') \to 0$, and hence $f(x_i') \to \phi$ also. So we can unambiguously define $f(a) = \phi$.

In a similar manner, we can extend $f$ to $b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.