Let $E \subset F \subset X$ and $f:E\rightarrow Y$. We say that the function $g:F\rightarrow Y$ is an extension of $f$ if $g(x) = f(x)$ for all $x \in E$.
Let $f: (a, b) \rightarrow \mathbb{R}$. Show that $f$ has a continuous extension to $[a, b]$ if and only if $f$ is uniformly continuous on $(a, b)$.
$(X,d)$ is a metric space.
Hint: To prove $\Leftarrow$ start by showing that $f$ maps Cauchy sequences to Cauchy sequences. Then how should you define $g(a)$ and $g(b)$?
The $\Rightarrow$ part of the proof was quite simple, but even with the hint, the other way seems a bit difficult. Doesn't the first part of the "hint" follow from the way we defined $f$? If so, how do I prove this? Help/Hints appreciated!
Best Answer
A continuous function doesn’t necessarily map Cauchy sequences to Cauchy sequences; consider, for instance, the function $f(x)=\frac1x$ on $(0,1)$, which maps the Cauchy sequence $\langle 2^{-n}:n\in\Bbb Z^+\rangle$ to the very non-Cauchy sequence $\langle 2^n:n\in\Bbb Z^+\rangle$. But uniformly continuous functions do map Cauchy sequences to Cauchy sequences. To prove this, suppose that $\langle x_n:n\in\Bbb N\rangle$ in $(a,b)$ is Cauchy, and let $\epsilon>0$. There is a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $x,y\in(a,b)$ and $|x-y|<\delta$, and there is an $m\in\Bbb N$ such that $|x_k-x_n|<\delta$ whenever $k,n\ge m$, so ... ?
Now let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $(a,b)$ converging to $a$. Clearly $\sigma$ is Cauchy, so by the hint $\langle f(x_n):n\in\Bbb N\rangle$ is Cauchy. $\Bbb R$ is complete, so $\langle f(x_n):n\in\Bbb N\rangle$ converges to some $c\in\Bbb R$; this $c$ is the natural candidate for $g(a)$. In similar fashion you can find a natural candidate $d$ for $g(b)$, and all that remains is to show that the function
$$g:[a,b]\to\Bbb R:x\mapsto\begin{cases} c,&\text{if }x=a\\ f(x),&\text{if }x\in(a,b)\\ d,&\text{if }x=b \end{cases}$$
is continuous at $a$ and $b$.