Show that the following are equivalent:

Question

If $f$ is a continuous function on a bounded set $S$, show that the following are equivalent:

(a) the function $f$ is uniformly continuous on $S$.

(b) it is possible to extend $f$ to a continuous function on the set $\text{cl}(S)$.

Answer 1

$$a) \Rightarrow b)$$ I assume we are working in normed spaces. The proof is completely the same for metric spaces. Let $f:S \rightarrow F$ where F is complete. The assumption that F be complete is necessary.

A) Construction of $g: cl(S) \rightarrow F $ such that $g|_{S} = f$.

Let $\epsilon >0$. $f$ is uniformly continuous so: $$\exists \delta >0: \forall ||y-x||< \delta \Rightarrow ||f(y)-f(x)||< \epsilon\: \: \text{for} \: \: x,y \in S \:\:\: (1)$$ Take a $u \in cl(S)$ and fix this element. We know that $S$ is dense in $cl(S)$. So there exists a sequence ${({a}_{n})}_{n \in \mathbb{N}}$ in $S$ such that $${a}_{n} \: {\longrightarrow} \: u \:\text{for} \: n \longrightarrow \infty$$ We know by convergence of ${a}_{n}$: $$\exists N \in \mathbb{N}, \forall p,q>N: ||{a}_{p}-{a}_{q}|| < \delta$$ f is continuous so for such $p$ and $q$ we know from (1): $$||f({a}_{p}) - f({a}_{q})||<\epsilon$$ To summarise: $$ \forall \epsilon>0,\exists N \in \mathbb{N}, \forall p,q>N: ||f({a}_{p})-f({a}_{q})|| < \epsilon$$ So ${(f({a}_{n}))}_{n \in \mathbb{N}}$ is a Cauchy sequence in $F$. So this has a limit let's call it: $g(u)$.

We have found for every $u \in cl(S)$ a $g(u)$.

B) We show that $g(u)$ is independent of the sequence ${({a}_{n})}_{n \in \mathbb{N}}$ we have chosen in part A)

Take another sequence ${({b}_{n})}_{n \in \mathbb{N}}$ in $S$ such that $${b}_{n} \: {\longrightarrow} \: u \:\text{for} \: n \longrightarrow \infty$$ By the same argument as before we see that ${(f({b}_{n}))}_{n \in \mathbb{N}}$ is a Cauchy sequence in $F$. We have: $$||g(u)-f({b}_{n})|| \leq ||g(u)-f({a}_{n})|| + ||f({b}_{n})-f({a}_{n})|| $$ for some $n \in \mathbb{N}$. We can find $M>0$ such that $$\forall n \geq M: ||{b}_{n}-u||<\frac{\delta}{2} \text{and} ||{a}_{n}-u||<\frac{\delta}{2}$$ So $$||{b}_{n}-{a}_{n}|| \leq ||{a}_{n}-u||+||{b}_{n}-u||<\delta$$ This implies by (1): $$||f({b}_{n})-f({a}_{n})||<\epsilon$$ We know that ${(f({a}_{n}))}_{n \in \mathbb{N}}$ converges to $g(u)$, so by taking $n>M $ big enough we get: $$||f({b}_{n})-g(u)|| \leq ||f({b}_{n})-f({a}_{n})||+||g(u) -f({a}_{n})||<2 \epsilon$$ So ${(f({b}_{n}))}_{n \in \mathbb{N}}$ converges to $g(u)$. So we can define: $$g: cl(S) \rightarrow F:u \rightarrow g(u)$$

C) g is continuous on cl(S)

We have proven in B) that for every sequence ${({b}_{n})}_{n \in \mathbb{N}}$ in $S$, ${(f({b}_{n}))}_{n \in \mathbb{N}}$ converges to $g(u)$. This shows that g is continuous on $S$. We still have to prove that $g$ is continuous on cl(S).

Take $x \in cl(S)$. We show that $g$ is continuous in $x$.
So take a sequence ${({x}_{n})}_{n \in \mathbb{N}}$ in $cl(S)$ wich converges to $x \in cl(S)$. Cause $S$ is dense in $cl(S)$, we know that for every ${x}_{n}$, there exists an element $y_{n} \in S$ such that $$||y_{n}-x_{n}||<\frac{\delta}{2}$$ And there is a $N \in \mathbb{N}$ such that $$\forall n>N: ||x_{n}-x||<\frac{\delta}{2}$$ From these inequalities one can see that $$\forall n> N: ||x-y_{n}||< \delta$$ From the continuity of $f$: $$\forall n> N: ||f(x)-f(y_{n})||< \epsilon$$ But also: $$\forall n> N: ||f(x_{n})-f(y_{n})||< \epsilon$$

The last step: $$||f(x_{n}) - f(x)|| \leq ||f(x_{n}) - f(y_{n})|| + ||f(x) - f(y_{n})||< 2\epsilon $$ So we have the following: $$(x_{n})_{n\in \mathbb{N}} \: \xrightarrow{n \longrightarrow \infty} \: x \:\Rightarrow \: (f(x_{n}))_{n\in \mathbb{N}} \: \xrightarrow{n \longrightarrow \infty} \: f(x)$$ So $g$ is continuous at $x$, hence $g$ is continuous.

D) The uniqueness

The uniqueness follows from the construction of $g$. $g$ was constructed by limits of sequences. These limits are unique so $g$ is unique to.