A Theorem regarding Uniform Continuity states :
A function $f(x)$ is uniformly continuous on $(a,b)$ $\iff$ It can be defined on the endpoints $a$, $b$ such that $f$ is continuous on the $[a,b]$.
Now I have a doubt regarding this ,especially regarding the use of ''$\iff$ '' here.
Suppose we have the function $f:\Bbb R\to[-1,1]$ $$f(x)=sgn(x)=\begin{cases} 1,&x>0 \\0,&x=0\\-1,&x<0\end{cases}$$
So here I'm choosing a open interval $A$ from it's domain, which is $(0,1)$. So the endpoints of $A$ are $0$ and $1$ , (i.e. $\bar A=[0,1])$ .
Here f(x) is uniformly continuous on $(0,1)$ . But at the endpoint 0 is it really continuous ? We can see the $sgn(x)$ function is continuous everywhere except at $x=0$.So instead of not being continuous everywhere in $[0,1]$ , $f(x)$ is uniformly continuous on$(0,1)$.
So now my question is that how can the statement__ $\text{if $f$ can be defined on the endpoints a,b such that $f$ is continuous on the [a,b] } \implies $$\text{ $f(x)$ is uniformly continuous on (a,b)} $
?
Thank You.
N.B: First of all , all kinds of responses are welcome . Secondly I don't know whether this question has asked before in this platform of not but I have no intention to copy or duplicate any one's question .
Best Answer
Another way to read the theorem is:
$f:(a, b) \to \Bbb R$ is uniformly continuous $\iff \exists \tilde f:[a, b] \to \Bbb R$ such that $\tilde f$ is continuous and $\forall x \in (a, b) f(x)=\tilde f(x)$.
In other words, $\tilde f$ extends $f$ to a function that's continuous on $[a, b]$.
The function you're defining isn't limited to being continuously defined on $(0,t)$, so the theorem has nothing to say about it. The theorem does speak about restrictions of your function to non-negative domains.