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Suppose I want to check if $f(x)$ is uniform continuous on a bounded interval $I$ (for eg open interval $(0,1)$), given that it is continuous on $I$. How do I do that?

My approach: Take $bar{I}$, then two case can happen:

Case I: If I can continuously extend the function, then $f(x)$ is uniformly continuous on $I$.

Case II: If I cannot extend the function continuously, then two sub cases are possible

Subcase II a: $f(x)$ is tends to an infinite limit i.e. it shoots up/down arbitrarily for eg functions like $frac{1}{x}$. In which case I conclude that $f$ is not uniformly continuous on $I$.

Subcase II b: $f(x)$ doesn't have a limit i.e. function of the type sin$frac{1}{x}$. In this case as well $f(x)$ is not uniform continuous on $I$.

So is my above classification of continuous function sufficient to determine which functions are uniform continuous and which are not? So far it had worked well for me.

Yes, that is correct. In fact, assuming that the domain of $f$ is $(a,b)$:

1. If both limits $lim_{xto a^+}f(x)$ and $lim_{xto b^-}f(x)$ exist, then $f$ is uniformly continuous, because you can define$$begin{array}{rccc}Fcolon&[a,b]&longrightarrow&mathbb R\&x&mapsto&begin{cases}lim_{xto a^+}f(x)&text{ if }x=a\f(x)&text{ if }xin(a,b)\lim_{xto b^-}f(x)&text{ if }x=b.end{cases}end{array}$$Then $F$ is continuous and, since its domain is a closed and bounded interval, $F$ is unifomly continuous. In particular, $f$ is uniformly continuous.


2. If the limit $lim_{xto a^+}f(x)$ doesn't exist, then $f$ cannot be uniformly continuous because then either $lim_{xto a^+}bigllvert f(x)bigrrvert=+infty$ or there will two real numbers $m$ and $M$, with $m<M$, such that the inequalities $f(x)>M$ and $f(x)<m$ will occur for values of $x$ arbitrarily close to $a$. It is easy to prove that each possibility is incompatible with the fact that $f$ is uniformly continuous.


3. The case in which the limit $lim_{xto b^-}f(x)$ doesn't exist is similar.

Let $f:(0,1)rightarrow Bbb R$ be uniformly continuous and consider a sequence ${x_n}$ converging to $0$ , so that ${x_n}$ is cauchy. Now a uniformly continuous function sends Cauchy sequence to Cauchy sequence i.e. ${f(x_n)}$ is also cauchy , hence ${f(x_n)}$ converges to some limit $lin Bbb R$. Again using uniform continuity you can show this limit is independent of choice of sequence converging to $0$ i.e. both $x_n,y_n$ converges to $0$ implies that $|x_n-y_n|$ can be made arbitrarily small for large $n$, so by uniform continuity $|f(x_n)-f(y_n)|$ can be made arbitrarily small for large $n$ and since both ${f(x_n)},{f(y_n)}$ are convergent ,the converges same limit. So that we can extend $f$ by defining $f(0)=l$. In a similar manner one can assign a value for $f$ on the point $1$. This gives a continuous extension of $f$ on $[0,1]$.

For the converse , let $tilde f$ be a continuous extension of $f$ on $[0,1$, then $tilde f$ is uniformly continuous as $[0,1]$ is compact , hence $f$ , being a restriction of $tilde f$ is uniformly continuous on $(0,1)$.

More general statement---
Let $D$ be a dense subset of $[0,1]$ and $f:Drightarrow Bbb R$ is uniformly continuous then $f$ can be extended to whole $[0,1]$ continuously.