# Not uniformly continuous

I know that a function $f:A→\mathbb{R}$ is not uniformly continuous iff there exists $ϵ>0$ and there exist sequences $(x_n)$ and $(y_n)$ in $A$ such that $\lim_{n→∞} |x_n-y_n|=0$ and $|f(x_n)-f(y_n)|≥ϵ$ for all $n∈\mathbb{N}$.

I want to know if there is any other method to disprove the uniform continuity of a function, either by boundlessness of the function (or its derivative), or by using the graph of the function.

Thank you.

• You can also find a Cauchy sequence $(x_n)_n$ such that $(f(x_n))_n$ is not Cauchy. Jul 30, 2016 at 4:32
• Yes but any method by graph or boundlessness Jul 30, 2016 at 4:34

On an open and bounded interval $(a,b)$:

Theorem. If $f\colon(a,b)\to\mathbb{R}$ is uniformly continuous, then it can be extended by continuity into a continuous (and thus uniformly continuous) function $\tilde{f}\colon[a,b]\to\mathbb{R}$.

As contrapositive, we get that if $f$ is unbounded on $(a,b)$, then it cannot be uniformly continuous. (Otherwise, it could be extended into a continuous $\tilde{f}$ bounded on $[a,b]$.)

This fails if you consider an unbounded interval, of course. $f\colon x\in[0,\infty)\mapsto \sqrt{x}$ is uniformly continuous, but not quite bounded.

• Sin (1/x) on (0,1) can be the counter example for this as it is boundede on [0,1] but not uniformly continuous Jul 30, 2016 at 5:04
• @AshokSinghBhandari How is it a counterexample? It is simply a case where it does not apply. Jul 30, 2016 at 5:10
• So can we say that theorem is not applicable on every function? Jul 30, 2016 at 5:15
• If it is not bounded, then it is not uniformly continuous. That's an implication, not an equivalence; if the $f$ is bounded, you cannot say anything with this... Jul 30, 2016 at 5:17
• But f(x)= x is not bounded on R but still uniformly continuous Jul 30, 2016 at 5:20