# Uniform Continuity

• Oct 15th 2009, 09:25 AM
redsoxfan325
Uniform Continuity
A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably $\sqrt{x}$) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

Theorem: An everywhere continuous function $f$ is uniformly continuous on $\mathbb{R}$ if its derivative is bounded on $(-\infty,-m]$ and $[n,\infty)$ ( $m,n\in\mathbb{R}$).

Proof: On $[-m,n]$ we have a continuous function defined on a compact set, so it is uniformly continuous and $\exists~\delta_1$... On $(-\infty,-m]$ and $[n,\infty)$, $f$ has a bounded derivative so there exist $\delta_2$ and $\delta_3$ such that... Therefore taking $\delta=\min\{\delta_1,\delta_2,\delta_3\}$ proves that the entire function is uniformly continuous on $\mathbb{R}$.

So what's my question? My question is whether the above theorem is an if-and-only-if statement or whether the implication only goes in one direction. If not, what hypotheses can we add so that it does go both ways?

Thoughts? Counterexamples?
• Oct 15th 2009, 11:36 AM
Laurent
Quote:

Originally Posted by redsoxfan325
A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably $\sqrt{x}$) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

Theorem: An everywhere continuous function $f$ is uniformly continuous on $\mathbb{R}$ if its derivative is bounded on $(-\infty,-m]$ and $[n,\infty)$ ( $m,n\in\mathbb{R}$).

Proof: On $[-m,n]$ we have a continuous function defined on a compact set, so it is uniformly continuous and $\exists~\delta_1$... On $(-\infty,-m]$ and $[n,\infty)$, $f$ has a bounded derivative so there exist $\delta_2$ and $\delta_3$ such that... Therefore taking $\delta=\min\{\delta_1,\delta_2,\delta_3\}$ proves that the entire function is uniformly continuous on $\mathbb{R}$.

About this proof: perhaps you know that, but for a "full" proof you should take care of the case when the two points $x,y$ such that $|x-y|<\delta$ lie in different intervals (like $x). Either you say $|f(x)-f(y)|\leq |f(x)-f(n)|+|f(n)-f(y)|\leq 2\varepsilon$. Or you use the following trick: you define $\delta_1$ with uniform continuity on $[-m-1,n+1]$ and choose $\delta=\min(\delta_1,\delta_2,\delta_3,1)$ so that $|x-y|<\delta$ implies that $x,y$ are in the same interval (either $(-\infty,-m],$, $[-m-1,n+1]$ or $[n,+\infty)$).

Quote:

So what's my question? My question is whether the above theorem is an if-and-only-if statement or whether the implication only goes in one direction. If not, what hypotheses can we add so that it does go both ways?

Thoughts? Counterexamples?
Counterexample: you will be delighted to hear that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2\sin\frac{1}{|x|^{4/3}}$ ( $x\neq 0$), $f(0)=0$, is derivable on $\mathbb{R}$, yet its derivative is unbounded on any neighbourhood of 0.

In particular, $f$ is uniformly continuous and derivable on $[-1,1]$ while its derivative is unbounded (but finite, that's the difference with $\sqrt{x}$).

You can then define a derivable periodic function $g$ that looks like $f$ repeatedly. So that the derivative of $g$ is unbounded on any $[m,+\infty)$, while $g$ is uniformly continuous (it is continuous and periodic, so...).
• Oct 15th 2009, 01:17 PM
redsoxfan325
So what can we add to the theorem I proposed so that it is an iff statement?