# Uniform Continuity

• Oct 15th 2009, 08:25 AM
redsoxfan325
Uniform Continuity
A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably $\displaystyle \sqrt{x}$) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

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

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

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, 10:36 AM
Laurent
Quote:

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

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

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

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

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 $\displaystyle f:\mathbb{R}\to\mathbb{R}$ defined by $\displaystyle f(x)=x^2\sin\frac{1}{|x|^{4/3}}$ ($\displaystyle x\neq 0$), $\displaystyle f(0)=0$, is derivable on $\displaystyle \mathbb{R}$, yet its derivative is unbounded on any neighbourhood of 0.

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

You can then define a derivable periodic function $\displaystyle g$ that looks like $\displaystyle f$ repeatedly. So that the derivative of $\displaystyle g$ is unbounded on any $\displaystyle [m,+\infty)$, while $\displaystyle g$ is uniformly continuous (it is continuous and periodic, so...).
• Oct 15th 2009, 12:17 PM
redsoxfan325
So what can we add to the theorem I proposed so that it is an iff statement?