1. ## 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?

2. 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 $\displaystyle [n,+\infty)$).

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...).

3. So what can we add to the theorem I proposed so that it is an iff statement?