# Simple Construction of a Uniformly Continuous Real Valued Function With No Derivative Anywhere In The Domain?

Alright,every kid who's taken and passed undergraduate real analysis knows there are functions defined on subsets of the real line that are continuous everywhere and not differentiable everywhere (i.e.the Weierstrass function and its many descendants and bretheren). But I was going through my old analysis texts and was wondering: Is there an example of a uniformly continuous everywhere function whose domain is a well-defined nonempty subset of the real line that has no derivative anywhere?

My first response would be yes: Take the restriction of the Weierstrass function to any closed and bounded subset of the real line [a,b]. Them since this function is continuous everywhere and defined on a compact subset of the real line,then this restriction is uniformly continuous on [a,b]. Since the derivative doesn't exist anywhere on the domain of the original Weierstrass function, it doesn't exist anywhere on [a,b] either. Granted,it's not a genius construction, but as far as I can see, there are no logical errors here.

Are there? More importantly,if this example is correct, can anyone give a more creative example, one that isn't obvious? If so, I'd love to see them posted here.

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Weierstrass, with two e's and no u. – Rahul Mar 12 '12 at 4:00
I don't understand what you're suggesting. You say you want an example of a uniformly continuous but nowhere differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, but the example you give is of a function with domain $[a,b]$. How can this be what you want? – Pete L. Clark Mar 12 '12 at 4:04
Also, what is the "Bolzano-Weirstrauss function"? – Pete L. Clark Mar 12 '12 at 4:09
e.g., i.e. - whatever :) – The Chaz 2.0 Mar 12 '12 at 4:31
@Mathemagician1234: I am pretty sure The Chaz is not accusing you of ignoring anyone. He's just pointing out that you wrote "i.e." where you probably should have written "e.g." (@The Chaz: in the realm of pedantic -- but of course correct -- comments on people's writing, this is a pretty "advanced" one. If you have the energy to spare and want to help, I suggest you get on the OP to put spaces after his commas and doublecheck "ie" versus "ei" in word spellings.) – Pete L. Clark Mar 13 '12 at 5:29

Hints:

1) If $f: [0,1] \rightarrow \mathbb{R}$ is any continuous nowhere differentiable function, then by subtracting off a suitable linear function we get a continuous nowhere differentiable function $g: [0,1] \rightarrow \mathbb{R}$ with $g(0) = g(1) = 0$.

2) Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $f(x+n) = f(x)$ for all $x \in \mathbb{R}$ and $n \in \mathbb{Z}$. Then $f$ is uniformly continuous.

Added: When I wrote this answer I didn't have firmly in mind exactly which function Weierstrass constructed. But having checked this I see that it is $2$-periodic and therefore itself uniformly continuous on the entire real line. In fact, see this previous math.SE question.

It follows of course that for any subset $S \subset \mathbb{R}$, the restriction of Weierstrass's function to $S$ is uniformly continuous. As for nowhere differentiable -- here one should be careful, as it is not so clear what differentiability should mean for a function defined on a subset of $\mathbb{R}$ which is not an interval. The usual definition of differentiability at $x \in S$ makes sense for any limit point $x$ of a subset $S$ but may behave strangely. It is not clear to me that, for instance, the restriction of the Weierstrass function to every dense subset of $\mathbb{R}$ is nowhere differentiable.

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I'll mull this over tomorrow when I got a rested brain.But I already got an idea how to proceed. Thanks,Pete! – Mathemagician1234 Mar 12 '12 at 4:32
@Mathemagician: I somehow forgot to address part of your question: yes, restricting any continuous nowhere differentiable function to a closed, bounded interval gives you a uniformly continuous nowhere differentiable function. As you say, this is not exactly profound, but it is absolutely correct. – Pete L. Clark Mar 13 '12 at 5:24