Must a uniformly continuous function from $\mathbb{R}$ to $\mathbb{R}$ have a finite modulus of continuity?

I searched for similar questions and I could find a duplicate for this.

NOTE: I am using the definition of "modulus of continuity" as found in Wikipedia for example ( http://en.wikipedia.org/wiki/Modulus_of_continuity ). Under this definition, any uniformly continuous function has many "moduli of continuity". Some people use a different definition, under which one refers to "the modulus of continuity of $f$", and that modulus of continuity satisfies all the requirements that Wikipedia, for example, requires of a modulus of continuity. It appears that different people use the term slightly differently, and I can't make everyone 100% happy. I am not claiming Wikipedia is the ultimate authority for math matters, it is just a convenient reference for me at the moment.

The title is the question: lacking a good reference, and unable to find one online, I was proving to myself that a uniformly continuous function $f$ with domain and codomain $\mathbb{R}$ has a modulus of continuity $\omega$. Using the definition of uniformly continuous, I constructed such an $\omega$, and it seems to me that I can prove that $\omega(t)$ is finite for all $t \geq 0$.

If this is indeed true, someone has undoubtedly discovered this before, probably one or two centuries ago. Can anyone provide a reference, or a counterexample in case I'm wrong?

If the answer to the title question is "yes", I would like some explanation why moduli of continuity are allowed to be infinite (finitivity is not part of any definition I have ever seen). Can anyone cite a good reason? Perhaps it is useful to allow $\omega$ to be infinite if the domain and/or codomain are infinite-dimensional normed linear spaces (?). I am not interested in silly, contrived examples, rather examples of uniformly continuous functions between metric spaces (such as familiar function spaces that analysts actually use) with no finite modulus of continuity that are of some interest for some real math problem (not just a counterexample for the sake of a counterexample).

EDIT: thinking about it some more, I am pretty sure any uniformly continuous function between any two "nice" metric spaces (is connectivity sufficient?) has a finite-valued modulus of continuity, but that leaves open the question of why the definition permits a modulus of continuity to be infinite-valued. Is it ever convenient to use one that takes infinite values?

• Do you define the modulus of continuity as $\omega_f(\delta) = \sup \{ \lvert f(x) - f(y)\rvert : \lvert x-y\rvert <\delta\}$ (or $\leqslant\delta$)? Or do you use some other definition? – Daniel Fischer Nov 7 '13 at 22:41
• Yes, if $f$ is continuous, $<$ and $\leqslant$ yield the same result. The uniform continuity says $\lim\limits_{\delta\searrow 0} \omega_f(\delta) = 0$, in particular, $\omega_f(\delta) < \infty$ for all small enough $\delta$. The subadditivity now says $\omega_f(t) < \infty$ for all $t\geqslant 0$. – Daniel Fischer Nov 7 '13 at 22:48
• For a uniformly continuous $f$, the modulus of continuity is (uniformly) continuous, since $\lvert \omega_f(t) - \omega_f(s)\rvert \leqslant \omega_f(\lvert t-s\rvert)$, and the continuity in $0$ is the definition of uniform continuity. – Daniel Fischer Nov 7 '13 at 23:04
• @JonathanY. Yup, will do. – Daniel Fischer Nov 7 '13 at 23:10
• Re: "why the definition permits a modulus of continuity to be infinite-valued" -- maybe it shouldn't. Opinion piece: The definition of uniform continuity is wrong – user103402 Nov 9 '13 at 22:01

Let us define for a function $f\colon X \to Y$ between metric spaces the minimal nondecreasing modulus of continuity as

$$\omega_f(\delta) := \sup \{ d(f(x),f(y)) : d(x,y) \leqslant \delta\}.$$

Among all monotonic functions that are admissible as a modulus of continuity of $f$, that is the smallest one. Non-monotonic moduli of continuity can be smaller at some values of course, consider a periodic function for example.

$f$ is uniformly continuous if and only if

$$\lim_{\delta \searrow 0} \omega_f(\delta) = 0.$$

The function $\omega_f$ is subadditive, $\omega_f(\delta + \varepsilon) \leqslant \omega_f(\delta) + \omega_f(\varepsilon)$ for functions whose domain is a convex domain in $\mathbb{R}^n$, hence $\omega_f(t)$ is finite for all $t \in [0,\infty)$ if $\omega_f(\delta) < \infty$ for some $\delta > 0$ then. For a space with infinitely many connected components that are at a positive distance from each other, $\omega_f(t)$ can be infinite for finite $t$ even for uniformly continuous $f$. We have

$$\omega_{\omega_f} \leqslant \omega_f,$$

so if $f$ is uniformly continuous, so is $\omega_f$, for functions with good domain.

Thus the question in the title can be answered affirmative, a uniformly continuous function $\mathbb{R}\to \mathbb{R}$ always has a (uniformly continuous) modulus of continuity that attains only finite values on $[0,\infty)$.

Note that $\lim\limits_{t\searrow 0} \omega(t) = \omega(0) = 0$ is part of the definition of a modulus of continuity, so if a function $f$ admits a global modulus of continuity, it must be uniformly continuous, and then allowing moduli of continuity that attain the value $\infty$ for finite $t$ is somewhat pointless for functions on good domains, but for functions on spaces with separated components, it can be necessary. And when one considers only local moduli of continuity (at a fixed point $x_0$), allowing infinite values is a simple way to not care about the behaviour far enough away from $x_0$.

• Actually, it just occurred to me that when you have a space with infinitely many connected components at a positive distance from each other, even a uniformly continuous function can have $\omega_f(t) = \infty$ for finite $t$. – Daniel Fischer Nov 8 '13 at 0:01
• Hmm, right, that isn't strong enough. I need almost straight lines for that, the sum of the distances must be about $d(x,y)$. Or perhaps even equal? I'll make it convex sets in $\mathbb{R}^n$ for now, and think about what conditions would be sufficient. – Daniel Fischer Nov 8 '13 at 0:12
• I think that for a metric space it is sufficient if the intrinsic metric is Lipschitz. – Niels J. Diepeveen Nov 8 '13 at 0:44
• What I mean is that $\hat{d}(x, y) \le Md(x, y)$ for some constant $M$, where $d$ is the existing metric and $\hat{d}$ is the intrinsic metric it induces. – Niels J. Diepeveen Nov 8 '13 at 1:55
• @StefanSmith: I don't know a name for such spaces. To answer your question to Daniel Fischer: Of course any bounded function has a bounded modulus of continuity. Any function with a bounded codomain is obviously bounded. A uniformly continuous function with a totally bounded domain (such as any bounded subset of $\mathbb{R}^n$) is also bounded. – Niels J. Diepeveen Nov 16 '13 at 20:41