Characterization of Uniform Continuity

Question

If $f:X \to Y$ is uniformly continuous, then there exists $\omega : (0, +\infty) \to (0, +\infty)$ with $\lim \limits_{t \to 0^+} \omega(t) = 0$ such that $d_Y(f(x), f(x')) < \omega(d_X(x, x'))$.

I am trying to come up with such an $\omega$ function.

\begin{equation*} \omega(t) =t + \sup \limits_{x \in X, x' \in X} \{d_Y(f(x), f(x')) : d_X(x, x') < t \} \end{equation*}

I don't know if it makes sense. There will probably be some issues with $\sup$.

Can anyone suggest improvements/corrections?

Thanks.

Answer 1

What you are looking for is called the modulus of continuity.

Recall that if $f: X \to Y$ is uniformly continuous, then $d_X(u, v) \leq \delta(\epsilon) < +\infty$ implies $d_Y(f(u), f(v)) \leq \epsilon$. Let $$ \omega(\delta) = \sup\{d_Y(f(u), f(v)) : d_X(u, v) \leq \delta\}. $$ By assumption of uniform continuity, $\omega: \mathbb{R}_+ \to \mathbb{R}_+ \cup \{+\infty\}$ and by construction $$d_Y(f(u), f(v)) \leq \sup_{d_X(u', v') \leq d_X(u, v)} d_Y(f(u'), f(v')) = \omega(d_X(u, v)), \quad \text{for all}~u, v \in X $$.

We cannot expect that $\omega$ is finite in general. Indeed, consider $f: \mathbf{R} \to \mathbf{R}$, where in the preimage it is given the discrete metric $(d_{\rm disc})$, and in the image, it is given the Euclidean metric. Take $f(x) = x$. Then if $|f(x) - f(y)| \leq \omega(1\{x \neq y\})$, we have $\omega(1) \geq |f(x)|$ for all $x \neq 0$. Take $x \to +\infty$, and note then $\omega(1) = +\infty$. On the other hand, if $\epsilon > 0$, then if $d_{\rm disc}(u, v) \leq 0.5$, then $|f(u) - f(v)| = 0 < \epsilon$. Hence, $f$ is uniformly continuous.

Answer 2

Such a function need not exist. Consider $\mathbb N$ with the discrete metric and let $f(n)=n$ for all $n$. Then $f$ is uniformly continuous. But $|n-1| <\omega (1) \vee \omega (0)$ does not hold for all $n$ whatever your function $\omega$ is.

Here $Y$ is the real line with the usual metric.