# Is the following function uniformly continuous on the closed and bounded interval?

Let us consider the identity function $$f:(\mathbb{R},d)\to (\mathbb{R},d_{usual})$$ $$f:x \to x$$ Here we are considering $$d(x,y)=|(x)^3-(y)^3|$$

Is the function $$f$$ uniformly continuous on closed and bounded interval?

I am looking for an example of function $$f$$ which is not uniformly continuous on a closed and bounded interval but it is continuous.

• Any continuous function on a closed bounded interval is uniformly continuous. Feb 24, 2022 at 9:06
• That is in R under usual metric.What if the metric is different? Feb 24, 2022 at 9:07
• Any continuous function on a compact metric space into any metric space is uniformly continuous. Feb 24, 2022 at 9:12
• If you are using the metric $d$ on the domain also you should specify it. Also, a closed and bounded interval is compact in the metric $d$ also. Feb 24, 2022 at 9:15
• No. It doesn't. On a closed and bounded interval $x^{1/3}$ is continuous hence also uniformly continuous. That makes your map uniformly continuous. Feb 24, 2022 at 11:35

In this answer I assume that $$\tan^{-1}$$ denotes the arctangent $$\arctan$$.
No, your example does not work, because $$\arctan$$ is Lipschitz continuous. Indeed for all $$x, y \in \Bbb R$$ $$|\arctan x - \arctan y| \le |x-y|$$
this implies that the identity map $$f: \Bbb R \to ( \Bbb R , d)$$ is Lipschitz (hence uniformly continuous).
• How about $(R,d_{usual}) \to (R,d)$ where $d(x,y)=|x^3-y^3|$. Will this work if I take $f(x)=x$? Feb 24, 2022 at 9:10