$u: \mathbb{R} \to \mathbb{R}$: does bounded and uniformly continuous imply Lipschitz?

Question

Let $u: \mathbb{R} \to \mathbb{R}$: if $u$ is bounded and uniformly continuous, then is it Lipschitz continuous? Or is it at least locally Lipschitz continuous?

Answer 1

No, consider the function $f:\mathbb R\rightarrow \mathbb R$ defined as $f(x)=\min(1,\sqrt{|x|})$. This function is not locally lipschitz at $0$.

Answer 2

Take

• $n=1$

• $f(x)=\arcsin(x)$

$f$ is continuous at $[-1,1]$, thus it is uniformly continuous by Heine's Theorem. Also, it is bounded .

But it is not Lipschitz cause its derivative $\frac{1}{\sqrt{1-x^2}}$ is not bounded.

Answer 3

Consider the function $$f(x) = \tan^{-1}(\sqrt[3]x)$$ $f(x)$ is bounded and uniformly continuous but is not Lipschitz.

None of the earlier answers are bijective functions from $\Bbb R^1 \to \Bbb R$; this one is.