Asking for clarification of a Wikipedia article on uniform continuity

Question

A portion of Wikipedia article on uniform continuity

Functions that have slopes that become unbounded on an infinite domain cannot be uniformly continuous. The exponential function $ x\mapsto e^{x}$ is continuous everywhere on the real line but is not uniformly continuous on the line, since its derivative tends to infinity as $x\to \infty$

1. We know that a differentiable function $f:\Bbb{R}\to\Bbb{R}$ is Lipschitz iff $\|f'\|<\infty$.

2. $f$ Lipschitz implies $f$ is uniformly continuous.

3. $f$ uniformly continuous but $f$ may not be differentiable $( x\mapsto{|x|})$. Even when $f'$ exists,$f'$ need not be bounded $(x\mapsto{\sqrt{x}} $ on $(0, \infty) ) $

Please elaborate the quoted portion.

Answer 1

The thing is that functions that are continuously differentiable are locally (i.e. on all compact subsets) Lipschitz due to the mean value theorem: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuously differentiable, then for any compact subset $K\subset \mathbb{R}$ and $x,y\in K$, there exists a $\xi\in K$ so that $$|f(x)-f(y)| = |f'(\xi)|\cdot |x-y|.$$ Since $K$ is compact and $f'$ is continuous, $f'$ takes on its maximum and minimum value on $K$, i.e. there exists a $L\in [0,\infty)$, so that $L = \max\{|f'(\xi)| : \xi \in K\}$. This means that $|f'(\xi)|\leq L$ for all $\xi\in K$, so using the above formula gives us that $f$ is Lipschitz continuous on $K$ with Lipschitz constant $L$ (and thus also uniformly bounded).

However, we fundamentally used here that $K\subset \mathbb{R}$ is compact (i.e. closed and bounded due to Heine-Borel). If $K$ was for example unbounded, then $x\mapsto|f'(x)|$ could just as well be an unbounded function, so the above argument would not work any longer.

One example for this is $K=\mathbb{R}$ and $f(x) = e^x$.

Answer 2

The wording in the article has been corrected as of June 26, 2023. It now reads

Functions whose derivative tends to infinity as x grows large cannot be uniformly continuous.

citing (in the edits page) $ \dfrac{\sin(x^4)}{1+x^2} $ as an example. The wording "unbounded in an infinite domain" in not precise. The author meant (I assume) that the set $ \{ x : |f'(x)| > M \} is unbounded for all M. The $ \sqrt{x} $ is not a valid counter example in this sense.