Neccesary condition for uniform continuity under norm induced metric spaces.

Question

For a $f:X \rightarrow Y$
Where $(X, d)$ and $(Y, \rho)$ are metric spaces.
Standard $\epsilon \text{-} \delta$ definition of uniform continuity:
Suppose $\emptyset \ne A \subset X$, $f$ is said to be uniformly continuous on $A$ if $\forall \epsilon \gt0 \exists \delta \gt0$ s.t $\forall x \in A, y \in X$ $d(y,x) \lt \delta \implies \rho(f(y), f(x)) \lt \epsilon$.

While perusing through a problem in math StackExchange, I came to this result :-
If for every $h>0$ we have that $|f(x+h)−f(x)|$ is unbounded on $I$, then $f$ is not uniformly continuous on $I$.

How to prove the above statement.
Thanks in advance!!

Answer 1

Suppose $f$ is u.c. Then for small $h > 0$ we have $|f(x + h) - f(x)| < \frac{1}{n}$.

But for $h > 0$ and $n \ge 1$ there is a $x_n \in I$ such that $n < |f(x_n + h) - f(x_n)|$.

It follows that if $h>0$ is small, then $n < |f(x_n + h) - f(x_n)| < \frac{1}{n}$.

Answer 2

A bounded continuous function can fail to be uniformly continuous. Let $X=\Bbb R^+$ and $Y=\Bbb R$ and $f(x)=\cos 1/x.$ Then $|f(x)|\le 1$ and $|f(x+h)-f(x)|\le |f(x+h)|+|f(x)|\le 2.$ But $f$ is not uniformly continuous. For if $0<\epsilon<2$ then for any $\delta >0,$ take $n\in \Bbb N$ with $2n\pi >1/\delta,$ and we have

$|1/2n\pi -1/(2n+1)\pi|<\delta$

but $|f(1/2n\pi)-f(1/(2n+1)\pi)|=2>\epsilon.$