TITLE: Let $f: \Bbb R \to \Bbb R$ be a differentiable function such that $\sup_{x \in \Bbb R}|f'(x)| \lt \infty$. Then
QUESTION [9 upvotes]: (UGC CSIR-2015, DECEMEMBER, MATHEMATICAL SCIENCES)


$f$ maps a bounded sequence to a bounded sequence.
$f$ maps a Cauchy sequence to a Cauchy sequence.
$f$ maps a convergent sequence to a convergent sequence.
$f$ is uniformly continuous.


I choose all of the options as possible answers because the condition $\sup_{x\in \Bbb R}|f'(x)| \lt \infty$ forces $f$ to be uniformly continuous.(Because $f$ becomes Lipschitz and Lipschitz condition implies uniform continuity)
i.e. $\frac {|f(x)-f(y)|}{|x-y|} \le \sup_{x\in \Bbb R}|f'(x)|$
$ \forall x,y$.
Hence all other options are bound to be true.
Am I correct?

REPLY [3 votes]: All are correct.

If $\{x_n\}_{n\in\mathbb N}\subset\mathbb R$ is bounded, i.e., $\lvert x_n\rvert\le M<\infty$, then
$$
\lvert\,f(x_n)-f(x_1)\rvert=\lvert x_n-x_1\rvert\lvert f'(y_n)\rvert,
$$
for some $y_n\in(x_1,x_n)$, by virtue of the Mean Value Theorem, and hence
$$
\lvert\,f(x_n)\rvert\le \lvert\,f(x_1)\rvert +\lvert x_n-x_1\rvert\lvert f'(y_n)\rvert\le \lvert\,f(x_1)\rvert +2M \|f'\|_\infty.
$$
If If $\{x_n\}_{n\in\mathbb N}\subset\mathbb R$ is Cauchy,
then
$$
\lvert\,f(x_m)-f(x_n)\rvert=\lvert\,f'(y_{m,n})\rvert\lvert x_m-x_n\rvert\le\|f'\|_\infty \lvert x_m-x_n\rvert,
$$
hence $\{f(x_n)\}_{n\in\mathbb N}$ Cauchy.
If $x_n\to x$, then 
$$
\lvert\,f(x_n)-f(x)\rvert=\lvert\,f'(y_n)\rvert\lvert x_n-x\rvert\le\|f'\|_\infty \lvert x_n-x\rvert,
$$
where $y\in(x,x_n)$, and hence $f(x_n)\to f(x)$.
If $x,y\in\mathbb R$, then 
$$
\lvert\,f(x)-f(y)\rvert\le\|f'\|_\infty \lvert x-y\rvert,
$$
etc...