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Let $\delta>0$ be given and $f: A \subset \mathbb{R} \rightarrow \mathbb{R}$ an uniformly continuous function (with $A$ a nonempty interval (?)).

Is it possible to find a $c >0$ such that for all $x,y \in A$ we have

$|x-y| < \delta \Rightarrow |f(x) - f(y) | < c$?

This holds for example for each linear function (which are uniformly continuous). But it does not hold for the exponential function, which is not uniformly continuous.

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    The uniform continuity of $f$ ensures that there exists $a$ such that if $|x-y|$c=1+(\delta/a)$. – 2017-01-03
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    So let $\delta>0$ and set $c$ as above. Then $|x-y| < \delta$ implies $|x/(c-1) - y/(c-1)| < a$, thus $ |f(x/(c-1)) - f(y/(c-1))| <1$. How can we get now $|f(x) - f(y)| < c$? – 2017-01-03
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    No, rather there exists some intermediary points linking $x$ to $y$, each at distance at most $a$ from the next one, the images of the intermediary points are at most at distance $1$ and the triangular inequality completes the job. Note that for some sets $A$ such intermediary points may not exist -- but then the conclusion does not hold anyway. – 2017-01-03
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    I don't understand. – 2017-01-03

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To elaborate Did's comment:

Let $A\subseteq \Bbb R$ be an interval $f\colon A\to\Bbb R$ be uniformly continuous. By uniform continuity, there exists $a>0$ such that for all $x,y\in A$ with $|x-y|

For given $\delta>0$, let $n=\left\lceil \frac{\delta}a\right\rceil$ (so that $n-1<\frac \delta a\le n$). Then we succeed with $c=n$:

Let $x,y\in A$ with $|x-y|<\delta$. For $k=0,\ldots, n$, let $t_k=x+\frac kn(y-x)$. Then $t_0=x$, $t_n=y$ and $|t_{k+1}-t_k|=\frac{|y-x|}{n}<\frac \delta n\le a$ for $0\le k