# Finding an $c>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<c$ for uniformly continuous $f$

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.

• The uniform continuity of $f$ ensures that there exists $a$ such that if $|x-y|<a$ then $|f(x)-f(y)|<1$. Try $c=1+(\delta/a)$. – Did Jan 3 '17 at 18:06
• 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 $|f(x) - f(y)| < c$? – CHwC Jan 3 '17 at 18:55
• 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. – Did Jan 3 '17 at 19:26
• I don't understand. – CHwC Jan 3 '17 at 19:42

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 $|x-y|<a$, we have $|f(x)-f(y)|<1$.
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 $0\le k<n$. Note that $t_k\in A$ for all $k$ (this is where we use that $A$ is an interval!) We conclude $|f(t_{k+1})-f(t_k)|<1$ and so $$|f(y)-f(x)|\le |f(t_n)-f(t_{n-1})|+\ldots+|f(t_1)-f(t_0)|<n.$$