A property of uniformly continuous functions

Question

We know that if $f:\mathbb{R}\to\mathbb{R}$ is a function such that for all $x,y\in \mathbb{R}$ $$|f(x)-f(y)|\leq a(|x-y|),$$ where $a:\mathbb{R}\to\mathbb{R}$ is a function independent of $x$ and $y$ with $\displaystyle \lim_{h \to 0}a(h)=0$. Then $f$ is uniformly continuous. Now conversely, if $f$ is a uniformly continuous function. Do we have the existence of a function $a:\mathbb{R}\to\mathbb{R}$ which satisfies the above inequality ?

Answer 1

My previous deleted answer was based on a misunderstanding of your question. I thought you were asking about a uniform modulus of continuity, whereas you wanted to bound $|f(x)-f(y)|$ by a function of $|x-y|$ which goes to zero as $|x-y|$ goes to zero. These are similar but not quite the same.

For your problem you had the right idea. If $f$ is uniformly continuous and you define $a(\delta)=\sup_{x,y \in \mathbb{R} : |x-y| \leq \delta} |f(x)-f(y)|$, then $|f(x)-f(y)| \leq a(|x-y|)$ and $\lim_{\delta \to 0^+} a(\delta)=0$.

The proof is just carrying through definitions: given $\varepsilon$, choose $\delta$ from the uniform continuity of $f$, then every $x,y$ with $|x-y| \leq \delta$ satisfy $|f(x)-f(y)| \leq \varepsilon$. So when you take the supremum the inequality is preserved and you get $a(z) \leq \varepsilon$ for $z \in (0,\delta]$, as desired.