# A function is uniformly differentiable if its derivative is uniformly continuous?

Suppose $$I$$ is an open interval and $$f:I\rightarrow\mathbb{R}$$ is a differential function. We can say $$f$$ is uniformly diferentiable if for every $$\epsilon> 0$$ there exists $$\delta> 0$$ such that

$$x,y\in I$$ and $$0\lt|x-y|<\delta \Rightarrow \Big| \frac{f(x)-f(y)}{x-y}-f'(x) \Big|\lt\epsilon$$

I would like to prove that, if and only if $$f'$$ is uniformly continuous, then $$f$$ is uniformly differentiable.

• Is $|x-y|<\delta$? Dec 8, 2017 at 5:41
• Yes, I left that out somehow. Thanks for pointing that out. Dec 8, 2017 at 5:59
• Maybe you want “f is uniformly differentiable if and only if f’ is uniformly continuous”?
– Dap
Dec 8, 2017 at 15:21

Abishanka proved that

(1) $f(x)$ uniformly differentiable implies $f'(x)$ is uniformly continuous

but not that

(2) $f'(x)$ is uniformly continuous implies $f(x)$ uniformly differentiable

Let's prove 2. From the uniform continuity of $f'(x)$, we have a $\delta > 0$ such that for every $x$ and $y$,

$|f'(y) - f'(x)| < \epsilon$ if $|y - x| < \delta$

Note that from the mean value theorem, we have, for any $x$

$\frac{f(x+h) - f(x)}{h} = f'(\tilde{x})$

for $\tilde{x}$ in between $x$ and $x+h$. Then, for any $x$ and $h < \delta$, $|\tilde{x} - x| < \delta$, and so

$\left| \frac{ f(x+h) - f(x) }{ h } - f'(x)\right| = |f'(\tilde{x}) - f'(x)| < \epsilon$

which proves the result.

Interchanging $x$ and $y$ we get $\Big| \frac{f(x)-f(y)}{x-y}-f'(y) \Big|\lt\epsilon$ and hence your condition implies $|f'(x)-f'(y)|<2\epsilon$ for all $x,y\in I$ such that $|x-y|<\delta$, or equivalently, $f'$ is uniformly continuous. But there are unifromly continuous functions whose derivative is not uniformly continuous.

• I think I understand what you are saying, this helps a lot. However, why is it that you can interchange f'(x) and f'(y)? Dec 8, 2017 at 6:12
• Because the condition was symmetric in $x$ and $y$, you can obviously change the variable names.
– QED
Dec 8, 2017 at 6:13
• You have proved that $f$ uniformly differentiable implies $f'$ uniformly continuous, but not the "only if" part. May 30, 2018 at 23:32