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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.

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

2 Answers 2

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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.

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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.

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  • $\begingroup$ 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)? $\endgroup$ Dec 8, 2017 at 6:12
  • $\begingroup$ Because the condition was symmetric in $x$ and $y$, you can obviously change the variable names. $\endgroup$
    – QED
    Dec 8, 2017 at 6:13
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    $\begingroup$ You have proved that $f$ uniformly differentiable implies $f'$ uniformly continuous, but not the "only if" part. $\endgroup$ May 30, 2018 at 23:32

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