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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.

David C. Ullrich
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Chemical Engineer
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2 Answers2

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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.

Guillaume F.
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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.

QED
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