Here's a statement in Zygmund's Measure and Integral on page 17:

If $f$ has a continuous derivative on $[a,b]$, then (by the mean-value theorem) $f$ satisfies a Lipschitz condition on $[a,b]$.

This does not seem obvious to me. How can I show it?

Also, what does a continuous derivative imply? Can we conclude the function is differentiable? If so, how can I prove it?

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    $\begingroup$ If $f'$ is continuous, it is bounded on the compact interval $[a,\,b]$. A bound of $\lvert f'\rvert$ is a Lipschitz constant of $f$ by the MVT. $\endgroup$ – Daniel Fischer Aug 25 '13 at 20:37
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    $\begingroup$ "Also, may I ask what does continuous derivative imply? Can we conclude the function is differentiable?" If a functions has "continuous derivative" it already is differentiable. What this adds is that $f'$ is continuous where it is defined. $\endgroup$ – Pedro Tamaroff Aug 25 '13 at 20:46
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    $\begingroup$ I see, thank you @DanielFischer. $\endgroup$ – 1LiterTears Aug 25 '13 at 20:53

By the mean value theorem,

$$f(x) - f(y) = f'(\xi)(x-y)$$ for some $\xi \in (y,x)$. But since $f'$ is continuous and $[a,b]$ is compact, then $f'$ is bounded in that interval, say by $C$. Thus taking absolute values yields

$$\lvert f(x) - f(y)\rvert \le C \lvert x-y\rvert$$

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  • $\begingroup$ Thank you Deven. Does this depend on $|x-y| \geq 1$? $\endgroup$ – 1LiterTears Aug 25 '13 at 20:52
  • $\begingroup$ @Jellyfish no not at all. It simply relies on the mean value theorem, which is true no matter how close $x$ and $y$ are and the fact that the derivative is bounded. $\endgroup$ – Deven Ware Aug 25 '13 at 20:57
  • $\begingroup$ Oh I see how you defined $C$, it is depend on $x-y$. Got it, thanks! $\endgroup$ – 1LiterTears Aug 25 '13 at 21:16

Since $f'$ is continuous it is bounded on $[a,b]$. Choose $M\in \Bbb R$ such that $|f'(x)|\le M$ for all $x\in [a,b]$. If $x,y\in [a,b]$, then by the MVT there is $\xi\in [a,b]$ with $f(x)-f(y)=f'(\xi)(x-y)$, thus $|f(x)-f(y)|\le M|x-y|$. In fact you can show that a differentiable function on an open interval (not necessarily a bounded interval) is Lipschitz continuous if and only if it has a bounded derivative. This is because any Lipschitz constant gives a bound on the derivative and conversely any bound on the derivative gives a Lipschitz constant.

To your other question: Continuous derivative does not imply twice differentiable. In fact, we know that there are functions $f$ (such as the Weierstrass-function) which are continuous but not differentiable. Taking the antiderivative $F$ gives us a differentiable function with derivative $f$ (by the Fundamental Theorem of Calculus), so it has a continuous derivative, but as $f$ is not differentiable, $F$ is not twice differentiable.

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  • $\begingroup$ Did you mean "you can show that a continuous differentiable function..."? $\endgroup$ – philmcole Jan 16 '18 at 17:02

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