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Let $f$ be continuous on $\mathbb R$ and differentiable with derivative $f'$ on $\mathbb R \setminus \{t_0, t_1, \dots \}$. Let $\sup | f'(t) | < \infty$, then $f$ is Lipschitz continuous with $L=\sup |f'(t)|$.

Does this hold? How could one prove it?

Johannes L
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2 Answers2

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Since $-L \leq f'(x) \leq L $ except on a set of only measure zero, we may integrate this between $x_1$ and $x_2$ and the desired $ -L(x_2-x_1) \leq f(x_2) - f(x_1) \leq L(x_2-x_1) $ pops right out.

Note to all: The following is what was my intial answer, but it is faulty.

First do the problem for each segment that the function is differentiable (so $(-\infty,t_0), (t_0,t_1) $ etc): By the mean value theorem, for $x_1 , x_2 \in (t_k, t_{k+1}) $ we get $$ \frac{f(x_1)-f(x_2)}{x_1-x_2} = f'(c) $$ for some $c\in (x_1,x_2)$.

This means for any $x_1,x_2 \in \mathbb{R}$ we have in $(t_k, t_{k+1})$ that $$ |f(x_1)-f(x_2) | \leq |f'(c)||x_1-x_2| \leq L|x_1-x_2|. $$

which makes it Lipschitz continuous in each segment with Lipschitz constant $L$. Can you see how to prove that if we stitch together Lipschitz continuous functions like this, it remains so?

Ragib Zaman
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  • From the notion I think I can really imagine it, $|\frac{f(x_1)-f(x_2)}{x_1-x_2}|$ as the slope of a secant, and this can be at tops as big as the maximum of the derivates between $x_1$ and $x_2$, and the non-differentiable points $t_0, t_1, \dots$ don't do any harm because with the continuity the function doesn't "rise" (or fall)... But in the moment I don't manage to make a straight argument out of this idea
    • ($x_1, x_2$ in different intervals)
     – Johannes L Sep 06 '11 at 11:19
  • Wonderful! This is such a nice and clean argument, it's perfect. (( and now I could even go back to "almost everywhere, because for this proof we actually use $f'$ almost everywhere defined.)) So, if $f$ has almost everywhere defined and bounded derivate the statement also holds. I think the continuity is still needed, when asserting $\int_{x_1}^{x_2} f'(x)=f(x_2)-f(x_1)$ when $x_2=t_k$ for example. – Johannes L Sep 06 '11 at 12:54
  • @Johannes L, I thought of that as well, and it worried me as my entire argument fails if the fundamental theorem of calculus required continuity. Luckily, a less well known version does not require it: See http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Second_part – Ragib Zaman Sep 06 '11 at 13:23
  • @Ragib : There is another problem with your first argument, when the set ${t_0, t_1, \dots }$ is dense in $\mathbb{R}$. The second argument (proof) got rid of this. – Rajesh D Sep 06 '11 at 13:26
  • @all I'm unsure of the etiquette in this situation? It turns out the first part of my answer is quite useless, but the 2nd proof is fine. Should I just delete the first part? – Ragib Zaman Sep 06 '11 at 13:28
  • @Ragib : I think its better to keep it as scratched out, although i've seen this kind before but i do not knoe the latex code for this. – Rajesh D Sep 06 '11 at 13:32
  • @Ragib: I guess you could put it in parenthesis after the correct argument saying that this is the old "first part" (so the comments stay understandable for later readers), or but the right part in bold? – Johannes L Sep 06 '11 at 13:34
  • @Ragib and i just looked, the code for scratching out is < strike > funny < / strike > . – Johannes L Sep 06 '11 at 13:37
  • @all I just decided to put it at the end, I think there is a bit too much written to be all scratched out, it may just be annoying for other readers. Plus, it seems the strike does not work on LaTeX code. – Ragib Zaman Sep 06 '11 at 13:42
  • The first part of your answer is wrong as stated (you're implicitly assuming absolute continuity of $f$). Let $\phi$ be the Cantor-Lebesgue function and put $$f(x) = [0,1)\phi(x) + \sum_{n=1}^{\infty} (n,n+1](2^n + 2^n \phi(x-n))$$ Then $f$ is continuous, $f'(x) = 0$ a.e. but as $f(n) = 2^n$ this function isn't Lipschitz. – t.b. Sep 06 '11 at 13:59
  • Sorry about posting the wrong link for the Cantor-Lebesgue function. To make my complaint more explicit: You're assuming that $f(x_2) = f(x_1) + \int_{x_1}^{x_2} f'(t),dt$ (which is equivalent to absolute continuity of $f$ by Lebesgue's differentiation theorem). The Cantor-Lebesgue function is the standard counterexample. – t.b. Sep 06 '11 at 14:22
  • @Theo, I clearly have much to learn. May I ask though, what is wrong with me using $f(x_2)-f(x_1) = \int^{x_2}_{x_1} f'(t) dt $ ? How have I misapplied the theorem here?: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Second_part . Also, is there any way to salvage or fix the current proof, or does it require a new approach? – Ragib Zaman Sep 06 '11 at 14:42
  • I see what you mean. Note, though, that $g' = f$ needs to hold everywhere, so $g$ is assumed differentiable everywhere and $f$ to be everywhere defined in the Wikipedia-notation. You can then extend the FTC in the Riemann-setting to piecewise differentiable functions, but not much further AFAIK. I'd have to think about a proper counterexample but I'm not terribly motivated (in view of the OPs other questions what you did is good enough). – t.b. Sep 06 '11 at 15:03
  • @Theo I put the current proof in an extra-answer.. So maybe to become sure what's happening with those different version of FTC exactly – Johannes L Sep 06 '11 at 15:09
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I try a correct fix of @Ragib's answer, then we can continue to discuss here

Let $A=\{t_0, t_1, \dots \}$. Define

$$\tilde{f}'(x) = \begin{cases} 0 & x \in A\\ f'(x) & \text{else} \end{cases}$$ $A$ has measure 0 (in other words $f$ almost everywhere differentiable). Then $\sup|\tilde{f}'(t)| \equiv \sup |f(t)| =L$.

$\tilde{f}'(t)$ has an anti-derivative $\forall x \in R$ and is Lebesgue-integrable, hence with the Fundamental Theorem of Analysis for Lebesgue-integral: $\int_{x_1}^{x_2} \tilde{f}'(x)=f(x_2)-f(x_1)$

In Wikipedia under generalizations it says "Part II of the theorem is true for any Lebesgue integrable function ƒ which has an antiderivative F (not all integrable functions do, though)."

Johannes L
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