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?

  • $\begingroup$ @Theo Oh. You're right, so I should rephrase like "at only countable points not differentiable", it's a bit due to translating to English while writing.. And thanks for correcting my title $\endgroup$
    – Johannes L
    Sep 6, 2011 at 11:01
  • $\begingroup$ You should also change the definition of L to $ L = \sup |f'(t)| $. $\endgroup$ Sep 6, 2011 at 11:03
  • $\begingroup$ @all: I cast a vote to close this question because of my misreading it. Please ignore that vote. Johannes: sorry about that. $\endgroup$
    – t.b.
    Sep 6, 2011 at 11:19
  • $\begingroup$ @Joh The title still says differentiable at only countably many points. $\endgroup$
    – Srivatsan
    Sep 6, 2011 at 12:31
  • $\begingroup$ @Srivatsan I want to expres that f is differentiable on $\mathbb{R} \setminus \{ t_0, t_1, \dots\}$ where $\{t_0, t_1, \dots \}$ is countable, so this should be okay. Before it said "almost everywhere differentiable" which Theo pointed out to not to be what I meant. $\endgroup$
    – Johannes L
    Sep 6, 2011 at 12:44

2 Answers 2


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?

  • $\begingroup$ 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) $\endgroup$
    – Johannes L
    Sep 6, 2011 at 11:19
  • $\begingroup$ 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. $\endgroup$
    – Johannes L
    Sep 6, 2011 at 12:54
  • $\begingroup$ @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 en.wikipedia.org/wiki/… $\endgroup$ Sep 6, 2011 at 13:23
  • $\begingroup$ @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. $\endgroup$
    – Rajesh D
    Sep 6, 2011 at 13:26
  • $\begingroup$ @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? $\endgroup$ Sep 6, 2011 at 13:28

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)."


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