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I have this excersize:

Let $I$ be an interval in $\Bbb R$ and $f:I\to \Bbb R$ a differentiable function such that $sup_{x\in I}|f'(x)|<\infty$. Show that $f$ is Lipschitz continuous.

Well, I know that I have to show a $c>0$ such that $|f(x)-f(y)|\le c|x-y|$, $\forall \;x,y\in I$, also we have that $|f'(t)|\le k$, for some $k\in \Bbb R^+$. I want to use this: $f(a)-f(b)=\int _a^b f'(t)dt$, that way the proof writes itself: $$f(a)-f(b)=\int _a^b f'(t)dt$$ $$\Rightarrow |f(a)-f(b)|=|\int _a^b f'(t)dt|\le |\int _a^b |f'(t)|\;dt| \le |\int _a^b k\;dt|=|k(b-a)|$$ $$\Rightarrow |f(a)-f(b)| \le k\;|b-a|$$ However, I don't know if I can use this, I feel something's missing.

Ana Galois
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  • @njguliyev Really??? shouldn't I ask for more to the $f$ function? like continuity or something? – Ana Galois Sep 03 '13 at 18:18
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    A differentiable function is already continuous. Bu if you mean the fundamental theorem of calculus, then use the mean value theorem instead. – njguliyev Sep 03 '13 at 18:19
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    If we're picky: Do you know that $f'$ is locally integrable, and that $f(b) - f(a) = \int_a^b f'(t),dt$? In what you've quoted, you have no continuity properties of $f'$ given. So that the given premises imply that $f$ is the integral of its derivative is not trivial. – Daniel Fischer Sep 03 '13 at 18:24
  • Somebody, I think it was @njguliyev, suggested earlier that you use the mean value theorem (of differential calculus, no integration). I think that's the way to go, justifying integrability of $f'$ from the given premises is probably beyond what has already been covered. – Daniel Fischer Sep 03 '13 at 18:35

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I think that your proof is not correct, because you are using that the derivative is Riemann integrable. If $f'$ were continuous, then you could apply the fundamental theorem of calculus (like you did) and the result would follow.

But there are examples of differentiable functions with bounded derivative, such that their derivative is not Riemann integrable. One of these examples is Volterra's function.

Aldo Guzmán Sáenz
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  • well the only thing that i got for this problem is that $sup_{x\in I}|f'(x)|<\infty$ but other than that I don't have anything else. Would that also mean that what they're asking me is not true? or not enough? – Ana Galois Sep 03 '13 at 18:29
  • Perhaps you should state explicitly that you're referring the the Riemann integral there. A bounded derivative that exists everywhere is (locally) Lebesgue integrable. – Daniel Fischer Sep 03 '13 at 18:30
  • @AnaGalois The result is true, but using the integral is not the way (with these hypothesis). An option for your problem is to use the mean value theorem, as in this answer: http://math.stackexchange.com/a/382820/92139 . – Aldo Guzmán Sáenz Sep 03 '13 at 18:32
  • @DanielFischer, you are right, I'll edit the answer. – Aldo Guzmán Sáenz Sep 03 '13 at 18:32
  • @AnaGalois Note that the answer I linked to deals with a closed interval, not the whole $\mathbb{R}$. – Aldo Guzmán Sáenz Sep 03 '13 at 18:39
  • @AldoGuzmánSáenz but I'm also working in an interval, the definition that we use in class, is that $I$ is an interval if $(1-t)x+ty\in I$ with $x,y\in I$ and $t\in [0,1]$, or what do you mean? – Ana Galois Sep 03 '13 at 18:47
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    @AnaGalois I just wanted you to remember that the defintion of Lipschitz continuity requires a single constant that works for the whole domain ($\mathbb{R}$), sometimes we forget that kind of thing. – Aldo Guzmán Sáenz Sep 03 '13 at 18:53