# Show that this real function is Lipschitz continuous

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.

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@njguliyev Really??? shouldn't I ask for more to the $f$ function? like continuity or something? – Ana Galois Sep 3 '13 at 18:18
A differentiable function is already continuous. Bu if you mean the fundamental theorem of calculus, then use the mean value theorem instead. – njguliyev Sep 3 '13 at 18:19
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 3 '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 3 '13 at 18:35

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.
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 3 '13 at 18:29
@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 3 '13 at 18:53