# $f'$ is bounded $\Rightarrow$ $f$ is bounded?

Let $I$ be a bounded connected subset in $\mathbb{R}$ and $f: I\rightarrow \mathbb{R}^k$ be a differentiable function.

Does boundedness of $f'$ imply boundedness of $f$?

(I edited this post after I realized that I didn't actually write what i wanted, after i saw Gautam's post.)

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Bounded interval would have sufficed. I'll edit my answer after some thought. – Gautam Shenoy Nov 22 '12 at 11:06

If you mean by interval a bounded interval, then the answer is yes because of the mean value theorem. If not then the identity function provides a counterexample.

Added explanation of the bounded interval case. We may assume (by adding a constant which does not affect boundedness of $f$ and even less that of $f'$) that $f(x_0)=0$ for some $x_0\in I$. The interval has some finite length $l$, and the mean value theorem implies that whenever $|f(x)|>lC$ for some $C$ then there exists $x'\in I$ such that $|f'(x')|=\frac{|f(x)|}{|x-x_0|}>C$, so that if $f$ is unbounded, $f'$ must also be unbounded.

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Yes i meant this. I know mean value theorem, but i don't know how to derive boundedness of $f$.Would you help? – Katlus Nov 22 '12 at 11:00
The mean value theorem he is referring to is the Lagrange's MVT. Before that, Rolle's MVT is derived and Lagrange's MVT is a special case of it. – Gautam Shenoy Nov 22 '12 at 11:08
@GautamShenoy: Your comment confuses me. The link I provided makes perfectly clear which theorem I'm referring to, and I would say that Rolle's theorem is a special case of it (which does not exclude it being used in a proof of the MVT) rather than the other way around. – Marc van Leeuwen Nov 22 '12 at 11:15
@Marc: You are right. What I meant was, Rolle's theorem was independently proved. And by a clever substitution, you get the generalization that is Lagrange MVT. If you see the proof, you'll see what I mean. – Gautam Shenoy Nov 22 '12 at 11:18
It's great. Thank you – Katlus Nov 22 '12 at 11:26

Consider f(x)=x. Over $\mathbb{R}$, it is unbounded but the derivative is constant(and hence bounded). However if you take a finite interval (or an interval contained in a compact set), then the function will be bounded NOT because the derivative is bounded but because the function is continuous.

Edit: I stand corrected. Given continuity and an open interval, a function can be unbounded. Differentiability, same story. Bounded derivative on the other hand will indeed imply boundedness of the function.

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Intervals can be open, and continous functions on them unbounded (think $\tan$ on $(-\frac\pi2,\frac\pi2)$). – Marc van Leeuwen Nov 22 '12 at 10:50
Same goes for differentiable functions. Additionally tan does not have a bounded derivative. – Gautam Shenoy Nov 22 '12 at 10:52
@Marc: If a function is differentiable in I, I is open and if it is unbounded, like your example, can it have a bounded derivative? – Gautam Shenoy Nov 22 '12 at 10:54
My first comment was (only) because you wrote "the function will be bounded NOT because the derivative is bounded but because the function is continuous". Answering your second comment, a function with bounded derivative on a bounded interval will be bounded, see my answer. – Marc van Leeuwen Nov 22 '12 at 10:56
@Gautam Would you give me another example? I know it is a theorem that if $I$ is compact and $f$ is continuous,then $f(I)$ is bounded. What if domain of $f$ is $(a,b)$ and $f'$ is bounded? ($-\infty<a<b<\infty$) – Katlus Nov 22 '12 at 10:58