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I have the question:

Let $I=[0,1]$ be the closed interval, $f:I\to\mathbb{R}^n$ and $g:I\to\mathbb{R}$ differentiable, with $|f(t)|\le g'(t)$, for all $t\in I$. Show that $|f(1)-f(0)|\le g(1)-g(0)$

The book suggests to use the same trick in proving the Mean Value Theorem: for any $\epsilon>0$, define $X=\{t\in I:|f(t)-f(0)|\le g(t)-g(0)+\epsilon t+\epsilon\}$ and prove that if $\alpha\in X$, with $\alpha<1$, then there exists $\delta>0$, such that $\alpha+\delta\in X$ (so, since it's easy to prove that $X$ is a interval and $\sup X\in X$, we have the question).

Sincerely I have no progress...
The condition $|f(t)|\le g'(t)$ is quite annoying, and I have no idea how to manipulate things to it be useful...
I tried the basic: let $\alpha\in X,\alpha<1$, so, since $f$ is differentiable, we can find $\delta>0$ and write $|f(\alpha+\delta)-f(\alpha)|\le|f'(\alpha)\delta+r(\alpha)|\le|f'(\alpha)|\delta+\epsilon\delta$, where $\lim_{\delta\to0}r(\alpha)/|\delta|=0$. If we guarantee that $|f'(\alpha)|\delta\le g(\alpha+\delta)-g(\alpha)$, we prove the suggestion...
Also, if we assume that $|f(\alpha)| (and see what we do with the case $|f(\alpha)|=g'(\alpha)$ later), then, we can find $\delta>0$ such that $|f(\alpha)|\delta, so, in this case, we have a relation between $|f|$ and $g$ (but doesn't seem to help...)

Any hint will be appreciated! Thanks!

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    Thanks @JoelCohen! Ow! I was thinking in this condition (|f'|, the best I could), but your solution is better and beautiful and simple! =p2012-06-10

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Take $f,g:I \to \mathbb{R}\,\,,\,f(t):=t^2\,,\,g(t)=\frac{t^2}{2}$Then, for $\,t\in I\,\,,\,|f(t)|=t^2\leq t=g'(t)\,$ , but nevertheless $f(1)-f(0)=1\rlap{/}{\leq} \frac{1}{2}=g(1)-g(0)$So either you ommited some condition or the above contradicts the claim (or, of course, I missed something)

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    Thanks very much, @DonAntonio!! And these are exactly the conditions (I've checked now!). Maybe the book was wrong... (should I change the question for "prove or find a counterexample?")2012-06-10