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 $|f(\alpha)|<g'(\alpha)$ (and see what we do with the case $|f(\alpha)|=g'(\alpha)$ later), then, we can find $\delta>0$ such that $|f(\alpha)|\delta<g(\alpha+\delta)-g(\alpha)$, so, in this case, we have a relation between $|f|$ and $g$ (but doesn't seem to help...)
Any hint will be appreciated! Thanks!