# Does this use the mean value theorem?

Let $f:[a,b] \to \mathbb{R}^n$ be a smooth curve such that the estimate $\|f'(t)\| \leq f'(t)$ holds. Show that the estimate $\|f(b) - f(a)\| \leq f(b) - f(a)$ also holds.

I was started to use the mean value theorem because this appears to prove this straight away but then I remembered that we only defined the mean value theorem for a function $f:D \to \mathbb{R}$ where $D \subset \mathbb{R}^n$, is there another way of doing this?

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For $v$ in $\mathbb{R}^n$, what does $||v|| \leq v$ mean? –  Pete L. Clark May 19 '12 at 13:31
Now you mention it I'm not entirely sure, I'm directly copying the question, would it make more sense if the function was from $f:[a,b] \to \mathbb{R}$? –  user26069 May 19 '12 at 13:42
More sense, certainly: for a real number $x$, $|x| \leq x$ holds iff $x \geq 0$ (in which case $|x| = x$). Then the hypothesis would be that the derivative is non-negative at every point, hence (yes, by the Mean Value Theorem) $f$ is weakly increasing, and in particular $f(b) \geq f(a)$, so $f(b) - f(a) \geq 0$, so $|f(b)-f(a)| \leq f(b) - f(a)$. Still, this seems like a weird way of writing things... –  Pete L. Clark May 19 '12 at 13:54
Yeah that is odd, thank you. So does the mean value theorem work for a function $f:[a,b] \to \mathbb{R}^n$? –  user26069 May 19 '12 at 14:31
The statement you are trying to copy might be that $\|f'\|\leqslant g'$ for $f:[a,b]\to\mathbb R^n$ and $g:[a,b]\to\mathbb R$ both smooth implies $\|f(b)-f(a)\|\leqslant g(b)-g(a)$. –  Did May 19 '12 at 19:27