# A particular proof of the mean value inequality for vector valued functions

I am asked to prove the mean value inequality for vector valued functions using the following fact:

Let $x,y,z\in \mathbb{R}^{m}$ and $a<b<c$ be real numbers. Then $\frac{z-x}{c-a}$ belongs to the closed line segment joining $\frac{y-x}{b-a}$ and $\frac{z-y}{c-b}$.

I must use this fact to show that if $a,b\in \mathbb{R}^{m}$, $f$ is a differentiable function on a neighbourhood of $[a,b]$ into $\mathbb{R}^{n}$, then there is an $x\in[a,b]$ such that $\|f(b)-f(a)\|\leq \|b-a\|\|f'(x)\|$.

I know how to prove the mean value inequality using the mean value theorem for functions of a single variable and the Cauchy Schwarz inequality (as in Rudin, for example). However, I do not know where the above fact (which is easy to prove) is used. I would be grateful for a hint. Thanks.

• Are you sure that the domain of $f$ is an interval (and not a multidimensional set)? In this case, the first part of your question might make a bit more sense. Also, you write $\|b-a\|$ and not $|b-a|$. – PhoemueX Feb 11 '18 at 8:11
• $[a,b]$ is an interval in $\mathbb{R}^{m}$, that is, all vectors in $\mathbb{R}^{m}$ of the form $(1-t)a+tb, t\in [0,1]$. – Arundhathi Feb 11 '18 at 10:26
• At least this is what I understood! Is there a different possible meaning for $[a,b]$ as a subset of $\mathbb{R}^{m}$? – Arundhathi Feb 11 '18 at 10:31