Is this an analog of the mean value theorem for vector-valued functions? I was reading through various proofs of the multi-dimensional analogues of the mean value theorem. Suppose we have a $C^1$ function $f: \mathbb{R}^n \supseteq U \to \mathbb{R}^m$. I had thought there was a theorem that


*

*Given a ball $B\subset U$,  $x,y\in B$, there exists a point $\xi \in B$ such that $f(x)-f(y)= Df(\xi)(x-y)$,


although in general $\xi$ will not lie on the line between $x$ and $y$. But the way I had remembered to prove it is incorrect. Is this a theorem at all?
Another related theorem is that 
\2. Suppose $C\subset U$ is convex. Given $x,y \in C$, suppose $\|Df(\xi)\| \leq \eta$ for all $\xi $ on the line between $x$ and $y$. Then $\|f(x) - f(y)\| \leq \eta \|x-y\|$.
The proofs I'm reading for #2 are somewhat complicated (e.g. Rudin PMA pp. 113 and 219), and I wondered if there was a simpler one just leveraging #1.
Thanks for clarifying this for me.

There is nothing like your proposed theorem (1).  In one dimension, the MVT says that if you cover $d$ miles in $t$ hours, you must at some time have been traveling at $\frac dt$ miles per hour. But in more than one dimension this is false, because you can drive all over town and return home, and so have a net velocity of zero, without ever stopping the car.  In one dimension, say on a long driveway, you have less room to maneuver, and have to stop the car in order to turn it around.
A specific counterexample is $f : \Bbb R \to \Bbb R^2$ given by $t\mapsto (\cos t, \sin t)$.  Then $f(2\pi)-f(0) = (0,0)$, but $(0,0)$ is nowhere in the range of $Df$, since $\left|Df\right|$ is everywhere $1$.  Here the car is driving around in circles.  Even though the car never stopped, its net velocity after a complete trip around is still zero.
