Mean value theorem for vectors I would like some help with the following proof(this is not homework, it is just something that my professor said was true but I would like to see a proof):
If $f:[a,b]\to\mathbb{R}^k$ is continuous and differentiable on $(a,b)$, then there is a $a< d < b$ such that $\|f(b)-f(a)\|\le\|f '(d)\|(b-a)$
Thanks for any help in advance.
EDIT: I forgot to mention that my professor said how it could be proven. He said that one could let u be a unit vector in the direction of f(b)-f(a) and go from there.
 A: Because $f$ is continuously differentiable,
$$
                 f(b)-f(a) = \int_{a}^{b}f'(t)\,dt
$$
and
$$
                   \|f(b)-f(a)\| \le \int_{a}^{b}\|f'(t)\|\,dt.
$$
The function $g(t)=\|f'(t)\|$ is a real, continuous function on $[a,b]$ and, therefore, by the mean value theorem for real functions, there exists $d \in (a,b)$ such that
$$
                     \int_{a}^{b}\|f'(t)\|\,dt = \|f'(d)\|(b-a).
$$
Therefore, there exists $d \in (a,b)$ such that
$$
                   \|f(b)-f(a)\| \le \|f'(d)\|(b-a).
$$
A: Consider 
$$
f(b) - f(a) = \int_a^b f'(t) dt.
$$
Take a dot product with $u$ on both sides, to get
$$
\| f(b) - f(a) \| = \left|\int_a^b u \cdot f'(t) \, dt \right|.
$$
Now suppose that $\|f'(d)\| < \|f(b) - f(a) \| / |b - a|$:
$$
\| f(b) - f(a) \| = \left|\int_a^b u \cdot f'(t) \, dt \right| \\
\le \int_a^b \|u\| \|f'(t)\| dt \\
=  \int_a^b \|f'(t)\| dt \\
< \int_a^b \|f(b) - f(a) \| / |b - a| dt \\
\le \|f(b) - f(a) \| / |b - a| \int_a^b  1 dt \\
= \|f(b) - f(a) \| 
$$
That's a contradiction.
A: Right.  Just define a new function $g(t)=f\bigl( a + t(b-a) \bigr)$ and use the mean value theorem for single-dimensional functions, noting that
$$g'(t) = \frac{\partial f}{\partial x^T}\bigl( a + t(b-a) \bigr) (b-a).$$
