countinuty & differentibility Show that if $F :\Bbb R\to \Bbb R^n$ is continuous on $[a,b]$ and diﬀerentiable on $(a,b)$, then there is a $c\in(a,b)$ such that $||F(b)−F(a)|| \le||F'(c)||(b−a)$.
I am not sure whether I should use mean value theorem or not.
 A: You cannot use the mean value theorem for vector valued functions.
But you can argue as follows: Let $v = F (b)-F (a) $. Then consider the function
$$
f (x) := \langle  F (x), v \rangle 
$$
for which you can apply the mean value theorem. 
For the resulting bound,  apply the Cauchy-Schwarz  inequality. 
Note though that the above assumes that you use the usual euclidean norm on $\Bbb {R}^n $. For other norms, you will have to use a (finite dimensional) variant of the Hahn-Banach theorem to obtain a functional $\phi $ on $\Bbb {R}^n $ with norm $\Vert \phi \Vert  \leq 1$ and $\phi (v)= \Vert v\Vert $.
A: I want to know are there any problems in my answer. Thank you!
Let ϕ(t)=(F(b)-F(a))⋅F(t) for t∈[a,b]
When t=a, ϕ(a)= (F(b)-F(a))⋅F(a) =F(b)⋅F(a)-||F(a)||^2
When t=b, ϕ(b)= (F(b)-F(a))⋅F(b) =||F(b)||^2-F(b)⋅F(a)
as F is continuous on [a,b] and diﬀerentiable on (a,b)
ϕ is continuous on [a,b] and diﬀerentiable on (a,b)
By mean value theorem, there exist c∈(a,b) such that
ϕ(b)-ϕ(a) = (ϕ'(c))(b-a)
Because ϕ(b)-ϕ(a) = ||F(a)||^2+||F(b)||^2 - 2F(a)⋅F(b)=(F(b)-(F(a))⋅(F(b)-(F(a))
=||F(b)-F(a)||^2 
and
(ϕ'(c))(b-a)=(F(b)-F(a))⋅F'(c)(b-a)
So||F(b)-F(a)||^2 =(F(b)-F(a))⋅F'(c)(b-a)
||F(b)-F(a)||^2 =||(F(b)-F(a))||||F'(c)||(cosθ)(b-a) where θ is the angle between F(b)-F(a) and F'(c)
Hence, ||F(b)-F(a)||=||F'(c)||(cosθ)(b-a)
and as -1≤cosθ≤1, (b-a)>0 
||F(b)-F(a)||≤||F'(c)||(b-a).
