Let $f$ and $g$ differentiables such that $|f'(t)| \le g'(t)$, for all $t \in [0,1].$ Prove that $|f(1)-f(0)| \le g(1)-g(0)$ Let $f:[0,1] \rightarrow \mathbb{R}^m $ and $g:[0,1] \rightarrow \mathbb{R}$ differentiables such that $|f'(t)| \le g'(t)$, for all $t \in [0,1].$ Prove that 
$$|f(1)-f(0)| \le g(1)-g(0)$$
Comments Chose $\epsilon > 0$ and define the set $X= \{ t \in [0,1]; |f(t)- f(0)|\le g(t)-g(0) + \epsilon t + \epsilon \}$.
Talking $0<a<1$, $a \in X$, I want to show that there $\delta >0 $ such that $a+ \delta < 1$ then $a+ \delta \in X$, soon we would have $1 = sup X$
To show I used the inequalities:
$|f(a)-f(0)| \le g(a) - g(0) + \epsilon a + \epsilon$
I want to prove that: 
$|f(a+ \delta)-f(0)| \le g(a+ \delta) - g(0) + \epsilon (a + \delta) + \epsilon$
But $|f(a+ \delta)-f(0)| \le |f(a+ \delta) - f(a)|+|f(a) - f(0)|$, therefore, it is shown that $|f(a+ \delta) - f(a)| \le g(a+ \delta) - g(a) + \epsilon \delta$  try the exercise.
Thanks for the help and excuse me any errors in traduação into English.

The key tool here is the mean value theorem. 
Let $h \in \mathbb{R}^m$ such that $\|h|| = 1$ and consider
$\phi_h(t) = g(t)-g(0)-\left( h^T (f(t)-f(0)) \right)$. Note that $\phi_h(0) = 0$ and $\phi_h$ is differentiable and
$\phi_h'(t) = g'(t)-h^T f'(t)$.
Note that $g'(t) \ge 0$ and so the
mean value theorem shows that
$g(t) \ge g(0)$ for all $t \in [0,1]$
Since $h^T f'(t) \le \|f'(t)\| \le g'(t)$ we see that $\phi_h'(t) \ge 0$ for all $t \in [0,1]$. The mean value theorem shows that $\phi_h(t) \ge 0$.
If $f(1)=f(0)$ there is nothing to show as the result is trivially true.
Otherwise, let $h = {f(1)-f(0) \over \|f(1)-f(0)\| }$, then
$\phi_h(1) \ge 0$ yields the desired result.

This was the answer to the original question:
This is not true.
Let $g(t) = t$, $f(t) = 2t-1$. Then $|f(t)| \le 1 = g'(t)$ for $t \in [0,1]$ but clearly $|f(1)-f(0)| = 2 > g(1)-g(0) = 1$.
