Equivalent statement for bounded derivative implies Lipschitz in $\mathbb{R^n}$. Let $B_r(x_0)$ be a ball in $\mathbb{R^l}$ and $f\in C^1(B_r(x_0)$,$\mathbb{R^m})$ such that $||Df(x)||_L\le M$ for all $x\in B_r(x_0)$. Then, I want to prove that for $x,y\in B_r(x_0)$,
$$|f(x) - f(y)| \le M|x-y|$$
So I want to use the function $F(t)=f(x+t(y-x))$ and the say that $F(1)-F(0)=\int_0^1F'(t)dt $ so the I have that 
$$f(y)-f(x)=\int_0^{1} f'(x+t(y-x))(y-x)dt$$
and we have that $||f'(x)||_E<||f'(x)||_L$ because by definition we get $||Tx||_L=\inf\{ M :|Tx|\le M|x|\}$ so,
$$|f(y)-f(x)| \le (M)(r)|(y-x)|$$
But I think there is something wrong in the derivative.
Can someone help me to fix the problems of my proof, or provide another proof unsing the same idea please?
Thanks a lot in advance.
 A: By chain rule, we have 
$$F'(t) = DF (x + t(y-x) ) \cdot (y-x).$$
Thus 
$$\|F'(t)\| = \|DF (x + t(y-x) ) \cdot (y-x)\| \le \|DF (x+ t(y-x))\|_L \ \|y-x\|$$ 
(see the remark) and 
$$\begin{split}
\|f(y)-f(x)\| &\le \int_0^1 \|F'(t)\| dt \\
& \le \int_0^1 \|DF (x+ t(y-x))\|_L \ \|y-x\| dt \\
&\le \int_0^1 M\|y-x\| dx \\
&= M\|y-x\|.
\end{split}$$
Remark Note that as 
$$\|DF(x)\|_L = \inf\{M : \|DF(x) (v)\|\le M \|v\|\  \forall v\}$$
we have 
$$\|DF(x) (v)\|\le M \|v\|$$
for all $M > \|DF(x)\|_L$. Now for any $v\neq 0$, 
$$\frac{\|DF(x) (v)\|}{\|v\|}\le M. $$
for all $M > \|DF(x)\|_L$. So we have 
$$\frac{\|DF(x) (v)\|}{\|v\|}\le \|DF(x)\|_L. $$
(To see this, try to show that $>$ is impossible). Multiply $\|v\|$ on both sides gives
$$\|DF(x) (v)\|\le \|DF(x)\|_L \ \|v\|$$
for all $v\neq 0$. But the same equality holds also when $v = 0$. Thus it holds for all $v$. In particular for $v = y-x$, we have 
$$\|DF(x) \cdot (y-x)\|\le \|DF(x)\|_L \ \|y-x\|.$$
A: By the chain rule you have
$$F'(t)=df\bigl(x+t(y-x)\bigr).(y-x)\qquad(0\leq t\leq1)\ ,$$
whereby $F'(t)$ is a vector. By definition of the norm $\|A\|$ of a linear map $A$ one has $|A.x|\leq\|A\|\>|x|$ for all vectors $x$. In the case at hand this implies that
$$\bigl|df\bigl(x+t(y-x)\bigr).(y-x)\bigr|\leq \|df\bigl(x+t(y-x)\bigr)\|\>|y-x|\ .$$ Given your assumption on $df$ we therefore may write
$$\bigl|F'(t)\bigr|\leq\|df\bigl(x+t(y-x)\bigr)\|\>|y-x|\leq M\>|y-x|\qquad(0\leq t\leq1)\ .$$
This implies
$$\bigl|f(y)-f(x)\bigr|\leq\int_0^1\bigl|F'(t)\bigr|\>dt\leq M\>|y-x|\ .$$
There is not more to it.
