I get stuck in the following exercise:
Let $A \subset \mathbb{R^n}$ an open set and $f:A\rightarrow\mathbb{R^m}$ differentiable. If $f$ is Lipschitz with $M$ constant then $||f'(x)|| \leq M $ for all $x \in A$.
My try:
Take $x, y \in A $ by MVT there is $c \in \overline{xy}$ such that
$$f(x) - f(y)= f'(c)(x-y)$$
Therefore
$$||f(x) - f(y)||= ||f'(c)(x-y)|| = ||f'(c)||\cdot||x-y|| $$
Here I get stuck, I'm not sure if it's the right way.
I would appreciate any help, Thank you.
On
I think I can give an argument if $f$ is $C^1$. Suppose $f'(q)>N$ for some $q$ in your domain. Since $U$ is open, there is a small ball $B_{\delta}(q)$ around $q$ and contained in $U$. We may take $\delta$ to be small enough so $f'(x)>N$ everywhere on $B_{\delta}(q)$, since $f'$ is continuous. Since balls are convex, take $x,y$ in $B_{\delta}(q)$ so that the line joining them passes through $q$. The function $g:\mathbb{R}\to\mathbb{R}$ by $g(t)=f((1-t)x+ty)$ is well defined, and $g(0)=f(x)$ and $g(1)=f(y)$. So by the one dimensional mean value theorem, there is a $\theta\in (0,1)$ such that $g'(\theta)=f(x)-f(y)$. But $g'(\theta)=Df((1-\theta)x+\theta y)\cdot(x-y)$. Since $(1-\theta)x+\theta y)$ is in the ball, taking absolute value gives $|f(x)-f(y)|>N|x-y|$, contradicting the our assumption on the constant $N$>
HINT: Work with the definition of the derivative instead and use the reverse triangle inequality. Then you might show that (for any $\varepsilon>0$) $\|f'(x)h\| \le (M+\varepsilon)\|h\|$ for $h$ sufficiently small.