I am working through the proof of finding the lipschitz of a continuously differentiable function. Specifically, trying to prove that it is equal to
$Lip(F) = \max \{ \lVert \nabla F(x) \rVert | x \in \Omega \} $
I am able to prove that $Lip(F) \leq \max \{ \lVert \nabla F(x) \rVert | x \in \Omega \} $ but having a hard time understanding the other part of the proof. How do I prove that $Lip(F) \geq \max \{ \lVert \nabla F(x) \rVert | x \in \Omega \} $ so that $Lip(F) = \max \{ \lVert \nabla F(x) \rVert | x \in \Omega \} $ ?
Hint: The mean value theorem gives one side:
$$|f(x)-f(y)|\leq |f'|_{C^0}|x-y| \implies \operatorname{Lip}(f)\leq |f'|_{C^0}.$$
For the other side it suffices to show that if $c>0$ is such that $|f(x)-f(y)|\leq c\,|x-y|$, then $|f'|_{C^0}\leq c$. Choosing $y=x+h$, and dividing both sides by $|h|$, one obtains
$$\dfrac{|f(x+h)-f(x)|}{|h|}\leq c.$$
One can then take the limit as $h$ goes to zero.