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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 \} $ ?

Prabhjot Singh Rai
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  • You want $\lVert \nabla F\rVert$ not $\nabla\lVert F\rVert$ and sup rather than max. Give yourself some wiggle room and prove $\operatorname{Lip}F\geq\sup_{x\in\Omega}\lVert\nabla F(x)\rVert -\varepsilon$, arbitrary $\varepsilon>0$. – user10354138 Oct 28 '23 at 22:22
  • Sorry the former was a typo. The sup is I believe the largest value (similar to max)? The equation you mention is exactly what I am pondering how to prove. – Prabhjot Singh Rai Oct 28 '23 at 22:43

1 Answers1

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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.

Alp Uzman
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