When can we assume a function is Lipschitz-continuous? I'm working with $f \colon X \subset \mathbb{R}^{N} \longrightarrow \mathbb{R}$ for $X$ compact. With a little search, seems that 2 results are the most usual:
- If $f$ is continuous differentiable, $f$ is Lipschitz-continuous. (Demostration for 1 variable is easy to find, for example at Continuous differentiability implies Lipschitz continuity, but not sure how it would work on several variables). It shouldn't be a tough task but I'm having doubts.
My attempt of prove it.
As $D_{f}(x)$ is continous and $X$ is compact, $D_{f}(x)$ is bounded. Being so, if $D_{f}(x)$ is also its gradient, proof becomes the same as made in 2-.
So the question would be: is the linear function $D_{f}(x)$ its gradient too? Or not the same?
- If $\nabla f$ is well defined and it's bounded, $f$ is Lipschitz-continuous. As Mean Value Theorem applies with several variables, I guess 2-. can be prooved using it.
My try
$f(x) - f(y) = \nabla f(\xi) \cdot (x - y)$ where $\xi \in [x, y] \subset X$. By Cauchy-Schwartz' inequality, it is hold that $|f(x) - f(y)| \leq ||\nabla f(\xi)|| \ ||x - y||$ and, as gradient is bounded, it is possible to conclude that $|f(x) - f(y)| \leq k ||x - y||$.
However, MVT applies for open sets, so I don't know if it would be correct for the kind of functions I'm working with.
Are those proofs right? Am I missing something? Is there more sufficient conditions?
Thanks.
Edit: Answer given in $C^1$ function on compact set is Lipschitz by user RobertLewis looks like a proof without MVT.