I have a couple of questions regarding Lipschitz continuity:
1- If a function f is smooth, does it imply that \(\nabla f(x)\) is Lipschitz continuous?
2- If a function f is Lipschitz continuous, does it imply that \(\nabla f(x)\) is bounded ?
I think for both of them the inverse hold unconditionally, but I am not sure if this direction always hold. I believe for at least one of them there must be a condition on the open and convex set X
Thanks in advance
1 is true. for convex and differentiable functions, the two properties (smoothness and lipschitz gradients) are equivalent. see Lemma 2.4 in the notes.
(unbounded gradients are ok, they can still be lipschitz, as in the x^2 example)
2 is true, as mentioned in the other answer also and theorem 1.9 in lecture notes
1 is false. See
$$ f(x) = x^2 $$
Which is smooth with smoothness parameter 2 and has unbounded gradients.
2 true, see theorem 1.9 in lecture notes
Regarding 1st, I've mixed up the gradient of the function and function itself being Lipschitz. In the second case the gradients would need to bounded, so the function does not necessarily has to be Lipschitz when it's smooth. Sorry for that.