The problem I am Having
Suppose $f:\mathbb E\mapsto \mathbb{\bar R}$, mapping from some euclidean space to augmented real with a well-defined gradient defined everywhere, and it has the smooth property:
$$ \exists L > 0 \text{ s.t: } |f(y) - f(x) - \langle \nabla f(x), y - x\rangle| \le \frac{L}{2}\Vert x - y\Vert^2 \quad \forall x, y\in \mathbb E, $$
then, is it possible that the function's gradient is globally Lipschitz? For generality, we may assume $\Vert \cdot \Vert$ is any norm, but proving it only for Euclidean norm suffices the argument.
For context, I am just curious because the converse of the statement is a simple proof, but this statement is not; I assume smoothness and then get a globally Lipschitz gradient without convexity assumption, and it makes me feel unease.
What I had Tried
A list of things I tried to solve it.
- Simple algebraic manipulations, thinking that taking supremum or Cauchy Schwartz inequality might help.
- Googling and some UCLA lecture slides are what I found to be the closest: here; however, they assume the convexity of the function to prove the statement.
- Looking for counter-examples, where some non-convex/non-concave function might be smoothly bounded above and below two quadratics and have non- globally Lipschitz gradient but failed.
Some help and pointer would be appreciated.