# Is a Lipschitz continuous gradient equivalent to this condition?

I know if a function $$f: \mathbb{R}^n \to \mathbb{R}$$ is $$L$$-smooth, i.e. its gradient $$\nabla f$$ is $$L$$-Lipschitz continuous, then it satisfies the following inequality for any $$x, x_0 \in \mathbb{R}^n$$: $$\left| f(x) - \left( f(x_0) + \nabla f(x_0)^\top (x - x_0)\right) \right| \leq \frac{L}{2} \lVert x - x_0 \rVert_2^2 \text{.}$$

Here's my question: is the converse true? I know that, provided $$f$$ is convex, then we can conclude that $$f$$ satisfying the above inequality must be $$L$$-smooth. But what if $$f$$ is not necessarily convex? If the converse is false, can you give some counterexamples?

• I don't know off the top of my head, but this post xingyuzhou.org/blog/notes/Lipschitz-gradient does explicitly say that this implication does not hold without convexity (but does not give a counterexample).
– Dirk
Dec 20, 2022 at 16:04
• @Dirk Still very helpful, thank you!
– aest
Dec 20, 2022 at 17:29

In fact, your condition yields $$| (\nabla f(x_0) - \nabla f(x))^\top (x_0 - x) | \le L \| x_0 - x \|_2^2$$ and the linked answer then gives the Lipschitzness of $$\nabla f$$.