$C^\infty$ Lipschitz function with non-Lipschitz derivative

Question

Let $F \in C^\infty(\mathbb{R}^d, \mathbb{R}^d)$ be smooth and Lipschitz continuous. Is it true that $\nabla F$ is also Lipschitz continuous? The answer is negative if we only assume $F$ to be $C^1$ (see Lipschitz function with non-lipschitz derivative), but what about smooth functions?

Answer 1

The answer is still negative for smooth maps.

Consider $f(x)=\frac{\cos(x^2) +1}{2}$, $F(x) = \int_0^x f(t) \ dt$

You have

$$\begin{cases} F^\prime(x) &= f(x) = \frac{\cos(x^2) +1}{2}\\ F^{\prime \prime}(x) &= -x \sin(x^2) \end{cases}$$

$F^\prime$ is bounded, hence $F$ is Lipschitz. However $F^{\prime \prime}$ is unbounded and $F^\prime$ not Lipschitz. And $F$ is a smooth map.