My question is:
Which of the following is more restrictive? $\nabla f$ Lipschitz & $f$ Lipschitz
I think each one cannot imply the other.
For example ($1$D): $$f(x) = \frac {x^2}{3}$$
In this case, $f(x)$ is not Lipschitz; however, $\nabla f = \frac{2}{3}x $ is Lipschitz.
For example ($1$D):
$$f(x) = |x|$$
In this case, $f(x)$ is Lipschitz; however, $\nabla f$ is not.
Is this correct?
Here, I just give two example. Can I say this just by two examples or is there any deeper discussion in this question?
Locally, the continuity of $\nabla f$ implies that $f$ is Lipschitz. For example, the Mean Value Theorem implies that in on a convex set, $$ f(x)-f(y)=(x-y)\cdot\nabla f(\xi)\tag{1} $$ for some $\xi$ on the line between $x$ and $y$. Thus, $$ |f(x)-f(y)|\le|x-y|\sup_\xi|\nabla f(\xi)|\tag{2} $$
So in a bounded subset of $\mathbb{R}^n$, $\nabla f$ being Lipschitz implies that $\nabla f$ is bounded and $(2)$ implies that $f$ is Lipschitz. Therefore, $\nabla f$ being Lipschitz is more restrictive.
However, as your first example shows on an unbounded subset of $\mathbb{R}^n$, even a $C^\infty$ function need not be Lipschitz.