I want to know $\|f\cdot \nabla^2f\|\leq \|\nabla f + \epsilon\|^2$ is true or not, where $f:\mathbb{R}^n\rightarrow\mathbb{R}$, $\|\cdot\|$ is norm operator, and $\epsilon$ is a small positive constant.
If it is true, please help me prove it. Otherwise, please give an counter-example.
Update: Any norm (2, Frobenius, ...) works for me. I am trying to prove $\left\Vert\nabla \frac{f}{\|\nabla f\|+\epsilon}\right\Vert\leq 1$. During the proof, I found that it is equivalent to prove $\|f\cdot \nabla^2f\|\leq \|\nabla f + \epsilon\|^2$.
No, such an inequality cannot hold for general functions.
There is a second-order term on the LHS, which has no reason to be bounded by the first-order one of the RHS. You can refer to my answer to your other question for a precise construction of a counter-example with $n=1$. The counter-example I constructed there also contradicts this inequality, as both are roughly equivalent as you established.