TITLE: Prove or disprove by counterexample $\sup_{x\in \mathbb{R}^n}\frac{[\nabla (\nabla \cdot f)]\cdot f}{(1+|\nabla\cdot f|)^2}<\infty$
QUESTION [8 upvotes]: Let $f\colon \mathbb{R}^n\to \mathbb{R}^n$ be twice continuously differentiable. Prove or disprove by counterexample that
\begin{equation}
\sup_{x\in \mathbb{R}^n}\frac{[\nabla (\nabla \cdot f)]\cdot f}{(1+|\nabla\cdot f|)^2}<\infty,
\end{equation}
where $``\cdot"$ denotes the dot product, $``\nabla"$ denotes gradient, and $``\nabla \cdot"$ denotes divergence.
Any help is appreciated.
Here is my attempt: If we know that the Taylor series of $f(x)$ converges to $f(x)$, then I can show the claim at least for the scalar case. But I do not know how to do this rigorously for a general case.

REPLY [2 votes]: In the case $n=1$ we can take
$$f(x) = \int\limits_0^x\sin\left(t^2\right)\mathbb{d}t$$
for all $x \in \mathbb{R}$. For $k \in \mathbb{N}$ we get
$$\frac{ f''(\sqrt{2\pi k})f(\sqrt{2\pi k})}{\left(1+\vert f'(\sqrt{2\pi k})\vert\right)^2} = \frac{2 \sqrt{2\pi k}\cos \left(2\pi k\right)f(\sqrt{2\pi k})}{\left(1+\vert \sin\left(2\pi k\right)\vert\right)^2} = 2 \sqrt{2\pi k}f(\sqrt{2\pi k}),$$
and since
$$\lim_{x \to \infty} f(x) = \sqrt{\frac{\pi}{8}},$$
we get
$$\lim_{k \to \infty}2 \sqrt{2\pi k}f(\sqrt{2\pi k}) = \infty$$
and therefore
$$\sup_{x \in \mathbb{R}} \frac{ f''(x)f(x)}{\left(1+\vert f'(x)\vert\right)^2} = \infty.$$
For $n > 1$ you can get a counterexample by taking this function as the first component and $0$ as the other components.