If $f>0$ and $(\ln f)'=f'/f$ is Lipschitz continuous, are we able to conclude that $(\ln f)'''$ is bounded?

Question

Let $f\in C^2(\mathbb R)$ be positive and $g:=\ln f$. Assume $$g'=\frac{f'}f$$ is Lipschitz continuous and hence $g''$ is bounded.

Is $g'''$ bounded too?

Answer 1

No, of course. Let $$u(x)=\begin{cases} e^{-x^2}x^2\sin x^{-2}&\text{if }x\ne 0\\ 0&\text{if }x=0\end{cases}$$ and consider $$f(x)=\exp\left(\int_0^x\,dt\int_0^t u(s)\,ds\right)$$

Then, $(\ln f)''=u$, which is bounded and continuous, but $$u'(x)=\begin{cases} 2(1-x^2)xe^{-x^2}\sin x^{-2}-\frac2xe^{-x^2}\cos x^{-2}&\text{if }x\ne 0\\ 0&\text{if }x=0\end{cases}$$ is unbounded.