Let $f(\theta, \epsilon)$ be smooth on $[0,2\pi] \times [0,\infty)$ such that

- $f(\theta, \epsilon)$ converges to $f(\theta, 0)$ uniformly as $\epsilon \rightarrow 0$.
- $f(\theta, \epsilon) > 0$ for $\epsilon > 0$, $f(\theta, \epsilon)$ has a unique minimum $\theta_{\epsilon}$ with $\frac{\partial^2 f}{\partial \theta^2}(\theta_{\epsilon}, \epsilon) > 0$ for every $\epsilon > 0$.
- $f(\theta, 0) = (\theta - \theta_0)^2 h(\theta)$ with $h(\theta) > 0$ for all $\theta \in [0,2\pi)$.

Then $f(\theta_{\epsilon}, \epsilon) \rightarrow f(\theta_0, 0) = 0$ and $\theta_{\epsilon} \rightarrow \theta_0$ as $\epsilon \rightarrow 0$.

Is it true that for every such function $f$ one can find a function $g: \mathbb{R} \rightarrow \mathbb{R}$, with $g(0)=0$ and $g(x) > 0$ for $x > 0$, such that

$$ \limsup_{n \rightarrow \infty} \frac{g(|\theta_{\epsilon_n} - \theta_0|)}{f(\theta_{\epsilon_n}, \epsilon_n)} <C<\infty? $$ for every sequence $\epsilon_n\rightarrow 0$, for some constant $C$ depending only the function $f$, and not the sequence $\epsilon_n$.

This is a follow up question to my previous post. In the previous post, which is fully answered, the question is wether a subsequence $\epsilon_n \rightarrow 0$ exists such that the limit is infinity for any such $g$. This question asks wether for any function $g$ there exists a sequence $\epsilon_n \rightarrow 0$ such that the limit is infinity. The question in this post seems more challenging to me. Intuitively, I think the answer is no, but I could be wrong.