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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 $\epsilon \rightarrow 0$. I wonder if $\theta_{\epsilon} - \theta_0$ can decay arbitrarily slow compared to $f(\theta_{\epsilon}, \epsilon)$ or not. In other words, does there exist a function $f$ satisfying the assumptions above and a subsequence $\epsilon_n \rightarrow 0$ such that

$$ \lim_{n \rightarrow \infty} \frac{g(|\theta_{\epsilon_n} - \theta_0|)}{f(\theta_{\epsilon_n}, \epsilon_n)} = \infty $$

for any function $g: \mathbb{R} \rightarrow \mathbb{R}$ with $g(0)=0$ and $g(x) > 0$ for $x > 0$?

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$\newcommand\ep\epsilon\newcommand\th\theta$Yes, $|\th_\ep-\th_0|$ can decay arbitrarily slowly compared to $f(\th_\ep,\ep)$.

Indeed, let $g\colon\Bbb R\to\Bbb R$ be any smooth function such that $g(0)=0$ and $g(x)>0$ for $x>0$. Suppose that $f(\th,\ep)=(\th-\ep)^2+\ep g(\ep)$ for $\th\in[0,2\pi]$ and real $\ep\ge0$. Then all your conditions on $f$ hold, with $\th_\ep=\ep$ and $f(\th_\ep,\ep)=\ep g(\ep)$, so that $$\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}=\frac{g(\ep)}{\ep g(\ep)}\to\infty$$ as $\ep\downarrow0$.

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  • $\begingroup$ Thank you for your answer. But the questions asks for the existence of an $f$ for which the limit is infinity for any $g$. In your example $f$ depend on $g$, and the limit will not be infinity if one takes $g'=\sqrt{g}$. $\endgroup$
    – MathLearner
    Commented Apr 16 at 21:37
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    $\begingroup$ @MathLearner : Then the question would not make any sense. For instance, one can always let $g$ be such that $g(|\theta_{\epsilon_n}-\th_0|)=f(\theta_{\epsilon_n},\epsilon_n)$ for some $\epsilon_n\to0$. $\endgroup$
    – Iosif Pinelis
    Commented Apr 17 at 1:36
  • $\begingroup$ I don't think so. Please note that $f(\theta_{\epsilon_n}, \epsilon_n)$ is not a function of the form $g(|\theta_{\epsilon_n}-\theta_0|)$ for some real valued function $g$. $\endgroup$
    – MathLearner
    Commented Apr 17 at 3:02
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    $\begingroup$ @MathLearner : Think again. You have $\theta_\epsilon\to\theta_0$ as $\epsilon\downarrow0$. So, you can pick a sequence $(\epsilon_n)$ converging down to $0$ such that $\theta_{\epsilon_n}$ converges to $\theta_0$ monotonically. So, for each large enough $n$ you can define $g(|\theta_{\epsilon_n}-\theta_0|)$ as $f(\theta_{\epsilon_n},\epsilon_n)$ and then extend this definition of $g$ to an entire right neighborhood of $0$ by (say) linear interpolation. $\endgroup$
    – Iosif Pinelis
    Commented Apr 17 at 3:40
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    $\begingroup$ @MathLearner : At this point, I don't know an answer to this latter question, but I will have it in mind. You may want to post this latter question separately. And I think this edit is not appropriate, after both versions of your question (the one initially intended and the one perceived) have been answered. $\endgroup$
    – Iosif Pinelis
    Commented Apr 17 at 15:20

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