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.

  • $\begingroup$ Perhaps, it makes sense to highlight the difference between this question and the previous one. $\endgroup$ Commented Apr 17 at 15:31
  • $\begingroup$ I think this should be further clarified. Of course, the matter here is the clear placement of quantifiers $\exists$ and $\forall$. $\endgroup$ Commented Apr 17 at 16:49
  • $\begingroup$ Note that the first assumption is superfluous . Since $f$ is continuous , it is uniformly continuous on $[0,2\pi]\times[0,T]$, so $\epsilon\mapsto f(\cdot,\epsilon)$ is continuous wrto the uniform norm. $\endgroup$ Commented Apr 17 at 17:27
  • $\begingroup$ @IosifPinelis It should be very clear now. $\endgroup$ Commented Apr 17 at 17:57

1 Answer 1


$\newcommand\ep\epsilon\newcommand\th\theta\newcommand\de\delta$The answer is yes, even with $C=1$ for all such $f$.

Indeed, let $f$ satisfy all your conditions on $f$. Let $g(0):=0$. For real $x>0$, let $$g(x):=1\wedge\inf\{f(\th_\de,\de)\colon\de\in E_x\},$$ where $1\wedge u:=\min(1,u)$ and $$E_x:=\{\de\in(0,1]\colon|\th_\de-\th_0|=x\}; $$ recall that $\inf\emptyset=\infty$.

Then $g(x)>0$ for all real $x>0$. Indeed, suppose the contrary: that $g(x)=0$ for some real $x>0$. Then $E_x\ne\emptyset$ and, moreover, there exist a sequence $(\de_n)$ in $E_x$ and some real $\ep\ge0$ such that $\de_n\to\ep$ and $f(\th_{\de_n},\de_n)\to0$ and $\th_{\de_n}\to\th$ for some $\th\ne\th_0$. So, then $$0=\lim_n f(\th_{\de_n},\de_n)=f(\th,\ep)>0,$$ since $\th\ne\th_0$ and $f=0$ only at $(\th_0,0)$. So, we get $0>0$, which proves that $g(x)>0$ for all real $x>0$.

Finally, take any $\ep\in(0,1]$. Then $\ep\in E_{|\th_\ep-\th_0|}$ and hence $g(|\th_\ep-\th_0|)\le f(\th_\ep,\ep)$, which implies $$\limsup_{\ep\downarrow0}\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}\le1.\quad\Box$$

Replacing $g(x)$ by $xg(x)$, we can even get $$\lim_{\ep\downarrow0}\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}=0.$$

  • $\begingroup$ This is great. Thank you so much for your help. $\endgroup$ Commented Apr 17 at 21:25

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