# Rate of convergence of the minimum point over a product space

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

$$\limsup_{n \rightarrow \infty} \frac{g(|\theta_{\epsilon_n} - \theta_0|)}{f(\theta_{\epsilon_n}, \epsilon_n)} for every sequence $$\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.

• Perhaps, it makes sense to highlight the difference between this question and the previous one. Commented Apr 17 at 15:31
• I think this should be further clarified. Of course, the matter here is the clear placement of quantifiers $\exists$ and $\forall$. Commented Apr 17 at 16:49
• 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. Commented Apr 17 at 17:27
• @IosifPinelis It should be very clear now. Commented Apr 17 at 17:57

$$\newcommand\ep\epsilon\newcommand\th\theta\newcommand\de\delta$$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 $$\lim_{\ep\downarrow0}\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}=0.$$