# Decay rate of 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 $$\epsilon \rightarrow 0$$. 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 $$x > 0$$?

$$\newcommand\ep\epsilon\newcommand\th\theta$$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 $$\ep\downarrow0$$.
• 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 $g'=\sqrt{g}$. Commented Apr 16 at 21:37
• @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 $\epsilon_n\to0$. Commented Apr 17 at 1:36
• 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 $g$. Commented Apr 17 at 3:02
• @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. Commented Apr 17 at 3:40