Timeline for Decay rate of minimum point over a product space
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17 at 15:27 | comment | added | MathLearner | Here is the new post: mathoverflow.net/questions/469362/… | |
| Apr 17 at 15:21 | comment | added | MathLearner | Thank you. I will remove the edit and make a new post. | |
| Apr 17 at 15:20 | comment | added | Iosif Pinelis | @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. | |
| Apr 17 at 14:51 | comment | added | MathLearner | Thank you losif. I had a follow up question which I added in the body of the question. I would appreciate if you could take a look at it and let me know what you think. Thank you so much. I truly appreciate the time that you have spent answering my question:) | |
| Apr 17 at 13:56 | vote | accept | MathLearner | ||
| Apr 17 at 3:40 | comment | added | Iosif Pinelis | @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. | |
| Apr 17 at 3:02 | comment | added | MathLearner | 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 $g$. | |
| Apr 17 at 1:36 | comment | added | Iosif Pinelis | @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 $\epsilon_n\to0$. | |
| Apr 16 at 21:37 | comment | added | MathLearner | 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 $g'=\sqrt{g}$. | |
| Apr 16 at 20:37 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |