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May 5, 2020 at 15:07 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 5, 2020 at 14:57 vote accept Hussain Rashed
Apr 5, 2020 at 14:57
Apr 3, 2020 at 17:24 comment added Zach Teitler * For $f(x) = 1/(1+x^2)$, $\operatorname{Var}(f)(x) = \max\{1/(1+x^2), 1-1/(1+x^2)\}$ for all $x$, and the maximum is $1-1/(1+x^2)$ for $|x| \geq 1$. The point is the same though.
Apr 3, 2020 at 14:37 comment added Zach Teitler At this point it may be a question for the authors.
Apr 3, 2020 at 1:36 comment added Hussain Rashed I am reading this paper arxiv.org/pdf/1711.06836.pdf, and it stated on page 16 that $D_0(X)$ is a subset of $D_h(X)$, and that is not clear to me.
Apr 3, 2020 at 1:22 comment added Zach Teitler Try an example: $X = E = \mathbb{R}$, $f(x) = 1/(1+x^2)$, so $\operatorname{Var}(f)(x) = 1 - 1/(1+x^2)$ for all $x$. Now whatever $K$ is, if $x \notin K$, then $\operatorname{Var}(f)|_{\mathbb{R}-K}(x)$ is still $1-1/(1+x^2)$. The problem is that you shouldn't just take all values $y$. You need to restrict to $y \in X-K$ as well. ... Can I ask you where this question came from?
Apr 3, 2020 at 1:04 history edited Hussain Rashed CC BY-SA 4.0
added 43 characters in body
Apr 3, 2020 at 1:01 comment added Hussain Rashed $\operatorname{Var}_E(f)(x)$ where $x \in X-K$ is the largest distance between $f(x)$ and $f(y)$ where $(x,y)\in E$, while $\operatorname{Var}_E(f)|_{X-K}(x)$ is the largest distance between $f(x)$ and $f(y)$ where $(x,y)\in E \cap (X-K)\times (X-K)$. How would that help?
Apr 3, 2020 at 0:03 comment added Zach Teitler Difference between $\operatorname{Var}(f)|_{X-K}$ and $\operatorname{Var}(f|_{X-K})$.
Apr 2, 2020 at 20:48 history edited Wlod AA CC BY-SA 4.0
a nasty typo corrected
Apr 2, 2020 at 20:41 history edited YCor CC BY-SA 4.0
formatting, changed tag
Apr 2, 2020 at 20:21 history asked Hussain Rashed CC BY-SA 4.0