User safoura - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:28:53Z http://mathoverflow.net/feeds/user/17503 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130016/regularizing-a-convex-function-with-itself/130023#130023 Answer by Safoura for Regularizing a Convex function with itself Safoura 2013-05-07T21:00:53Z 2013-05-07T21:00:53Z <p>It is clear that there is no meaningful bound from below except for $0$ (just take $f(x)=x^2$ for 1D case). As to the bound from above, the distance can be made arbitrary large. To see that take $\gamma=\frac{1}{2}$ and let $f$ to grow sufficiently fast for $x\ge 0$ and $f(0)=0.$ For $x\le 0,$ make $f=g$ to decrease very slowly and attain minimum at some point $x=x_0$ which is very far from the origin and $g(x)>-f(-x)$ for $x\ne 0.$ Then $f(x)+f(-x)\ge 0$ with the only minimum point $x=0.$ </p> http://mathoverflow.net/questions/129843/integral-inequality-for-convex-function/129892#129892 Answer by Safoura for Integral inequality for convex function Safoura 2013-05-06T21:29:26Z 2013-05-06T21:29:26Z <p>The inequality is false in general. To see this, take $a=0,$ $b=1,$ $f(x)=x^2.$ Let $u(0)=0$ and $u(1)=1.$ Then our problem can be reformulated as follows: given that $\int_{0}^1u(x)dx=\frac{1}{2}$ show that $\int_{0}^1u^2(x)dx\ge\frac{1}{3}.$ Now take $u(x)=x+th(x)$ where $h(0)=h(1)=0$ and $\int_{0}^1h(x)dx=0$ to end up with the inequality $$t^2\int_{0}^1h^2(x)dx+2t\int_0^1xh(x)dx\ge 0$$ for all $t\in\mathbb{R}.$ Taking $t$ sufficiently small we get $\int_0^1xh(x)dx\ge 0$ for all appropriately chosen $h(x).$ It is now easy to choose $h$ in way that last inequality is false (just take something antisymmetric with respect to $x=\frac{1}{2}$). </p> http://mathoverflow.net/questions/108448/two-metrics-and-a-sequence-converging-to-two-points/108452#108452 Answer by Safoura for Two metrics and a sequence converging to two points. Safoura 2012-09-30T08:16:33Z 2012-09-30T08:16:33Z <p>Take $\mathbb{Q}$ and consider two different norms on it: one is simply Euclidean and another is $p-$ adic. Take the sequence $x_n=\frac{p^n}{p^n-1}.$ Clearly $x_n\to 1$ in Euclidean norm and $x_n\to 0$ in $p-$ adic.</p> http://mathoverflow.net/questions/108384/beta-mathbbn-vs-beta-mathbbz $\beta\mathbb{N}$ vs $\beta\mathbb{Z}$ Safoura 2012-09-29T04:10:38Z 2012-09-29T05:20:26Z <p>Just started learning the Stone-Cech compactification of discrete groups this week. My motivation comes from a question on $\beta\mathbb{Z}$. Surprisingly, I realized there are muchhhh more literature devoted to $\beta\mathbb{N}$ than to $\beta\mathbb{Z}$. I wonder why is that? After all, algebraically $\mathbb{N}$ is a semigroup while $\mathbb{Z}$ is a group, and as discrete topological spaces they are homeomorphic. From your experience, how far $\beta\mathbb{N}$ and $\beta\mathbb{Z}$ are different in behaviour? Also, is $\beta\mathbb{N}$ ( or ($\beta\mathbb{N}\setminus\mathbb{N}$) easier to deal with?</p> http://mathoverflow.net/questions/108001/a-basic-question-on-stone-cech-compactification-of-mathbbz A basic question on Stone-Cech compactification of $\mathbb{Z}$ Safoura 2012-09-24T19:22:33Z 2012-09-25T02:42:54Z <p>Can the identity isomorphism on the additive group $\mathbb{Z}$ be extended to a non-identity semigroup isomorphism on $\beta\mathbb{Z}$, and still preserves $\beta\mathbb{Z}\setminus\mathbb{Z}$? Obviously this extension is discontinuous. </p> http://mathoverflow.net/questions/129969/estimate-for-i-th-smooth-number-gap-between-consecutive-smooth-numbers Comment by Safoura Safoura 2013-05-07T16:29:11Z 2013-05-07T16:29:11Z You might find useful to look at Granville's and Friedlander's paper MR1253496 (95b:11086) http://mathoverflow.net/questions/108384/beta-mathbbn-vs-beta-mathbbz/108386#108386 Comment by Safoura Safoura 2012-09-29T07:28:15Z 2012-09-29T07:28:15Z @unknow: It's not what I am looking for. I appreciate your comment though. Thanks! As you also mentioned, the extension of the homeomorphism between $'matbb{N}$ and $\mathbb{Z}$, shows that for the topologist these two spaces are the same. I am wondering how differently can these two objects $\beta\mathbb{N}$ and $\beta\mathbb{Z}$ behave? I consider the Stone-Cech compactification of a discrete semigroup as a special well-studied case of a more general notion namely compact Hausdorff right topological semigroups. So to me it's not only a topological object but also algebraic. http://mathoverflow.net/questions/108384/beta-mathbbn-vs-beta-mathbbz Comment by Safoura Safoura 2012-09-29T05:32:01Z 2012-09-29T05:32:01Z Hi Yemon. Yes! The library has a 2012 copy of it! In fact, it's open in front of me right now, working on chapter 3! Seems like I need a quite good deal of this book! Thanks for mentioning it!