# Mariano Suárez-Alvarez

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bio website location Buenos Aires, Argentina age 39 member for 3 years, 7 months seen 1 hour ago profile views 13,268

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 7h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? $\int_{-\infty}^\infty e^{-|x|}e^{-i0x}=2\int_0^\infty e^{-x}$ because the integrand in the first integral is an even function. You can surely compute this last integral. 9h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? That function is equal in $L^1$ to $e^{-|x|}$ (for exactly the same reason as before) and its transform is (a scalar multiple of) $1/(1+x^2)$. 9h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? Ok, I will simply stop now. 10h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? That function and $1/(1+x^2)$ have the same values, except on a countable set. They are therefore almost equal in the sense of measure theory, with respect to Lebesgue measure. It follows that one is in $L^1$ exactly when the other is, and $1/(1+x^2)$ is very much in $L^1$. 11h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? Well, I am somewhat fluent in questions and answers here, and to me it is not at all clear what relation that function has here and —much less, really— what is the purpose of «all math tricks acceptable» announcement. 11h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? That function is equal to $1/(1+x^2)$ in $L^1$. As I said above, it is not at all clear what exactly you are trying to do with it. 12h comment Isomorphic fundamental groups result in homeomorphism? You can wedge with a wedge of more-than-points-in-the-space intervals. 13h comment Sigma Algebra: Etymology Rather, $\sigma$ tends to stand for «countable» in this context. Other examples are $\sigma$-finite, $\sigma$-compact, and so on. 14h comment Construction of transitive group of degree $n$ Well... that is why I suggested you be more precise about what you want! 14h comment Construction of transitive group of degree $n$ For example: make a list of all subgroups of $S_6$ (this is a finite amount of work), now drop from the list those which are not transitive, and finally, drop those which do not have a special block system. 14h comment Construction of transitive group of degree $n$ You probably want to ask something more precise. Otherwise, the answer is yes: just make a list of all transitive groups of degree $n$ and filter out the ones you do not like. 18h answered In general is it true that $||v-u||= ||u-v||$? Thanks! 18h comment Why are we interested in irreducible representation but not faithful representation? Hm? I have no idea what you just wrote. The study of group representations considers all representations, because all representations arise in practice; even the trivial representation plays an extremely important role, although this may seem rather surprising to anyone starting. Maybe you can just withhold judgment until you know a little bit more about the subject? That plan usually does marvels! 18h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? @Darius, you should probably read carefully what saz has written before continuing to deduce strange things from it. 18h comment Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? I canno tmake sense of the last sentence. What possible relevace does that function have to this queestion? 18h comment Why are we interested in irreducible representation but not faithful representation? A non faithful representation of $G$ is the samee thing as a faithful representation of $G/\ker h$... 22h answered Ring Isomorphism of Subfield to Matrix Space 23h comment Why is $\pi_1(X,x_0)$ a group? Say $F$: it is a map defined on $[0,1]\times[0,1]$ by «by pieces». What can you do to show that such a map is continuous? 23h comment Why is $\pi_1(X,x_0)$ a group? Why can't you prove that the maps are contiuous? What have you tried? 1d comment Why is ring addition commutative? @MartinBrandenburg, I couldnot agree more. The periodic downvotes on this answer reflect that situation :-)