Yoni Rozenshein
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 Apr24 awarded Nice Answer Apr23 answered A problem that I'm not sure whether to use Weierstrass Approximation Theorem Apr22 accepted Haar's theorem for the rotation-invariant distribution on the sphere Apr22 answered Haar's theorem for the rotation-invariant distribution on the sphere Apr13 comment Haar's theorem for the rotation-invariant distribution on the sphere Thanks for your comment @user24142, you are right. In fact, the Lebesgue measure on $\mathbb R^n$ splits to $\mu \times$ (a measure on $\mathbb R$ with density $r^{n-2}$) or something like that. Apr13 comment Haar's theorem for the rotation-invariant distribution on the sphere Oh, I like this approach, but I don't understand something. Clearly $\mu \times m$ is invariant under orthogonal transformations. Why is it invariant under translations? Apr7 comment Haar's theorem for the rotation-invariant distribution on the sphere Thanks for the answer, but I am looking for something more elementary, i.e. without Lie theory or the general theory of compact groups, just this case of the sphere and orthogonal-matrix-invariance. Preferably in a book that I can just quote as in "it is a well-known fact that the Lebesgue measure is the unique rotation-invariant measure on the sphere [1]" :) Apr6 awarded Good Answer Apr6 answered Stuck on crucial step while computing $\int_{- \infty}^{\infty} e^{-t^2}dt$ Apr6 comment Stuck on crucial step while computing $\int_{- \infty}^{\infty} e^{-t^2}dt$ It's not a dirty trick, it's one of the most beautiful computations in mathematics :) Mar14 comment A functional equation: $4f(x)^3 +f(3x)=3f(x)$ Certainly there are more functions. For instance, $0$ on the rationals and $1/\sqrt 2$ on the irrationals. Mar12 revised Haar's theorem for the rotation-invariant distribution on the sphere added 3 characters in body Mar10 asked Haar's theorem for the rotation-invariant distribution on the sphere Mar4 comment If $X$ is a continuous random variable and $Y$ is a discrete random variable, is $P(X=Y) = 0$? @Lee You need some form of independence. Without independence, you cannot say much. It could be that they are not equal but still have a nonzero probability to be equal. Feb27 awarded Nice Question Feb26 comment Intuitive difference between a continuous map and a homeomorphism What does it mean that "$X$ has a singular point"? Singular points are defined for functions, not for spaces. If you mean isolated point - note that a simplex is connected, so its image is connected, therefore the image of the simplex must reside in one of the space's path-connected components. This leads to the very elementary (and important) fact that the singular homology of a space is the sum of homologies of its path-connected components. Particularly, if a continuous map from a simplex to $X$ touches an isolated point then it must be a constant function. Feb25 answered Intuitive difference between a continuous map and a homeomorphism Feb25 comment Show that $f\left(\lambda \right)=\lambda$ . The point here is that you have some sequence $x_n$ that converges to $\lambda$, and - never mind the recursion - you know that there is a function $f$ that when applied to the sequence, simply shifts the sequence to the left by one (i.e. $x_1, x_2, x_3\ldots$ becomes $x_2, x_3, x_4 \ldots$). Then of course $f(x_n)$ also converges to $\lambda$. Feb25 comment Show that $f\left(\lambda \right)=\lambda$ . It is defined in a recursive way. But it does not matter :) Feb25 answered Show that $f\left(\lambda \right)=\lambda$ .