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Jul
24
comment existence of a subgroup of a solvable group G of order d for any divisor d of |G|, with d<|G|^(1/2)
@OfirSchnabel I wasn't sure whether $6 < \sqrt{|A_4|5}$, but I was quite sure that $6 < \sqrt{|A_4|101}$.
Jul
24
answered existence of a subgroup of a solvable group G of order d for any divisor d of |G|, with d<|G|^(1/2)
Jul
10
awarded  Civic Duty
Jul
3
comment If $x_{n+1}\leq x_n + 1/n^2$ then $x_n$ converges
You should think of $x_n$ as approximately decreasing.
Jun
2
answered Extension of $|\cdot|_\infty$ on $\mathbb R$ to $\mathbb C$
May
12
comment Steinhaus theorem for topological groups
I just mean if you take any compact neighbourhood $K$ of the identity and let $H$ be the subgroup of $G$ generated by $K$ then $H$ is an open $\sigma$-compact subgroup of $G$. The Haar measure of $G$ is then just the sum of its restrictions to the cosets of $H$, so by translating $A$ and $B$ we may assume they intersect $H$ in sets of positive measure. In effect we may assume that $G$ is $\sigma$-compact so certainly $\sigma$-finite.
May
12
comment Steinhaus theorem for topological groups
The integrand is zero for $(x,y)$ outside the compact set $AB\times A$, so such problems do not arise. Alternatively you can always just restrict to a subgroup $H$ of $G$ generated by a compact neighbourhood of the identity: this is an open $\sigma$-compact subgroup.
May
11
answered Steinhaus theorem for topological groups
May
11
comment Steinhaus theorem for topological groups
@DavidChan Approximate $\chi_A$ and $\chi_B$ in $L^1$ norm by continuous and compactly supported functions. After convolution the $L^1$ error becomes an $L^\infty$ error, so $\chi_A\ast\chi_B$ is a uniform limit of continuous functions, so continuous. The convolution operation has many fundamental smoothing properties like this one.
May
7
answered Strengthening the intermediate value theorem to an “intermediate component theorem”
May
3
comment Steinhaus theorem for topological groups
This result is sometimes called Steinhaus's theorem. en.wikipedia.org/wiki/Steinhaus_theorem
May
3
comment Steinhaus theorem for topological groups
@WilliamCurtis Choose some $x_0$ such that $f(x_0)>0$. There is some neighbourhood $U$ of $x_0$ such that $f>0$ on $U$. That means that for every $x\in U$ there is some $y$ such that $xy^{-1}\in A$ and $y\in B$; in other words $x\in AB$.
May
3
revised Steinhaus theorem for topological groups
Corrected the definition of convolution.
Apr
26
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
Thanks, yes, I meant the modulus of $y$. It's fixed now.
Apr
26
revised The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
added 2 characters in body
Apr
24
answered The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
Apr
24
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
@G.Sassatelli You're probably right. If so then after rescaling this is the same as the limit of $\lambda_n(B_1(x)\setminus B_1(y))$ as $y\to x$, and now it's clear.
Apr
24
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
I think the statement as written is false. WLOG $x=0$ and $y=e_1$. Then we're measuring the set of points which are a distance $r$ from $0$ but not from $e_1$. This set contains a half ball of radius $1$ centred at $-(r-1)e_1$, so it has measure at least half the measure of the unit ball.
Apr
22
answered Show that there exists an entire function $h$ such that $\lim_{n\to\infty}{h(nz)}=0$ for all $z\ne0$
Apr
16
comment Difference between pushforward and differential
@IberêKuntz You can certainly use another map $g$ (instead of $f$, to reduce notation clash) for the identification $T_q\mathbb{R}\cong \mathbb{R}$ if you like, provided that $g'(q)=1$. Then you'll find that, at $q=f(p)$, $f_*(X)(g) = g'(q) X(f) = X(f)$.