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 Jun 16 comment Functions about integers which are divisible by all numbers less than or equal to a fractional power of the integer There's an obvious (and unique) answer to your question : $f(n)$ is the smallest $m$ such that all integers $\leq \sqrt[m]{n}$ divide $n$. Jan 28 comment Counting square-free numbers smoothly There must be some typo since the sum in the question is $+ \infty$. Jan 11 awarded Necromancer Jan 11 revised Elementary applications of Krein-Milman deleted 40 characters in body Jan 10 awarded Teacher Jan 10 answered Elementary applications of Krein-Milman Dec 20 comment Existence of non-trivial affine functions on Hadamard spaces ? Yes, the "good" question was indeed "which Hadamard spaces have this property?". The comment of Igor Belegradek gives exactly what I was looking for. Thanks all. Dec 19 asked Existence of non-trivial affine functions on Hadamard spaces ? Oct 28 comment Test functions with small support and nonnegative Fourier transform An unexpectedly simple solution ; thanks ! I think this still applies in the noncommutative case (the Haar measure is left and right invariant) : If $VV \subset U$, $g_1$ is any map vanishing outside $V$, with nonzero mean, and $g_2 : x \mapsto g_1^{*}(x^{-1})$, then $f = g_1 \star g_2$ should work. Oct 28 asked Test functions with small support and nonnegative Fourier transform Oct 24 accepted The identity element of a compact group is a limit point of any “polynomial sequence” Oct 24 comment The identity element of a compact group is a limit point of any “polynomial sequence” I accept this answer since this is close to what I was looking for. It seems reasonable to expect that your proof extends to the general case (without applying the reduction given by Terence Tao) : given a neighbourhood $U$ of the identity, all what we need is a function $f$ with real, nonnegative fourier transform, with nonzero mean, and which vanishes outside $U$. I don't know if this is always possible. Oct 23 comment The identity element of a compact group is a limit point of any “polynomial sequence” Thanks. There's an argument (also due to Bergelson I think !) which shows that Van der Waerden's theorem implies the theorem in my question : if $U$ is a small neighbourhood of the identity, $G$ has a finite covering by some $(g_iU)_i$, and any map $n\in \mathbb{N} \mapsto$ some $i$ with $g^{P(n)} \in g_iU$ defines a colouring of the integers to which Van der Waerden's theorem can be applied. On then concludes using equations like $(x+2y)^2-2(x+y)^2+x^2=2y^2$ and its generalizations. But since I've never considered Van der Waerden's theorem as "elementary" ... Oct 23 comment The identity element of a compact group is a limit point of any “polynomial sequence” @GH: when $G$ is a torus, one can conclude by a standard Weyl's sums argument. But by "a straightforward generalisation" I meant "replace the torus by $G$ throughout the proof" (since noncommutativity causes no trouble in our ultralimits). @Terry Tao: Thanks for the elegant reduction. Your argument using Peter-Weyl theorem works also in the noncommutative case (which could arise if one considers instead sequences $n \mapsto g^{P(n)}h^{Q(n)}$ - where the ultralimit argument still applies), in which case we only have to consider (a finite list of) compact Lie groups. Oct 23 comment The identity element of a compact group is a limit point of any “polynomial sequence” For example : this is a straightforward generalisation of theorem 7.2 in the following article of Vitaly Bergelson math.osu.edu/~bergelson.1/VBContempMathUltrafiltersEtc.pdf . Oct 23 asked The identity element of a compact group is a limit point of any “polynomial sequence” Jul 20 comment A mixing property of linear map over finite fields Thanks ! It's amusing that the same constant $\frac{q^2}{2q-1}$ appears here and also as the limit of the polynomial method in the Kakeya problem. Jul 20 awarded Scholar Jul 20 accepted A mixing property of linear map over finite fields Jul 20 awarded Editor