User km - MathOverflow

Answer by km for Elementary applications of Krein-Milman

2013-01-10T22:21:28Z

One can prove by using repeatedly the Krein-Milman theorem that

If $T : C[0,1] \rightarrow X$ is an operator of norm at most one, with $T$ isometric on the $2$-dimensional subspace spanned by $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is an isometry.
If $T$ is an endomorphism of $C[0,1]$ of norm at most one, which fixes the functions $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is the identity operator.

Existence of non-trivial affine functions on Hadamard spaces ?

2012-12-19T22:54:41Z

Let $X$ be a Hadamard space. Any two points $x$ and $y$ of $X$ have a unique midpoint $m = m(x,y)$.

Given $x$ and $y$ any two points of $X$, is it always possible to find an affine function $f : X \rightarrow \mathbb{R}$ with $f(x) \neq f(y)$ ? (where by "affine" I mean $\forall x',y' : f(m(x',y')) = \frac{f(x') + f(y')}{2})$ ?

Test functions with small support and nonnegative Fourier transform

2012-10-28T06:38:49Z

The following problem arose in a question I recently asked : given a (possibly non abelian) compact group $G$ and a neighbourhood $U$ of the identity in $G$, can we always find a function $f : G \mapsto \mathbb{R}$, which vanishes outside $U$, whose Fourier transform is nonnegative, and which satisfies $\hat f(1) \neq 0$ ?

The identity element of a compact group is a limit point of any "polynomial sequence"

2012-10-23T12:10:03Z

Is there an "elementary" (say ultrafilter-free) proof of the following fact: if $G$ is a compact (Hausdorff) topological group, if $g \in G$ is any element from this group, and if $P$ is a polynomial with integer coefficients without constant term, then the identity element of $G$ is a limit point of the sequence $n \mapsto g^{P(n)}$.

An other question: for which integer-valued sequences $u_n$ is the result above still true with $P(n)$ replaced by $u_n$, whatever $G$ and $g$ are?

A mixing property of linear map over finite fields

2012-07-20T08:35:23Z

Let $F$ be a finite field of odd size $q$, and $\phi_0 : F \mapsto F$ be any map from $F$ to itself. For each $a \in F$, set $\phi_a : x \in F \mapsto \phi_0 (x) + ax $.

When $\phi_0 : x \mapsto x^2 $ , each image $\phi_a (F)$ has size $\frac{q+1}{2}$. It turns out that for any $\phi_0$, there's always some $a \in F$ such that $| \phi_a(F) | \geq \frac{q+1}{2}$.

But the only proof I know is somewhat artificial : it relies on the observation that $ K =\bigcup_{a \in F} \phi_a(F)^n $ is a Kakeya set of dimension $n$ over the finite field $F$. By subsequent improvements of Dvir's work on such sets, it is known that $K$ must have $\geq \left( \frac{q^2}{2q-1} \right)^n $ elements. As $n$ is arbitrary and $\frac{q^2}{2q-1} > \frac{q+1}{2} -1 $, this yields the above claim.

I've found no direct proof so far. It seems MO might be the right place to ask for such a proof.

Comment by km

2013-01-28T15:24:09Z

There must be some typo since the sum in the question is $+ \infty$.

Comment by km

2012-12-20T07:45:46Z

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.

Comment by km

2012-10-28T16:04:01Z

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.

Comment by km

2012-10-24T06:41:23Z

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.

Comment by km

2012-10-23T16:31:45Z

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" ...

Comment by km

2012-10-23T16:09:51Z

@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.

Comment by km

2012-10-23T13:10:27Z

For example : this is a straightforward generalisation of theorem 7.2 in the following article of Vitaly Bergelson math.osu.edu/~bergelson.1/… .

Comment by km

2012-07-20T13:04:45Z

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.