15 votes
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Steinhaus theorem and Hausdorff dimension

The answer to the question is negative. Körner in Hausdorff dimension of sums of sets with themselves and Schmeling-Shmerkin in On the dimension of iterated sumsets showed that for any increasing ...
Jarosław Błasiok's user avatar
14 votes
Accepted

Are there extremally disconnected groups?

Yes: every locally compact ED group is discrete. Indeed, for topological groups, ED passes to quotients and open subgroups. Let by contradiction $G$ be a non-discrete ED locally compact group. Passing ...
YCor's user avatar
  • 60.1k
8 votes

Uniqueness of the uniform distribution on hypersphere

$\newcommand\S{\mathbb S}$The answer is yes, and it is enough to assume that $f$ is continuous. It is convenient to renormalize the uniform distribution on the sphere so that it be the uniform ...
Iosif Pinelis's user avatar
7 votes

Moments of Plücker coordinates on complex Grassmannian

Notation: $\mathcal{CN}(0,1)$ denotes the complex standard normal distribution: if $Z \sim \mathcal{CN}(0,1)$, then $\sqrt{2} \operatorname{Re}(Z), \sqrt{2} \operatorname{Im}(Z)$ are independent real ...
user42355's user avatar
  • 1,521
7 votes
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Use of Stein's maximal principle in Bourgain's paper on Besicovitch sets

This may be a slight misattribution. It is the implication $(1.22) \implies (1.21)$ which is essentially in Stein's paper; the implication $(1.21) \implies (1.9)$ is much simpler, following from ...
Terry Tao's user avatar
  • 108k
6 votes
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Is there an operation in topology analogous to the operation of averaging over a compact subgroup in harmonic analysis?

As stated (i.e., without assuming metrizability of $X$), the answer is negative: Let $Y=\alpha\mathbb N$ be the Alexandrov (one point) compactification, $X=\beta\mathbb N$ the Stone-Cech ...
Jochen Wengenroth's user avatar
6 votes
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How to give a counterexample of this estimate related to Paley-Littlewood theorem?

This question is really about non-coincidence of different function spaces: the right hand side in your inequality is equal to the the square of the norm in (homogeneous) Besov space $\dot{B}_{p}^{0,...
Anton Tselishchev's user avatar
6 votes
Accepted

If $f$ is bounded, decays fast enough at infinity and $\int f=0$, does this imply that $f$ is in the Hardy space $\mathcal H^1(\mathbb R^n)$?

Edit. For the sake of improving the quality of the post, I modified the proof to make it work for all $M>n$ after Prof. Tao’s comments (the previous version was admittedly way too loose). In the ...
Lorenzo Pompili's user avatar
6 votes
Accepted

How to get $\int_{\mathbb R^d} |\partial_i\partial_j(1-\Delta)^{-\frac{\delta}{2}}p_t(\cdot-y)(x)| \, \mathrm d x \lesssim t^{\frac{\delta}{2}-1}$?

This is a duality argument (the author is really invoking the adjoint of Lemma 5.2(2), rather than Lemma 5.2(2) directly). We can write $$ I = \sup_g \left|\int_{{\bf R}^d} \partial_i \partial_j (1-\...
Terry Tao's user avatar
  • 108k
5 votes
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Every locally compact group gives rise to a locally compact quantum group

In the general, not necessarily $\sigma$-compact setting, one has to interpret $L^\infty(G,\lambda)$ as the von Neumann algebra of locally measurable functions that are bounded outside a locally null ...
Stefaan Vaes's user avatar
  • 4,011
5 votes

Moments of Plücker coordinates on complex Grassmannian

Since the bounty is about to expire, I am posting my findings so far. Using the method mentioned by user42355, one can prove the following more general result. Let $\lambda=(\lambda_1,\ldots,\lambda_{\...
Abdelmalek Abdesselam's user avatar
5 votes

Steinhaus theorem and Hausdorff dimension

Here is a quite short example to show that you question cannot have a positive answer. Assume that $V$ is a Sacks extension of constructible universe $L$. Then the set of constructible reals $A=(\...
喻 良's user avatar
  • 4,191
5 votes

The integrability of $\widehat{e^{-|x|^a}}$, $a>0$

The Fourier transform of $|x|^b$, $b\notin\mathbb Z$, is the function $c|\xi|^{-1-b}$ away from $\xi =0$. See entry 313 of the table here and the discussion in the last column. Moreover, the large $\...
Christian Remling's user avatar
5 votes

Possible research directions in analysis?

I find that students often come in with these kinds of ideas, that almost everything is known, it's hard to do anything, the problems are mostly solved. Then they ask how to best position themselves ...
user378654's user avatar
5 votes

Completeness of exponentials with frequencies in a periodic set

Under your conditions, the system $e^{2\pi i\ell_nx}$ is complete. Moreover, it is a Riesz basis for $L^2[0,1]$. This follows from the result of B. Ya. Levin in Interpolation by entire functions of ...
Alexandre Eremenko's user avatar
4 votes
Accepted

A lower bound for the $L^1$ norm of real trigonometric polynomials

This is not true. Consider the Fejer Kernel $F_n$. From the explicit formula for $F_n$ we see that $$\|F_n\|_{L^1} \leq C $$ independent of $n$. On the other hand since $ F_n = 2 \sum_{0\leq k \leq n-...
Mark Lewko's user avatar
  • 11.7k
4 votes
Accepted

Calculation of Riesz energy for balls

This is just an extended comment to Iosif Pinelis's answer above, which provides an answer in terms of an unknown constant $C_{d,s}$. Here we evaluate this constant. Let $B$ be the unit ball. If $f(x)=...
Mateusz Kwaśnicki's user avatar
4 votes
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Is the Hardy Littlewood “minimal function” comparable to the original function in $L^1$ norm?

This is not true. Take $\epsilon=1$ and, on the real line, $f_\delta=\delta^{-1} \chi_{(0, \delta)}$, so that $\|f_\delta\|_1=1$. Then $0 \leq m_1 f_\delta \leq \chi_{(0, \delta)}$, by choosing small ...
Giorgio Metafune's user avatar
4 votes

Uniqueness of the uniform distribution on hypersphere

Here is an alternative approach (in addition to the great answer by Iosif Pinelis), assuming only that $f \in L_2(\mathbb{S}^{d-1})$ --- since the post is tagged harmonic analysis, this idea might ...
Jarosław Błasiok's user avatar
4 votes
Accepted

Fourier coefficients of Selberg polynomials

As it is odd, we can write the Vaaler polynomial $V_K(x)=\sum_{1 \le k \le K}c_{k,K} \sin 2\pi kx$ and its fundamental property is that $|c_{k,K}| \le \frac{1}{\pi k}$. This follows easily from its ...
Conrad's user avatar
  • 1,854
4 votes
Accepted

When is $W^{1,p}(\Omega)$ a Banach algebra?

I'll just complete the answers in the comments showing that for $p\le d$ it is not a Banach algebra. Let me take $\Omega=B_1(0) \subset \mathbb{R}^n$ and $p=d$. Assume that $W^{1,d}(B_1) $ is a Banach ...
Michele Caselli's user avatar
4 votes
Accepted

Does this dyadic sum converge?

$\newcommand\ep\epsilon\newcommand\Ga\Gamma\newcommand\Z{\mathbb Z}$The answer is yes. Indeed, let $b:=d/p$, so that $1>a>b>0$. For real $\ep>0$, let $$K(\ep):=\int_0^\infty e^{-\ep s}\...
Iosif Pinelis's user avatar
4 votes
Accepted

Examples of non-discrete, cocompact subgroups

You can find many such examples among groups acting on trees. Let $T$ be a $k$-regular tree and let $G$ be the subgroup of $\operatorname{Aut}(T)$ of automorphisms stabilizing each of the 2 parts of ...
Tom De Medts's user avatar
  • 6,494
4 votes
Accepted

Average size of the Fourier--Stieltjes transform of the fractal measures

No, we cannot have $\int_{-T}^T |\widehat{\mu}|^2 \lesssim T^{1-s-\epsilon}$. This would imply that $$ I_{s+\epsilon/2}(\mu) = \int d\mu(x)\int d\mu(y) |x-y|^{-s-\epsilon/2} = c \int |t|^{s+\epsilon/2-...
Christian Remling's user avatar
4 votes
Accepted

In what sense does the Hermite expansion of a bounded smooth function converge?

According to the conclusion at the bottom of p. 603 of (say) Uspensky's paper, the Hermite expansion of $f$ converges pointwise to $f$ (under conditions much less restrictive that yours). If $f$ is (...
Iosif Pinelis's user avatar
4 votes
Accepted

Any references for generalised square functions?

For $0<p<\infty$, $0<q\le\infty$ and $s\in\mathbb R$, the Triebel-Lizorkin space $F_{pq}^s(\mathbb R^n)$ is the set of all (tempered distributions) $f$ such that $\big\||\{2^{js}\cdot P_jf\}...
Liding Yao's user avatar
4 votes
Accepted

Equivalent Littlewood-Paley-type decompositions

The norms should be equivalent with a constant depending only on the ratio between the two bases (in this case, 2 and 3). I'll just consider the Besov case. I'll adopt the notation $\hat{f}_{n,b}(\xi)=...
Ben Johnsrude's user avatar
4 votes
Accepted

An integrable estimate of the Hölder constant of the map $x \mapsto \int_{\mathbb R^d} f(y) \partial_1 \partial_1 g_t (x-y) \, \mathrm d y$

If $T_tf=g_t*f$ is the heat semigroup you are asking for the norm of $D_{ij}T_t$ from $C^\alpha$ to itself (endowed with the Holder seminorm). Let $I_\lambda f(x)=f(\lambda x), \lambda >0$. Then $[...
Giorgio Metafune's user avatar
4 votes
Accepted

On a density property of signed singular measures

Yes, this works. Write $d\mu = s\, d|\mu|$, with $|\mu|$ denoting the total variation of $\mu$ and $s(x)=\pm 1$. We can recover $s(x)$ for $|\mu|$-a.e. $x$ as the derivative $s(x)=\lim_{|I|\to 0} \mu(...
Christian Remling's user avatar
4 votes
Accepted

Schroedinger operator in 2 dimensions with singular potential

Taking advantage of the spherical symmetry to decompose this into a sum of one-dimensional problems sounds like the right approach. I will probably just be redoing what Reed-Simon had in mind here. ...
Christian Remling's user avatar

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