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2 votes

Existence of a positive measurable set with disjoint preimage under iterated transformation

If $(X,\mathcal{B},\mu)$ is a Lebesgue probability space, only the aperiodicity is needed by a simple application of Rokhlin's lemma: There exists $B\in\mathcal{B}$ such that $B,\dots,T^{-k}(B)$ are ...
KhashF's user avatar
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3 votes

Existence of a positive measurable set with disjoint preimage under iterated transformation

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue ...
Saúl RM's user avatar
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1 vote

Entropy arguments used by Jean Bourgain

What follows is a bit of Computer-Sciency exposition, but since it proves the desired claim and it is an "elementary entropy consideration" I hope it answers the question. For convenience, ...
Jarosław Błasiok's user avatar
1 vote

Strict positive definite function gradient tuple

We need to show for all $a\in \mathbb{R}^n$ and all $v_1,\dots, v_n\in \mathbb{R}^d$ $$ \text{Var}\Bigl[\sum_{k=1}^n a_k f(x_k) + D_{v_k}f(x_k)\Bigr] > 0 $$ We assume $f$ to be centered by ...
Felix B.'s user avatar
  • 367
0 votes

Fourier series of smooth functions in infinitely many variables

Not sure why this is brought up after some years. But I think you should google "Fourier analysis on infinite torus" or something similar. The first result returns me a paper by Denis Fufaev....
Liding Yao's user avatar
3 votes

Does the complex interpolation space $(L^1(\mathbb{R}),W^{2,1}(\mathbb{R}))_{\frac{1}{2}}$ continuously embed into $L^\infty(\mathbb{R})$?

This is a very interesting question. My answer is no. In $\Bbb R$ the differentiation is "elliptic" and we have the isomorphism $I-\Delta:W^{2,1}(\Bbb R)\to L^1(\Bbb R)$. Therefore $\sqrt{I-\...
Liding Yao's user avatar
3 votes
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An asymptotic integral with complex phase

No, of course not. Essentially, you want to bound a $2$-dimensional object (the complex Fourier transform. I talk about the complex dimension here) from the at most $1$-dimensional bound (that is, ...
Aleksei Kulikov's user avatar
5 votes
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Prékopa-Leindler style inequality?

Is not it obvious (unlike Prékopa-Leindler)? We are given that for all $x_1,x_2$ we have $(f_1/g_1)^2 (x_1)\leqslant (g_2/f_2)^3(x_2)$, thus there exists $c>0$ such that $(f_1/g_1)^2 (x_1)\leqslant ...
Fedor Petrov's user avatar
1 vote
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Rate of convergence of the minimum point over a product space

$\newcommand\ep\epsilon\newcommand\th\theta\newcommand\de\delta$The answer is yes, even with $C=1$ for all such $f$. Indeed, let $f$ satisfy all your conditions on $f$. Let $g(0):=0$. For real $x>0$...
Iosif Pinelis's user avatar
5 votes
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$L^1$ convergence

No, this is not true. The following argument is about spaces of finite measure, but it can easily be adapted to your question about $\mathbb R$. Let $(X, m)$ be a space with $m(X) < \infty$. Take $...
Alex M.'s user avatar
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0 votes
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Decay rate of minimum point over a product space

$\newcommand\ep\epsilon\newcommand\th\theta$Yes, $|\th_\ep-\th_0|$ can decay arbitrarily slowly compared to $f(\th_\ep,\ep)$. Indeed, let $g\colon\Bbb R\to\Bbb R$ be any smooth function such that $g(0)...
Iosif Pinelis's user avatar
0 votes

Inverse Limit in the category of $C^{\ast}$-algebras or operator spaces

You can see in the paper "Some inverse limits of Cuntz algebras" when the inverse limits (Projective limits) exist in the category of C*-algebras.
user526378's user avatar
3 votes
Accepted

Hausdorff-Lipschitz continuity of cone correspondence

$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$A counterexample is as follows: $n=m=1$, $\X=[-1,1]$, $\P=[0,2]$, $f(x)=\sqrt2-1/\sqrt2+|x|^{3/2}$, $g(x)=1/\sqrt2$, and $h(x)=1=S(x)$ (for all $x\...
Iosif Pinelis's user avatar
5 votes
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Subspaces of $C_0$ on which $p$-norm are equivalent?

Use $\|f\|_p \leq \|f\|_\infty^{1- \frac 2p}\|f\|_2^{\frac 2 p}$ to get $\|f\|_\infty \leq C\|f\|_2$ from $\|f\|_\infty \leq C\|f\|_p$.
Giorgio Metafune's user avatar
3 votes
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On weighted Fourier transforms

The answer is yes. I will prove it using the assumption only for $\xi > 0$ (almost the same proof works if we only assume it for $\xi < 0$). For $z\in U = \{z: Im (z) \ge 0, |z| \ge 1\}$, ...
Aleksei Kulikov's user avatar
0 votes

References for well-posedness of weak solutions to Stefan problem

The existence and uniqueness theorems (and thus the well posedness) for weak solutions (in some suitable sense) to the Stefan problem were proved by Shoshana Kamin (neé Shoshana L'vovna ...
Daniele Tampieri's user avatar
1 vote
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A metric characterization of Hilbert spaces

Probably I.G. Nikolaev. See Theorem 10.10.13 in Burago, Dmitri; Burago, Yuri; Ivanov, Sergei "A course in metric geometry" [https://www.ams.org/books/gsm/033/][1] Hat tip to A. Eskenazis who ...
Manor Mendel's user avatar
2 votes
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Continuous extensions of tangent vector fields

I claim that $F$ exists if and only if $G$ has zero winding number around the boundary of $\Omega$. I will only give a complete proof of one direction. We can suppose that $\Omega$ is a plane domain, ...
Ben McKay's user avatar
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2 votes
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Continuous modification of tangent vector fields

Think of $\Omega$ as an open set in the plane. If $F$, restricted to the boundary, has a different winding number than $G$, then all deformations of $F$ through unit vector fields still have that same ...
Ben McKay's user avatar
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5 votes
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Is Toeplitz operator on the Bergman space bounded iff its symbol is bounded?

It is not neccesary in general that $\varphi \in L^\infty(\mathbb{D})$, but it is necessary and sufficient that in a certain sense $\varphi$ must be bounded ``on average in the hyperbolic sense''. The ...
an_ordinary_mathematician's user avatar
3 votes
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Is there any example of linear operator which is bounded on all Besov spaces but not on Triebel-Lizorkin spaces

Turns out it was not known. So I took a week and write it down. You can see in the arXiv for a complete proof. For convenience let $n=1$. Let $(\phi_j)_{j=0}^\infty$ be the family of Schwartz function ...
Liding Yao's user avatar
1 vote

Grothendieck on topological vector spaces

Maybe you will find interesting the following references Grothendieck's Theorem, past and present, Pisier or Grothendieck’s works on Banach spaces and their surprising recent repercussions A. ...
3 votes

Asymptotic behavior of the "Cauchy square" series

Thanks to Carlo, using Abel--Plana once again one finds the simple formula $$\lim_{n\to\infty}n^{3/2}S_n=2\zeta(3/2)$$
Henri Cohen's user avatar
  • 11.8k
5 votes

Asymptotic behavior of the "Cauchy square" series

You can reduce the evaluation of $S_n$ to a quadrature by means of the Abel-Plana formula, $$S_n=\sum _{s=1}^{n-1} g(s)=\int_1^{n-1}g(s)\,ds+\tfrac{1}{2}g(1)+\tfrac{1}{2}g(n-1)$$ $$\qquad\qquad -\,2\...
Carlo Beenakker's user avatar
2 votes

Analytic maps on Banach spaces: analyticity upgrade

I will summarize what has been said in the comments (thanks to Jochen Glueck for all his help). The answer to the question is no, in general. What is going on is actually very simple. Theorem. Let $E,...
Lorenzo Pompili's user avatar
0 votes

Fractional Sobolev spaces and interpolation in unbounded Lipschitz domains

In many cases people only consider bounded domain because of some convenience on using compactness and finiteness of partition of unity. This question is very important in the field of analysis on ...
Liding Yao's user avatar
2 votes
Accepted

A bilinear estimate with a simple one-dimensional oscillatory integral kernel

We can make more concrete what I suggested in my comment. Take $f(x)=x^{1/2p'}e^{-ix}$ on $1\le x\le 1/(10\epsilon)$ and $f=0$ otherwise. Then the LHS of (*) equals $$ \int_1^{1+\epsilon} |G(t)|^2\, \...
Christian Remling's user avatar
1 vote
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A strange functional inequality

Hmm. We have $(\cos t+\sin t)^2+(\cos t-\sin t)^2=2$, thus at least one of numbers $u:=\cos(x^{-1})+\sin(x^{-1})$, $v:=\cos(x^{-1})-\sin(x^{-1})$ have absolute value not less than 1. Since an even ...
Fedor Petrov's user avatar
3 votes
Accepted

Bounding supremum norm in terms of gradient L2-norm using a Poincare-like inequality

For $p>\max(2,d)$ you have by Sobolev embedding $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_p + \|\nabla f\|_p.$$ The interpolation $L^p(\Omega) = [L^2(\Omega),L^\infty(\...
Ayman Moussa's user avatar
  • 2,710
1 vote

Is every face exposed if all extreme points are exposed?

It is also possible to construct counterexamples from the idea that every extreme point of the convex hull of any subset of the unit sphere $S^2$ is an exposed point. Placing a convex set $C$ inside ...
Stephan's user avatar
  • 31
1 vote

Continuity upgrade for nonlinear maps

Hi Jan :) I was recently wondering whether the above works for analytic functions. I don't have a clean answer yet, but I can prove something. Proposition. Let $E,F,G$ be complex Banach spaces, $F\...
Lorenzo Pompili's user avatar
5 votes
Accepted

Weak convergence + convergence of the norm implies strong convergence in Orlicz spaces

The question essentially boils down to asking about sufficient conditions under which an Orlicz space $L_\Phi(\mathcal{X},\mu)$ over a nonatomic finite measure space $(\mathcal{X},\mu)$, equipped with ...
Ryszard Kostecki's user avatar
0 votes

Complex sum of squares of vector fields (hypoelliptic operators)

While this question is old, it's perhaps worth noting that something is known: It's known that the situation can be very complicated. For example, Kohn (Annals of Mathematics, 162 (2005), 943–986) ...
Brian Street's user avatar
0 votes

Is the difference between $\alpha$-Hölder constants of $f*\rho$ and $g*\rho$ controlled by $\|f-g\|_\infty$?

Let me add some general hints. You want a bound $[f-g]_\alpha\le C\|f-g\|_\infty$ for all non-negative $f$ and $g$ in $L^1(\mathbb R^n)\cap L^\infty(\mathbb R^n)$ with $\int_{\mathbb R^n}fdx= \int_{\...
Pietro Majer's user avatar
  • 56.6k
2 votes
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Lower bound the best $\alpha$-Hölder constant of a convolution

No. Take $f_1$ an $\alpha$-Hölder compactly supported convolution kernel (non-negative, with $\int_{\mathbb R}^nf_1dx=1$), and consider the usual approximation of identity by convolution with $f_\...
Pietro Majer's user avatar
  • 56.6k
2 votes

Is the difference between $\alpha$-Hölder constants of $f*\rho$ and $g*\rho$ controlled by $\|f-g\|_\infty$?

Not really, because the estimate "sees" the $L^1$ norm rather than the $L^\infty$ one. Notice first that $f\mapsto \sigma(\cdot,f)$ is linear, so you question amounts to asking whether $[\...
leo monsaingeon's user avatar
2 votes

Functions with asymmetrically decreasing Fourier transform?

I suddenly realized that there is a simple explicit way of getting the function $f_1=F$ in Alexandre Eremenko's answer. Just for record, I put it here as an answer. The idea is to take as $g$ the ...
TaQ's user avatar
  • 3,390
2 votes
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Can a lift satisfy Chen's relation, geometric condition but not be a rough path?

Indeed, this can fail. Take any geometric rough path lift $\mathbb X^{ij}$ and simply add in a non-symmetric way the increment of a function $F$ that fails to be $2\gamma$-Hölder. For example, take ...
Thomas Kojar's user avatar
  • 4,449
4 votes
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An asymmetric quadrilinear estimate

OK. Let's first figure out a necessary condition on $a_x>0; a_{xy}, a_{xz}, a_{yt}\in(\frac 1p,1)$ for the quantity $$ Q=\int_\Omega \frac{f_1(x)f_2(y)f_3(z)f_4(t)}{x^{a_x}(1-x^2-y^2)^{a_{xy}}|x-z|^...
fedja's user avatar
  • 60k
1 vote

Check an equation on the Heisenberg group $H_1$

First, the reference is: https://www.sciencedirect.com/science/article/pii/0022123682900787 Second, in the notation of the cited paper $H_n={\mathbb C}^n\times {\mathbb R}$. Hence, you must decide if ...
F Zaldivar's user avatar
  • 1,422
4 votes
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Functions with asymmetrically decreasing Fourier transform?

Edited. $f=u+iv$, where $u$ is even and $v$ is odd is equivalent to $f(-x)=\overline{f(x)}$. A function has this property if and only if its Fourier transform is real. Boundedness of support of $f$ is ...
Alexandre Eremenko's user avatar
1 vote
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Is the space $C_0^{k}(\Omega)$ a Montel space?

In the following I write $C^k_K(\Omega)$ for the space of all $C^k(\Omega)$ functions having support included in $K$, equipped with its "natural" norm $\|\cdot\|_{K,k}$ (uniform norm on $K$ ...
Ayman Moussa's user avatar
  • 2,710
4 votes
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Norm of a $2$-tuple of operators

$\newcommand\C{\Bbb C}\newcommand\om{\omega}$A counterexample is given by $E=\mathbb C^2$ and $K_1(x_1,x_2)=(x_1,0)$ and $K_2(x_1,x_2)=(0,x_2)$ for complex $x_1,x_2$. Indeed, then $\om(K_1)=\om(K_2)=1=...
Iosif Pinelis's user avatar
3 votes

Does the union of fractional Sobolev spaces fills $L^p$?

Let us assume that $p=2$, and let us consider $$ \cup_{s>0} H^s(\mathbb R^d)\subset H^0(\mathbb R^d)=L^2(\mathbb R^d). $$ The above inclusion is strict. Let us consider $u\in L^2(\mathbb R^d)$ ...
Bazin's user avatar
  • 15.2k
3 votes
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Example of a conditionally convergent series $\sum_{n=1}^\infty b_n$ such that $n^2(b_n-b_{n+1})$ is bounded

The answer is yes, such an example exists. For instance, let $b_1=0$ and $b_n=(-1)^j h_j(n)$ for $j=1,2,\dots$ and $n\in N_j:=\{2^j,\dots,2^{j+1}-1\}$, where $$h_j(n):=c_j\frac{\min(n-2^j,2^{j+1}-n)}{...
Iosif Pinelis's user avatar
3 votes

How to understand the unique continuation result

If I understand things correctly, $u$ is vanishing on some non-empty open subset. Moreover, you have pointwisely on $\mathbb R^3$ the differential inequality $$ \lvert \Delta u\rvert\le C\lvert u\...
Bazin's user avatar
  • 15.2k
2 votes

Combination of simple tensors - II

Take $X = Y= M_2$ and consider the tensor $e_1 \otimes e_1 + e_2 \otimes e_2$. You get the same result for any choice of orthonormal basis of $\mathbb{C}^2$, so there can already be two expressions ...
Nik Weaver's user avatar
  • 42.2k
4 votes

Simple proof that exactness implies strong mixing

I think the martingale theory really captures exactly what's happening. I don't know any other proof, but suspect that you would end up reproducing the backwards martingale theorem in some form? I am ...
Anthony Quas's user avatar
  • 22.5k
3 votes
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The cardinality of projections of subsets of the Hilbert cube by inner products

Consistently, there is no such set. Given $X\subset [0,1]^{\mathbb N}$ of cardinality continuum it suffices to find $z\in(0,1)$ such that $x\mapsto \sum_n x_n z^n$ is injective on some subset $X'\...
Colin McQuillan's user avatar
4 votes

Simple proof that exactness implies strong mixing

Not sure if this is what you're looking for, and I'm not an expert on exact systems, but: your definition of exact implies that $\bigcap_{n \geq 0} T^{-n} \mathcal{B} = \mathcal{N}$, where $\mathcal{B}...
Ronnie Pavlov's user avatar

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