Poincare inequality
It is known that this inequality is valid for bounded John domains if w ∈ Ap (see [DD]). As we will see, this result can be extended for more general weights. For example, for a class of weights introduced in [FKS] where the authors consider the classic Poincaré inequality in weighted norms, (1.6) kϕ−ϕΩ,wkLp w(Ω) ≤ Ck∇ϕkLp w(Ω)The topic of this thesis is a diffusion process on a potential landscape which is given by a smooth Hamiltonian function in the regime of small noise. The work provides a new proof of the Eyring-Kramers formula for the Poincaré inequality of the associated generator of the diffusion. The Poincaré inequality characterizes the spectral gap of the generator and establishes the exponential rate ...
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Since inequality (1) has significance for studies on partial differential equations, many researchers studied this type of Sobolev inequality and an explicit value of C p (Ω) (see, e.g., [7,11,23 ...Title: Poincaré inequality 3/2 on the Hamming cube. Authors: Paata Ivanisvili, Alexander Volberg. Download a PDF of the paper titled Poincar\'e inequality 3/2 on the Hamming cube, by Paata Ivanisvili and 1 other authors.In the present paper, we deal withthe weighted Poincark inequalitiesin weighted Sobolev spaces W"lP (fl;x0, xfl) and W"tP (Q; w, w), where R is one-dimensional unbounded domain, and give sufficient conditions for the weighted Poincare inequalities to hold. 2.We establish the Poincare-type inequalities for the composition of the homotopy operator, exterior derivative operator, and the projection operator with norm applied to the nonhomogeneous -harmonic equation in -averaging domains.AN OPTIMAL POINCARE INEQUALITY IN L1 FOR CONVEX DOMAINS GABRIEL ACOSTA AND RICARDO G. DURAN (Communicated by Andreas Seeger) Abstract. For convex domains ˆRnwith diameter dwe prove kuk L1(!) d 2 kruk L1(!) for any uwith zero mean value on!. We also show that the constant 1=2in this inequality is optimal. 1. Introduction Given a bounded domainPoincar e Inequalities in Probability and Geometric Analysis M. Ledoux Institut de Math ematiques de Toulouse, France. Poincar e inequalities Poincar e-Wirtinger inequalities from theorigintorecent developments inprobability theoryandgeometric analysis. workof Henri Poincar eThis paper deduces exponential matrix concentration from a Poincaré inequality via a short, conceptual argument. Among other examples, this theory applies to matrix-valued functions of a uniformly log-concave random vector. The proof relies on the subadditivity of Poincaré inequalities and a chain rule inequality for the trace of the matrixPoincaré inequality In mathematics, the Poincaré inequality [1] is a result in the theory of Sobolev spaces, named after the French mathematician Henri Poincaré. The inequality allows one to obtain bounds on a function using bounds on its derivatives and the geometry of its domain of definition.An optimal Poincare inequality in L^1 for convex domains. For convex domains Ω C R n with diameter d we prove ∥u∥ L 1 (ω) ≤ d 2 ∥⊇ u ∥ L 1 (ω) for any u with zero mean value on w. We also show that the constant 1/2 in this inequality is optimal.We establish the Sobolev inequality and the uniform Neumann-Poincaré inequality on each minimal graph over B_1 (p) by combining Cheeger-Colding theory and the current theory from geometric measure theory, where the constants in the inequalities only depends on n, \kappa, the lower bound of the volume of B_1 (p).derivation of fractional Poincare inequalities out of usual ones. By this, we mean a self-improving property from an H1 L2 inequality to an H L2 inequality for 2(0;1). We will report on several works starting on the euclidean case endowed with a general measure, the case of Lie groups and Riemannian manifolds endowed also with a generalIn view of our discussion of the Dirichlet integral, we call Inequality ♦ weak Hardy inequality if ker q ={0} and weak Poincaré inequality if ker q ={0}. In the case = 0, the function α becomes a constant and Inequality ♦is referred to as Hardy inequality if ker q ={0}, respectively Poincaré inequality if ker q ={0}.The assumption on the measure is the fact that it satisfies the classical Poincaré inequality, so that our result is an improvement of the latter inequality. Moreover we also quantify the tightness at infinity provided by the control on the fractional derivative in terms of a weight growing at infinity. The proof goes through the introduction ...1. Introduction The simplest Poincar ́ e inequality refers to a bounded, connected domain Ω ⊂ L2(Ω) n, and a function f L2(Ω) whose distributional gradient is also in ∈ (namely, f W 1,2(Ω)). While it is false that there is a finite constant S, ∈showed that it is implied by the Poincaré inequality (which means that the concen-tration consequences of that inequality are stronger than (1.3)). More precisely, Theorem 1.1 (Bobkov-Ledoux [8]). Let µbe a probability measure on Rd, which satisfies a Poincaré inequality with constant CP. Then for any c∈ (0,C −1/2 P), thereinequalities as (w,v)-improved fractional inequalities. Our first goal is to obtain such inequalities with weights of the form wF φ (x) = φ(dF (x)), where φ is a positive increasing function satisfying a certain growth con-dition and F is a compact set in ∂Ω. The parameter F in the notation will be omitted whenever F = ∂Ω.In many cases, people who have unequal opportunities in life often live in poverty, and people who live in poverty may be treated unequally. Although a person who experiences poverty may suffer from inequality, every person who faces inequa...The first part of the Sobolev embedding theorem states that if k > ℓ, p < n and 1 ≤ p < q < ∞ are two real numbers such that. and the embedding is continuous. In the special case of k = 1 and ℓ = 0, Sobolev embedding gives. This special case of the Sobolev embedding is a direct consequence of the Gagliardo–Nirenberg–Sobolev inequality.You haven't exactly followed the hint, but your proof seems correct. As pointed out by Chee Han, you could follow the hint by squaring the given identity (using the Cauchy-Schwarz inequality like you did), integrating from $0$ to $1$ and exchanging the order of integration.We derive bounds for the constants in Poincaré-Friedrichs inequalities with respect to mesh-dependent norms for complexes of discrete distributional differential forms. A key tool is a generalized flux reconstruction which is of independent interest. The results apply to piecewise polynomial de Rham sequences on bounded domains with mixed boundary conditions.Use Hoelder inequality. Share. Cite. Follow answered Dec 14, 2021 at 10:51. Son Gohan Son Gohan. 4,277 2 2 gold badges 5 5 silver badges 23 23 bronze badges $\endgroup$ 5 $\begingroup$ Can you elaborate some more? How would I use Hoelder? $\endgroup$ - Silver54.In the link above, the generalization of the Poincare inequality to general measure spaces is considered as well. I searched for papers myself but was not able to find anything specialized to Gaussian measures. Could anyone please help me? pr.probability; inequalities; gaussian; Share.
Bernoulli 25(3), 2019, 1794-1815 https://doi.org/10.3150/18-BEJ1036 On the isoperimetric constant, covariance inequalities and Lp-Poincaré inequalities in ...For other inequalities named after Wirtinger, see Wirtinger's inequality. In the mathematical field of analysis, the Wirtinger inequality is an important inequality for functions of a single variable, named after Wilhelm Wirtinger. It was used by Adolf Hurwitz in 1901 to give a new proof of the isoperimetric inequality for curves in the plane. inequality (2.4) provides a way to quantify the ergodicity of the Markov process. As it happens, the trace Poincaré inequality is equivalent to an ordinary Poincaré inequality. We are grateful to Ramon Van Handel for this observation. Proposition 2.4 (Equivalence of Poincaré inequalities). Consider a Markov process (Zt: t ≥ 0) ⊂ ΩHere we show existence of many subsets of Euclidean spaces that, despite having empty interior, still support Poincaré inequalities with respect to the restricted Lebesgue measure. Most importantly, ...Poincaré inequality in a ball (case $1\leqslant p < \infty$) There is a weaker inequality which is derived from \ref{eq:1} ...
Poincaré inequalities on graphs M. Levi, F. Santagati, A. Tabacco & M. Vallarino Analysis Mathematica 49 , 529-544 ( 2023) Cite this article 70 Accesses Metrics Abstract Every graph of bounded degree endowed with the counting measure satisfies a local version of Lp -Poincaré inequality, p ∈ [1, ∞].In this paper, we study the sharp Poincaré inequality and the Sobolev inequalities in the higher-order Lorentz-Sobolev spaces in the hyperbolic spaces. These results generalize the ones obtained in Nguyen VH (J Math Anal Appl, 490(1):124197, 2020) to the higher-order derivatives and seem to be new in the context of the Lorentz-Sobolev spaces defined in the hyperbolic spaces.Overall, the strategy of the proof is pretty similar to the one used in the proof of Theorem 3.20 in the aforementioned monograph, where a Gaussian Poincare inequality is demonstrated. I welcome any other approaches as well (either functional-analytic approach or geometric approach)!…
Reader Q&A - also see RECOMMENDED ARTICLES & FAQs. The purpose was to place the question in the right context, provi. Possible cause: Using the aforementioned Poincaré-type inequality on the boundary of the .
the improved Poincare inequality for any 3 > 0 (see Remark 3.11(4) and [BS,4(1)]). Our main theorems are Received by the editors May 4, 1992. 1991 Vathematics Subject Classification. Primary 46E35, 26D 10. Key words and phrases. Poincare inequality, Poincare domains, John domains, domains satisfy-ing a quasihyperbolic boundary condition.We show that certain functional inequalities, e.g. Nash-type and Poincaré-type inequalities, for infinitesimal generators of C 0 semigroups are preserved under subordination in the sense of Bochner. Our result improves earlier results by Bendikov and Maheux (Trans Am Math Soc 359:3085-3097, 2007, Theorem 1.3) for fractional powers, and it also holds for non-symmetric settings. As an ...This paper aims at proving new multipolar Hardy inequalities on negatively curved manifolds. To introduce the subject, let us recall the simplest form of the unipolar Hardy inequality on Riemannian manifolds, which is due to Carron [].If \((\mathcal {M},g)\) is an \(N\ge 3\) dimensional Cartan-Hadamard manifold, \(\mathrm{d}(., .)\) is the geodesic distance and \(x_0 \in \mathcal {M}\), the ...
The sharp Sobolev type inequalities in the Lorentz–Sobolev spaces in the hyperbolic spaces. Journal of Mathematical Analysis and Applications, Vol. 490, Issue. 1, p. 124197. Journal of Mathematical Analysis and Applications, Vol. 490, Issue. 1, p. 124197.derivation of fractional Poincare inequalities out of usual ones. By this, we mean a self-improving property from an H1 L2 inequality to an H L2 inequality for 2(0;1). We will report on several works starting on the euclidean case endowed with a general measure, the case of Lie groups and Riemannian manifolds endowed also with a generalsequence of this inequality, one obtains immediately the "existence" part of the Fredholm alternative for the positive Dirichlet Laplacian −Δ at the first eigenvalue λ1. In this article we replace the power 2 by p (2 ≤ p<∞) and thus extend inequality (1.1) to the "degenerate" case 2 <p<∞. A simplified version of
Remark 1.10. The inequality (1.6) can be viewed a Poincare inequality on balls to arbitrary open subset of manifolds. 4. A Poincaré-type inequality: proof or counterexample. 4. Cheeger inequality for measures. 3. Isoperimetric inequality for analytic functions on an annulus. 2. A simple 1-dimensional inequality, maybe Poincaré inequality or Hölder inequality? 4.The Poincaré inequality need not hold in this case. The region where the function is near zero might be too small to force the integral of the gradient to be large enough to control the integral of the function. For an explicit counterexample, let. Ω = {(x, y) ∈ R2: 0 < x < 1, 0 < y < x2} Ω = { ( x, y) ∈ R 2: 0 < x < 1, 0 < y < x 2 } Poincaré inequalities play a central role inNobody has time to read an 80 page paper [LE20]. Therefore I doubt The first nonzero eigenvalue of the Neumann Laplacian is shown to be minimal for the degenerate acute isosceles triangle, among all triangles of given diameter. Hence an optimal Poincaré inequality for triangles is derived. The proof relies on symmetry of the Neumann fundamental mode for isosceles triangles with aperture less than π / 3. THE EQUALITY CASE IN A POINCARE-WIRTINGER TYPE´ INEQUALITY Sobolev 空间: 庞加莱不等式 (Poincaré inequalities) - Sobolev 空间中的 Poincaré 不等式往往在微分方程弱解存在性的证明中扮演一个基础且关键的作用; 如典型的二阶椭圆方程. 我们将给出两种主要的 Poincaré 不等式并给出证明.Some generalized Poincaré inequalities and applications to problems arising in electromagneti. sm.pdf. Content available from CC BY 4.0: 02e7e52dffd36659c5000000.pdf. WEIGHTED POINCARÉ INEQUALITY AND RIGIDITY OF COMPLETCheeger, Hajlasz, and Koskela showed the importanceOct 12, 2023 · "Poincaré Inequality." Lecture Five: The Cacciopolli Inequality The Cacciopolli Inequality The Cacciopolli (or Reverse Poincare) Inequality bounds similar terms to the Poincare inequalities studied last time, but the other way around. The statement is this. Theorem 1.1 Let u : B 2r → R satisfy u u ≥ 0. Then | u| ≤2 4 2 r B 2r \Br u . (1) 2 Br First prove a Lemma. Article Poincaré and log-Sobolev inequalities for mixtures A We investigate links between the so-called Stein's density approach in dimension one and some functional and concentration inequalities. We show that measures having a finite first moment and a density with connected support satisfy a weighted Poincaré inequality with the weight being the Stein kernel, that indeed exists and is unique in this case. Furthermore, we prove weighted log-Sobolev ... Boundary regularity of the domain in the us[New inequalities are obtained which interpolate in a sharp wa2.1 Korn inequality from weighted Poincare inequality´ In this s Take the square of the inverse of (4a 2 r 2 + 1 e + 2)m (r − 1) as 1 2 β (s) for the desired conclusion. a50 In [24] Eberle showed that a local Poincaré inequality holds for loops spaces over a compact manifold. However the computation was difficult and complicated and there wasn’t an estimate on the blowing up rate.