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Конференция по комплексному анализу и математической физике, посвященная 70-летию А. Г. Сергеева
21 марта 2019 г. 18:10–18:40, г. Москва, МИАН, ул. Губкина, д. 8, конференц-зал
 


Hardy and Rellich type conformally invariant inequalities

Farit Avkhadiev
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Farit Avkhadiev
Фотогалерея



Аннотация: For real-valued test functions $u\in C_0^\infty (\Omega)$ we study conformally invariant integral inequalities on domains $\Omega$ of the Euclidean space $\mathbb{R}^n$, $n\geq 2$. To construct conformally invariant integrals we use the hyperbolic radius $R=R(x, \Omega)$ defined as follows. If $n=2$ then $1/R(x, \Omega)$ is the coefficient of the Poincaré metric with Gaussian curvature $\kappa =-4$. If $n=3$ then the hyperbolic radius is determined by $V=V(x):\equiv R^{1-n/2}(x, \Omega)$, where $V$ is the maximal solution of the Liouville differential equation $\Delta V= n(n-2) V^{(n+2)/(n-2)}$. In particular, $R(x, B)=1- |x|^2$ for the unit ball $|x|<1$ and $R(x, H^+)= 2 x_1$ for the half-space $x_1>0$. In the case $n\geq 3$ Loewner and Nirenberg are proved the existence of $R(x, \Omega)$ for domains $\Omega$ satisfying certain conditions on the boundary. If there exists $R(x, \Omega)$ we will say that $\Omega$ is a domain of hyperbolic type.
By $\Delta u$ and $\nabla u$ we denote the Euclidean Laplacian and the Euclidean gradient of the function $u$. It is very known that the Dirichlet integral $\int_{\Omega} |\nabla u(x)|^n dx$ is conformally invariant. We use some generalizations of this fact. Suppose that $p\in [1, \infty)$ is a fixed number. We show that the integrals $\int_{\Omega}|u|^p R^{-n}(x, \Omega)) dx$ and $\int_{\Omega}|\nabla u|^p R^{p-n} (x, \Omega) dx $ are conformally invariant for domains of dimension $n\geq 2 $ and that the integral $\int_{\Omega}|\Delta u|^p R^{2p-n}(x, \Omega) dx$ is conformally invariant for $n=2$, but it is invariant with respect to linear conformal mappings, only, in the case $n\geq 3$.
There are several results on the conformally invariant inequalities of Hardy and Rellich type in the case $n=2$ (see, for instance, Avkhadiev, F.G. "Integral inequalities in domains of hyperbolic type and their applications ". Sbornik: Mathematics, 2015, 206(12), 1657–1681). Now we present some results for the case $n\geq 2$ from our paper in press (see Avkhadiev, F.G. "Conformally invariant inequalities on domains of the Euclidean space" , to be published in Izvestiya: Mathematics, 2019). By $(\nabla u, \nabla R)$ we will denote the scalar product in $\mathbb{R}^n$.
Theorem 1. {\it Suppose that $n\geq 3$, $1\leq p< \infty$, and that $B=\{x\in \mathbb{R}^n: |x|<1\}$. Then
$$ \int_{B}\frac{\left|\nabla u(x)\right|^pdx}{(1-|x|^2)^{n-p}}\geq \frac {2^p (n-1)^{p}}{p^p}\int_{B}\frac{|u(x)|^p dx}{(1-|x|^2)^{n}}\quad \forall u\in C_0^{\infty}(B). $$
The constant ${2^p (n-1)^{p}}/{p^p}$ is sharp, i. e. it is the best possible one at this place.}
Theorem 2. {\it Suppose that $n\geq 3$ and that $\Omega \subset\mathbb{R}^n$ is a domain of hyperbolic type. Then
$$ \int_{\Omega}\frac{|\nabla u(x)|^2 dx}{R^{n-2}(x, \Omega)}\geq n (n-2)\int_{\Omega}\frac{|u(x)|^2 dx}{R^{n}(x, \Omega)} \quad \forall u \in C_0^{\infty}(\Omega), $$
and
$$ \int_{\Omega}|\nabla u(x)|^n dx \geq 2^n \left(1-{2}/{n}\right)^{n/2}\int_{\Omega}\frac{|u(x)|^n dx}{R^{n}(x, \Omega)} \quad \forall u \in C_0^{\infty}(\Omega). $$
}
Theorem 3. {\textit Suppose that $n\geq 2$, $1\leq p< \infty$, $1+n/2 \leq s <\infty$, and that $\Omega \subset\mathbb{R}^n$ is a domain of hyperbolic type. Then
$$ \int_{\Omega}\frac{\left|\left(\nabla u(x), \nabla R(x, \Omega)\right)\right|^p dx}{R^{s-p}(x, \Omega)}\geq \frac {2^pn^p}{p^p}\int_{\Omega}\frac{|u(x)|^p dx}{R^{s}(x, \Omega)} \quad \forall u\in C_0^{\infty}(\Omega). $$
If $\Omega$ is a half-space and $s=1+n/2$, then the constant ${2^pn^p}/{p^p}$ in this inequality is sharp}.
The integrals from theorem 3 are invariant with respect to linear conformal mappings.
This work was supported by the Russian Science Foundation under Grant no. 18-11-00115.

Язык доклада: английский
 
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