A geometrical version of Hardy's inequality
This paper contains a version of Hardy's inequality in a domain \(\Omega \subset \mathbf{R}^{n}\)\ which involves the distance to the boundary and the volume of \(\Omega \). A special case of the main result asserts that \[ \int_{\Omega }\left|\nabla u\right|^{2} dx\geq \frac{n}{4} \int_{\Omega }\int_{S^{n-1}}\frac{1}{\rho _{\nu }^{2}(x)}d\omega (\nu)\left|u(x)\right|^{2} dx+\frac{n^{(n-2)/n}s_{n-1}^{2/n}}{4} \int_{\Omega }\frac{\left|u(x)\right|^{2}}{\left( \text{vol} \Omega _{x}\right) ^{2/n}} dx \] for every \(u\in H_{0}^{1}(\Omega)\). Here \(s_{n-1}\) denotes the measure of the unit sphere \(S^{n-1}\) in \(\mathbf{R}^{n}\), \(d\omega (\nu)\) is the normalized measure on \(S^{n-1},\) \(\rho _{\nu }(x)\) is the distance of \(x\) to the nearest point of the boundary \(\partial \Omega \) in the directions \(\pm \nu \), and \(\Omega _{x}=\left\{ y\in \Omega ;\text{ }x+t(y-x)\in \Omega , \text{ }\forall t\in [0,1]\right\} .\) Notice that the domain \(\Omega \) can be unbounded and also a domain with fractal boundary. For convex domains, the above result leads to \[ \int_{\Omega }\left|\nabla u\right|^{2}dx\geq \frac{1}{4} \int_{\Omega }\frac{\left|u(x)\right|^{2}}{d(x,\partial \Omega)^{2}} dx+\frac{n^{(n-2)/n}s_{n-1}^{2/n}}{4\left( \text{vol} \Omega \right) ^{2/n}}\int_{\Omega }\left|u(x)\right|^{2} dx, \] a fact which gives a positive answer to a question raised by \textit{H. Brezis} and \textit{M. Marcus} [``Hardy's inequalities revisited'', Ann. Sc. Norm. Super. Pisa, Cl. Sci. (4) 25, No. 1-2, 217-237 (1997; Zbl 1011.46027)].
scientific article; zbMATH DE number 1797580