Bessel-Type Operators and a refinement of Hardy's inequality
arXiv:2102.00106 · doi:10.1007/978-3-030-75425-9_9
Abstract
The principal aim of this paper is to employ Bessel-type operators in proving the inequality \begin{align*} \int_0^πdx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^πdx \, \dfrac{|f(x)|^2}{\sin^2 (x)}+\dfrac{1}{4}\int_0^πdx \, |f(x)|^2,\quad f\in H_0^1 ((0,π)), \end{align*} where both constants appearing in the above inequality are optimal. In addition, this inequality is strict in the sense that equality holds if and only if . This inequality is derived with the help of the exactly solvable, strongly singular, Dirichlet-type Schrödinger operator associated with the differential expression \begin{align*} τ_s=-\dfrac{d^2}{dx^2}+\dfrac{s^2-(1/4)}{\sin^2 (x)}, \quad s \in [0,\infty), \; x \in (0,π). \end{align*} The new inequality represents a refinement of Hardy's classical inequality \begin{align*} \int_0^πdx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^πdx \, \dfrac{|f(x)|^2}{x^2}, \quad f\in H_0^1 ((0,π)), \end{align*} it also improves upon one of its well-known extensions in the form \begin{align*} \int_0^πdx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^πdx \, \dfrac{|f(x)|^2}{d_{(0,π)}(x)^2}, \quad f\in H_0^1 ((0,π)), \end{align*} where represents the distance from to the boundary of .
23 pages, updated references