Bessel-Type Operators and a refinement of Hardy's inequality

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 $1/4$ appearing in the above inequality are optimal. In addition, this inequality is strict in the sense that equality holds if and only if $f \equiv 0$. 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 $d_{(0,π)}(x)$ represents the distance from $x \in (0,π)$ to the boundary $\{0,π\}$ of $(0,π)$.

Theorems & Definitions (14)

• Remark 2.1
• Theorem 3.1
• proof
• Remark 3.2
• Remark 3.3
• Remark 3.4
• Remark 3.5
• Remark A.1
• Remark A.2
• Lemma B.1