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# Mathematics > Spectral Theory

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

(Submitted on 29 Jan 2021 (v1), last revised 16 Nov 2021 (this version, v5))

Abstract: The principal aim of this paper is to employ Bessel-type operators in proving the inequality \begin{align*} \int_0^\pi dx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^\pi dx \, \dfrac{|f(x)|^2}{\sin^2 (x)}+\dfrac{1}{4}\int_0^\pi dx \, |f(x)|^2,\quad f\in H_0^1 ((0,\pi)), \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 Schr\"{o}dinger operator associated with the differential expression \begin{align*} \tau_s=-\dfrac{d^2}{dx^2}+\dfrac{s^2-(1/4)}{\sin^2 (x)}, \quad s \in [0,\infty), \; x \in (0,\pi). \end{align*}

The new inequality represents a refinement of Hardy's classical inequality \begin{align*} \int_0^\pi dx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^\pi dx \, \dfrac{|f(x)|^2}{x^2}, \quad f\in H_0^1 ((0,\pi)), \end{align*} it also improves upon one of its well-known extensions in the form \begin{align*} \int_0^\pi dx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^\pi dx \, \dfrac{|f(x)|^2}{d_{(0,\pi)}(x)^2}, \quad f\in H_0^1 ((0,\pi)), \end{align*} where $d_{(0,\pi)}(x)$ represents the distance from $x \in (0,\pi)$ to the boundary $\{0,\pi\}$ of $(0,\pi)$.

## Submission history

From: Fritz Gesztesy [view email]**[v1]**Fri, 29 Jan 2021 23:19:28 GMT (26kb)

**[v2]**Sun, 21 Feb 2021 02:41:30 GMT (28kb)

**[v3]**Sat, 6 Mar 2021 20:26:54 GMT (28kb)

**[v4]**Wed, 17 Mar 2021 06:51:54 GMT (28kb)

**[v5]**Tue, 16 Nov 2021 00:09:19 GMT (29kb)

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