## Bessel-type operators and a refinement of Hardy’s inequality.(English)Zbl 07456826

Gesztesy, Fritz (ed.) et al., From operator theory to orthogonal polynomials, combinatorics, and number theory. A volume in honor of Lance Littlejohn’s 70th birthday. Cham: Birkhäuser. Oper. Theory: Adv. Appl. 285, 143-172 (2021).
Summary: The principal aim of this paper is to employ Bessel-type operators in proving the inequality \begin{aligned} \int_0^\pi dx |f'(x)|^2 \geq \frac{1}{4} \int_0^\pi dx \frac{|f(x)|^2}{\sin^2 (x)}+\frac{1}{4} \int_0^\pi dx |f(x)|^2,\quad f\in H_0^1 ((0,\pi)), \end{aligned} 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{aligned} \tau_s=-\frac{d^2}{dx^2}+\frac{s^2-(1/4)}{\sin^2 (x)}, \quad s \in [0,\infty), x \in (0,\pi). \end{aligned} The new inequality represents a refinement of Hardy’s classical inequality \begin{aligned} \int_0^\pi dx |f'(x)|^2 \geq \frac{1}{4}\int_0^\pi dx \frac{|f(x)|^2}{x^2}, \quad f\in H_0^1 ((0,\pi)), \end{aligned} and it also improves upon one of its well-known extensions in the form \begin{aligned} \int_0^\pi dx |f'(x)|^2 \geq \frac{1}{4}\int_0^\pi dx \frac{|f(x)|{}^2}{d_{(0,\pi)}(x)^2}, \quad f\in H_0^1 ((0,\pi)), \end{aligned} where $$d_{(0,\pi)}(x)$$ represents the distance from $$x \in (0,\pi)$$ to the boundary $$\{0,\pi\}$$ of $$(0,\pi)$$.
For the entire collection see [Zbl 1479.47003].

### MSC:

 26D10 Inequalities involving derivatives and differential and integral operators 47E05 General theory of ordinary differential operators 34L99 Ordinary differential operators

DLMF
Full Text:

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