Bessel-Type Operators and a refinement of Hardy's inequality
Fuente:
arXiv
Salvato in:
| Autori principali: | , , |
|---|---|
| Natura: | Preprint |
| Pubblicazione: |
2021
|
| Soggetti: | |
| Accesso online: | |
| Tags: |
Aggiungi Tag
Nessun Tag, puoi essere il primo ad aggiungerne!!
|
| _version_ | 1866909270850666496 |
|---|---|
| author | Gesztesy, Fritz Pang, Michael M. H. Stanfill, Jonathan |
| author_facet | Gesztesy, Fritz Pang, Michael M. H. Stanfill, Jonathan |
| contents | 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 $(0,π)$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2102_00106 |
| institution | arXiv |
| publishDate | 2021 |
| record_format | arxiv |
| spellingShingle | Bessel-Type Operators and a refinement of Hardy's inequality Gesztesy, Fritz Pang, Michael M. H. Stanfill, Jonathan Spectral Theory Mathematical Physics Primary: 26D10, 34A40, 34B20, 34B30, Secondary: 34L10, 34B24, 47A07 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 $(0,π)$. |
| title | Bessel-Type Operators and a refinement of Hardy's inequality |
| topic | Spectral Theory Mathematical Physics Primary: 26D10, 34A40, 34B20, 34B30, Secondary: 34L10, 34B24, 47A07 |
| url | https://arxiv.org/abs/2102.00106 |