Some results of Young-type inequalities

被引:0
作者
Yonghui Ren
机构
[1] Nanjing University of Aeronautics and Astronautics,Department of Mathematics
来源
Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas | 2020年 / 114卷
关键词
Arithmetic–geometric–harmonic; Kantorovich constant; Young-type inequalities; 15A45; 47A30; 15A60;
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摘要
In this paper, one of our main targets is to present some improvements of Young-type inequalities due to Alzer et al. (Linear Multilinear Algebra 63(3):622–635, 2015) under some conditions. That is to say: when 0<ν,τ<1,a,b>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0< \nu , \tau <1,\ a,b>0$$\end{document}, we have a∇νb-a♯νba∇τb-a♯τb≤ν(1-ν)τ(1-τ)and(a∇νb)2-(a♯νb)2(a∇τb)2-(a♯τb)2≤ν(1-ν)τ(1-τ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{a\nabla _{\nu }b-a\sharp _{\nu }b}{a\nabla _{\tau }b-a\sharp _{\tau }b}\le \frac{\nu (1-\nu )}{\tau (1-\tau )} \ \ { \mathrm {and}} \ \ \frac{(a\nabla _{\nu }b)^{2}-(a\sharp _{\nu } b)^{2}}{(a\nabla _{\tau }b)^{2}-(a\sharp _{\tau }b)^{2}}\le \frac{\nu (1-\nu )}{\tau (1-\tau )} \end{aligned}$$\end{document}for (b-a)(τ-ν)≥0;\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(b-a)(\tau -\nu )\ge 0;$$\end{document} and the inequalities are reversed if (b-a)(τ-ν)≤0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(b-a)(\tau -\nu )\le 0.$$\end{document} In addition, we show a new Young-type inequality (1-vN+1+vN+2)a+(1-v2)b≤vvN-(N+1)avb1-v+(a-b)2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} (1-v^{N+1}+v^{N+2})a+(1-v^{2})b\le v^{vN-(N+1)}a^{v}b^{1-v}+(\sqrt{a}-\sqrt{b} \ )^{2} \end{aligned}$$\end{document}for 0≤v≤1,N∈N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0\le v\le 1, N\in {\mathbb {N}}$$\end{document} and a,b>0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a,b>0.$$\end{document} Then we can get some related results about operators, Hilbert–Schmidt norms, determinants by these scalars results.
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