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We prove that u is constant if u is a bounded solution of Aαu(x)=Cn,αP.V.∫Rna(x-y)(u(x)-u(y))|x-y|n+αdy=0,x∈Rn,\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$\begin{aligned} A_{\alpha } u(x) = C_{n,\alpha } \text {P.V.} \int _{\mathbb {R}^n} \frac{a(x-y)(u(x)-u(y))}{|x-y|^{n+\alpha }} \mathrm{d}y=0, \;\; x \in \mathbb {R}^n, \end{aligned}$$\end{document}where the function a:Rn↦R\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$a:\mathbb {R}^n \mapsto \mathbb {R}$$\end{document} be uniformly bounded and radial decreasing. This result can be regarded as the generalization of usual Liouville theorem. To get the proof, we establish a maximum principle involving the nonlocal operator Aα\documentclass[12pt]{minimal}\usepackage{amsmath}\usepackage{wasysym}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{amsbsy}\usepackage{mathrsfs}\usepackage{upgreek}\setlength{\oddsidemargin}{-69pt}\begin{document}$$ A_{\alpha }$$\end{document} for antisymmetric functions on any half space.
Bulletin of the Malaysian Mathematical Sciences Society – Springer Journals
Published: Oct 23, 2020
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