Trudinger-Moser inequalities on a closed Riemannian surface with the action of a finite isometric group

Abstract

Let $(\Sigma,g)$ be a closed Riemannian surface, $W^{1,2}(\Sigma,g)$ be the usual Sobolev space, $\textbf{G}$ be a finite isometric group acting on $(\Sigma,g)$, and $\mathscr{H}_\textbf{G}$ be a function space including all functions $u\in W^{1,2}(\Sigma,g)$ with $\int_\Sigma udv_g=0$ and $u(\sigma(x))=u(x)$ for all $\sigma\in \textbf{G}$ and all $x\in\Sigma$. Denote the number of distinct points of the set $\{\sigma(x): \sigma\in \textbf{G}\}$ by $I(x)$ and $\ell=\inf_{x\in \Sigma}I(x)$. Let $\lambda_1^\textbf{G}$ be the first eigenvalue of the Laplace-Beltrami operator on the space $\mathscr{H}_\textbf{G}$. Using blow-up analysis, we prove that if $\alpha<\lambda_1^\textbf{G}$ and $\beta\leq 4\pi\ell$, then there holds $$\sup_{u\in\mathscr{H}_\textbf{G},\,\int_\Sigma|\nabla_gu|^2dv_g-\alpha \int_\Sigma u^2dv_g\leq 1}\int_\Sigma e^{\beta u^2}dv_g<\infty;$$ if $\alpha<\lambda_1^\textbf{G}$ and $\beta>4\pi\ell$, or $\alpha\geq \lambda_1^\textbf{G}$ and $\beta>0$, then the above supremum is infinity; if $\alpha<\lambda_1^\textbf{G}$ and $\beta\leq 4\pi\ell$, then the above supremum can be attained. Moreover, similar inequalities involving higher order eigenvalues are obtained. Our results partially improve original inequalities of J. Moser \cite{Moser}, L. Fontana \cite{Fontana} and W. Chen \cite{Chen-90}.