A Remark on Hardy-Trudinger-Moser Inequality

Abstract

Let $\mathbb{B}$ be the unit disc in $\mathbb{R}^2$, $\mathscr{H}$ be the completion of $C_0^{\infty}(\mathbb{B})$ under the norm

$$||u||_{\mathscr{H}}=\Big(\int_{\mathbb{B}}|\nabla u|^2\mathrm{d}x-\int_{\mathbb{B}}\frac{u^2}{(1-|x|^2)^2}\mathrm{d}x\Big)^{\frac{1}{2}},\quad \forall u\in C_0^{\infty}(\mathbb{B}).$$

Using blow-up analysis, we prove that for any $\gamma\leqslant 4\pi$, the supremum

$$\begin{align*}\sup_{u\in\mathscr{H},||u||_{1,h}\leqslant 1}\int_{\mathbb{B}}\mathrm{e}^{\gamma u^2}\mathrm{d}x\end{align*}$$

can be attained by some function $u_0\in \mathscr{H}$ with $||u_0||_{1,h}=1$, where $h$ is a decreasingly nonnegative, radially symmetric function, and satisfies a coercive condition. Namely there exists a constant $\delta>0$ satisfying

$$||u||_{1,h}^2=\|u\|_{\mathscr{H}}^2-\int_{\mathbb{B}}hu^2{\rm d}x\geq \delta\|u\|_{\mathscr{H}}^2,\quad\forall \, u\in\mathscr{H}.$$

This extends earlier results of Wang-Ye [1] and Yang-Zhu [2].