Aubin type almost sharp Moser-Trudinger inequality revisited

Fengbo Hang Courant Institute, New York University, 251 Mercer Street, New York NY 10012 fengbo@cims.nyu.edu

Abstract.

We give a new proof of the almost sharp Moser-Trudinger inequality on compact Riemannian manifolds based on the sharp Moser inequality on Euclidean spaces. In particular we can lower the smoothness requirement of the metric and apply the same approach to higher order Sobolev spaces and manifolds with boundary under several boundary conditions.

1. Introduction

Let $n\in\mathbb{N}$ , $n\geq 2$ . For $1<p<n$ , the classical Sobolev inequality states that there exists a positive constant $c\left(n,p\right)$ with

\left\|\varphi\right\|_{L^{\frac{np}{n-p}}}\leq c\left(n,p\right)\left\|\nabla\varphi\right\|_{L^{p}}

(1.1)

for any $\varphi\in C_{c}^{\infty}\left(\mathbb{R}^{n}\right)$ . We denote $S_{n,p}$ as the smallest possible choice for the constant $c\left(n,p\right)$ i.e.

S_{n,p}=\sup_{\varphi\in C_{c}^{\infty}\left(\mathbb{R}^{n}\right)\backslash\left\{0\right\}}\frac{\left\|\varphi\right\|_{L^{\frac{np}{n-p}}}}{\left\|\nabla\varphi\right\|_{L^{p}}}.

(1.2)

Let $\left(M^{n},g\right)$ be a smooth compact Riemannian manifold of dimension $n$ and $1<p<n$ . Aubin [Au1] showed that for any $\varepsilon>0$ , we have

\left\|u\right\|_{L^{\frac{np}{n-p}}\left(M\right)}^{p}\leq\left(S_{n,p}^{p}+\varepsilon\right)\left\|\nabla u\right\|_{L^{p}\left(M\right)}^{p}+c\left(\varepsilon\right)\left\|u\right\|_{L^{p}\left(M\right)}^{p}

(1.3)

for $u\in W^{1,p}\left(M\right)$ . We call (1.3) Aubin’s almost sharp Sobolev inequality. Aubin’s almost sharp Sobolev inequality and its closely related concentration compactness principle play fundamental role in the study of semilinear equations (see [He, Ln1, Ln2]).

On the other hand, let $B_{R}=B_{R}^{n}\subset\mathbb{R}^{n}$ be the open ball with origin as center and $R$ as radius. In [M], it is shown that for any $u\in W_{0}^{1,n}\left(B_{R}\right)\backslash\left\{0\right\}$ ,

\int_{B_{R}}\exp\left(a_{n}\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}\left(B_{R}\right)}^{\frac{n}{n-1}}}\right)dx\leq c\left(n\right)R^{n}.

(1.4)

Here

a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}.

(1.5)

$\left|\mathbb{S}^{n-1}\right|$ is the volume of $\mathbb{S}^{n-1}$ under the standard metric. Note that $a_{n}$ is sharp in the sense that if $a>0$ s.t.

\sup_{u\in W_{0}^{1,n}\left(B_{R}\right)\backslash\left\{0\right\}}\int_{B_{R}}\exp\left(a\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}\left(B_{R}\right)}^{\frac{n}{n-1}}}\right)dx<\infty\text{,}

then we have $a\leq a_{n}$ . For similar inequalities on general smooth compact Riemannian manifolds, in [F], it is shown that if $\left(M^{n},g\right)$ is a smooth compact Riemannian manifold, then for any $u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}$ with $\int_{M}ud\mu=0$ (here $\mu$ is the measure associated with $g$ ),

\int_{M}\exp\left(a_{n}\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}}^{\frac{n}{n-1}}}\right)d\mu\leq c\left(M,g\right).

(1.6)

The Moser-Trudinger inequality (1.6) and its analogy are closely related to spectral geometry through the classical Polyakov formula, Gauss curvature equation and Q curvature equations (see for example [Au2, CY1, CY2, M, OPS] and the references therein). A direct consequence of (1.6) is for $u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}$ with $\int_{M}ud\mu=0$ and $\varepsilon>0$ small,

\int_{M}\exp\left(\left(a_{n}-\varepsilon\right)\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}}^{\frac{n}{n-1}}}\right)d\mu\leq c\left(\varepsilon\right)<\infty.

(1.7)

We call (1.7) as Aubin type almost sharp Moser-Trudinger inequality. It is worth pointing out that for most applications, almost sharp Moser-Trudinger inequality is sufficient (using suitable approximation process when necessary, see for example Lemma 3.2 and the discussion after it). On the other hand, one can pass from (1.7) to (1.6) by blow-up analysis (see [Li1, Li2]).

To continue we observe that Aubin’s almost sharp Sobolev inequality follows from (1.2) by a smart but elementary application of decomposition of unit (see [Au1]). On the other hand, such passage from (1.4) to (1.7) is missing. Fontana’s proof for (1.6) uses potential theory approach by Adams [A] and requires accurate estimate of the Green’s function. This method does not work well for functions satisfying suitable boundary conditions (see [Y2]). The main aim of this note is to provide a direct and elementary method to deduce (1.7) from (1.4). Our approach is motivated from recent progress in concentration compactness principle in critical dimension in [CH, H], where Aubin’s Moser-Trudinger-Onofri inequality on $\mathbb{S}^{n}$ in [Au2] is generalized to higher order moments cases. The sequence of inequalities are motivated from similar inequalities on $\mathbb{S}^{1}$ (see [GrS, OPS, W]). The key point in [CH, H], which is different from [CCH, Ln1, Ln2], is recognizing the importance of the value of defect measure at a single point. This point of view will be crucial here as well (see Proposition 1.1 below). An interesting gain of our approach is we only need the metric to be continuous. Due to the elementary nature of our method, we can easily adapt it to higher order Sobolev spaces and functions satisfying various boundary conditions.

In the remaining part of this section, we will prove Theorem 1.1, which covers (1.7). In Section 2 we will adapt the approach to functions on manifolds with nonempty boundary, either under Dirichlet condition or no boundary condition at all. In Section 3, we prove similar results for second order Sobolev spaces under various boundary conditions and discuss its applications to Q curvature equations in dimension 4. In Section 4, we briefly describe what can be done for higher order Sobolev spaces.

The first fact is a qualitative property of Sobolev functions following from (1.4). It should be compared to [CH, Lemma 2.1] and [H, Lemma 2.1].

Lemma 1.1.

Let $u\in W^{1,n}\left(B_{R}^{n}\right)$ and $a>0$ , then

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-1}}}dx<\infty.

(1.8)

Proof.

We first treat the case $u\in W_{0}^{1,n}\left(B_{R}\right)$ . If $u$ is bounded the claim is clearly true. We assume $u$ is unbounded. For $\varepsilon>0$ , a tiny number to be determined, we can find $v\in C_{c}^{\infty}\left(B_{R}\right)$ such that

\left\|\nabla\left(u-v\right)\right\|_{L^{n}}<\varepsilon.

Let $w=u-v$ , then

\left|u\right|=\left|v+w\right|\leq\left\|v\right\|_{L^{\infty}}+\left|w\right|.

Hence

\left|u\right|^{\frac{n}{n-1}}\leq 2^{\frac{1}{n-1}}\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-1}}+2^{\frac{1}{n-1}}\left|w\right|^{\frac{n}{n-1}}.

It follows that

e^{a\left|u\right|^{\frac{n}{n-1}}}\leq e^{2^{\frac{1}{n-1}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-1}}}e^{2^{\frac{1}{n-1}}a\left|w\right|^{\frac{n}{n-1}}}\leq e^{2^{\frac{1}{n-1}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-1}}}e^{a_{n}\frac{\left|w\right|^{\frac{n}{n-1}}}{\left\|\nabla w\right\|_{L^{n}}^{\frac{n}{n-1}}}}

if $\varepsilon$ is small enough. Using (1.4) we see

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-1}}}dx\leq c\left(n\right)e^{2^{\frac{1}{n-1}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-1}}}R^{n}<\infty.

In general, if $u\in W^{1,n}\left(B_{R}\right)$ , then we can find $\widetilde{u}\in W_{0}^{1,n}\left(B_{2R}\right)$ such that $\left.\widetilde{u}\right|_{B_{R}}=u$ . Hence

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-1}}}dx\leq\int_{B_{2R}}e^{a\left|\widetilde{u}\right|^{\frac{n}{n-1}}}dx<\infty

by the previous discussion.

Next we localize [CH, Lemma 2.2] and [H, Proposition 2.1].

Proposition 1.1.

Let $0<R\leq 1$ , $g$ be a continuous Riemannian metric on $\overline{B_{R}^{n}}$ . Assume $u_{i}\in W^{1,n}\left(B_{R}\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(B_{R}\right)$ and

\left|\nabla_{g}u_{i}\right|_{g}^{n}d\mu\rightarrow\left|\nabla_{g}u\right|_{g}^{n}d\mu+\sigma\text{ as measure on }B_{R}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{1}{n-1}}$ , then for some $r>0$ ,

\sup_{i}\int_{B_{r}}e^{a_{n}p\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

(1.9)

Here

a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}.

(1.10)

Remark 1.1.

[CH, Lemma 2.2] and [H, Proposition 2.1] follows from Proposition 1.1. Hence to derive the main results in [CH, H], we may use Moser’s sharp inequality on Euclidean spaces instead of Fontana’s Moser-Trudinger inequality (1.6).

To prove Proposition 1.1, we first observe that by a linear changing of variable and shrinking $R$ if necessary we can assume $g=g_{ij}dx_{i}dx_{j}$ with $g_{ij}\left(0\right)=\delta_{ij}$ . Fix $p_{1}\in\left(p,\sigma\left(\left\{0\right\}\right)^{-\frac{1}{n-1}}\right)$ , then

\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}.

(1.11)

We can find $\varepsilon>0$ such that

\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}

(1.12)

and

\left(1+\varepsilon\right)p<p_{1}.

(1.13)

Let $v_{i}=u_{i}-u$ , then $v_{i}\rightharpoonup 0$ weakly in $W^{1,n}\left(B_{R}\right)$ , $v_{i}\rightarrow 0$ in $L^{n}\left(B_{R}\right)$ . To continue, we note that on $B_{R}$ we have two Riemannian metrics: The standard Euclidean metric and $g$ . For a function $f$ on $B_{R}$ , we write

\left\|f\right\|_{L^{n}}=\left\|f\right\|_{L^{n}\left(B_{R}\right)}=\left\|f\right\|_{L^{n}\left(B_{R},dx\right)},

(1.14)

here $dx$ is the standard Lebesgue measure. This should be compared with $\left\|f\right\|_{L^{n}\left(B_{R},d\mu\right)}$ . We write $\nabla f$ as the usual gradient of $f$ i.e. the gradient with respect to Euclidean metric and $\nabla_{g}f$ as the gradient of $f$ with respect to metric $g$ .

Let $0<R_{1}<R$ be a small number to be determined. For any $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ , we have

			$\displaystyle\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla v_{i}\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}+\left\\|\varphi\nabla u\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\nabla_{g}u_{i}\right\\|_{L^{n}\left(B_{R},d\mu\right)}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}.$

Here we have used the continuity of $g_{ij}$ , $g_{ij}\left(0\right)=\delta_{ij}$ and the smallness of $R_{1}$ . Hence

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla_{g}u\right\|_{g}^{n}d\mu+c\left(\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx.$

Since $\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}$ , we can choose $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ such that $\left.\varphi\right|_{B_{r}}=1$ for some $r>0$ and

\left(1+\varepsilon\right)\int_{B_{R}}\left|\varphi\right|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}}\left|\varphi\right|^{n}\left|\nabla u\right|^{n}dx<\frac{1}{p_{1}^{n-1}}.

(1.15)

Hence for $i$ large enough, we have

\left\|\nabla\left(\varphi v_{i}\right)\right\|_{L^{n}\left(B_{R}\right)}^{n}<\frac{1}{p_{1}^{n-1}}.

(1.16)

This implies

\left\|\nabla\left(\varphi v_{i}\right)\right\|_{L^{n}\left(B_{R}\right)}^{\frac{n}{n-1}}<\frac{1}{p_{1}}.

(1.17)

On the other hand,

\left|u_{i}\right|^{\frac{n}{n-1}}\leq\left(\left|v_{i}\right|+\left|u\right|\right)^{\frac{n}{n-1}}\leq\left(1+\varepsilon\right)\left|v_{i}\right|^{\frac{n}{n-1}}+c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-1}},

hence

e^{a_{n}\left|u_{i}\right|^{\frac{n}{n-1}}}\leq e^{\left(1+\varepsilon\right)a_{n}\left|v_{i}\right|^{\frac{n}{n-1}}}e^{c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-1}}}.

It follows from (1.4) that

\int_{B_{r}}e^{a_{n}p_{1}\left|v_{i}\right|^{\frac{n}{n-1}}}d\mu\leq c\int_{B_{R}}e^{a_{n}p_{1}\left|\varphi v_{i}\right|^{\frac{n}{n-1}}}dx\leq c\int_{B_{R}}e^{a_{n}\frac{\left|\varphi v_{i}\right|^{\frac{n}{n-1}}}{\left\|\nabla\left(\varphi v_{i}\right)\right\|_{L^{n}}^{\frac{n}{n-1}}}}dx\leq c.

Using $p<\frac{p_{1}}{1+\varepsilon}$ and Lemma 1.1, it follows from Holder’s inequality that $e^{a_{n}\left|u_{i}\right|^{\frac{n}{n-1}}}$ is bounded in $L^{p}\left(B_{r},d\mu\right)$ . This finishes the proof of Proposition 1.1.

Theorem 1.1.

Let $\left(M^{n},g\right)$ be a $C^{1}$ compact manifold with a continuous Riemannian metric $g$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ (here $u_{E}=\frac{1}{\mu\left(E\right)}\int_{E}ud\mu$ ), then for any $0<a<a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}$ ,

\int_{M}e^{a\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}}^{\frac{n}{n-1}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(1.18)

Remark 1.2.

It is interesting that we can find a continuous Riemannian metric on $M$ such that

\sup_{\begin{subarray}{c}u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}\\ u_{M}=0\end{subarray}}\int_{M}\exp\left(a_{n}\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}\left(M\right)}^{\frac{n}{n-1}}}\right)d\mu=\infty.

(1.19)

Indeed let $\left(U,\phi\right)$ be a coordinate on $M$ such that $\phi\left(U\right)=B_{2}$ . For convenience we identify $U$ with $B_{2}$ . Let

\alpha\left(r\right)=\frac{1}{\sqrt{\log\frac{1}{r}+4}}\quad\text{for }0<r<1\text{,}

(1.20)

then $0\leq\alpha\leq\frac{1}{2}$ and

\int_{0}^{1}\frac{\alpha\left(r\right)}{r}dr=\infty.

(1.21)

We can find a continuous Riemannian metric on $M$ such that on $B_{1}$ , $g=g_{ij}dx_{i}dx_{j}$ with

g_{ij}=\left(1-\alpha\left(\left|x\right|\right)\right)^{-\frac{2}{n}}\frac{x_{i}}{\left|x\right|}\frac{x_{j}}{\left|x\right|}+\left(1-\alpha\left(\left|x\right|\right)\right)^{\frac{2}{n\left(n-1\right)}}\left(\delta_{ij}-\frac{x_{i}}{\left|x\right|}\frac{x_{j}}{\left|x\right|}\right).

(1.22)

Hence

g^{ij}=\left(1-\alpha\left(\left|x\right|\right)\right)^{\frac{2}{n}}\frac{x_{i}}{\left|x\right|}\frac{x_{j}}{\left|x\right|}+\left(1-\alpha\left(\left|x\right|\right)\right)^{-\frac{2}{n\left(n-1\right)}}\left(\delta_{ij}-\frac{x_{i}}{\left|x\right|}\frac{x_{j}}{\left|x\right|}\right).

(1.23)

For $0<\varepsilon<1$ , we define

v_{\varepsilon}\left(x\right)=\left\{\begin{tabular}[]{ll}$\log\varepsilon,$&if $\left|x\right|\leq\varepsilon$;\\ $\log\left|x\right|,$&if $\varepsilon<\left|x\right|<1$;\\ $0,$&if $x\in M\backslash B_{1}$.\end{tabular}\right.

Then

$\displaystyle\left\\|\nabla v_{\varepsilon}\right\\|_{L^{n}}^{n}$	$\displaystyle=$	$\displaystyle\int_{\varepsilon<\left\|x\right\|<1}\left(g^{ij}\frac{x_{i}}{\left\|x\right\|^{2}}\frac{x_{j}}{\left\|x\right\|^{2}}\right)^{\frac{n}{2}}dx$
	$\displaystyle=$	$\displaystyle\int_{\varepsilon<\left\|x\right\|<1}\frac{1-\alpha\left(\left\|x\right\|\right)}{\left\|x\right\|^{n}}dx$
	$\displaystyle=$	$\displaystyle\left\|\mathbb{S}^{n-1}\right\|\left(\log\frac{1}{\varepsilon}-\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr\right).$

On the other hand, $v_{\varepsilon,M}=O\left(1\right)$ as $\varepsilon\rightarrow 0^{+}$ . Letting $u_{\varepsilon}=v_{\varepsilon}-v_{\varepsilon,M}$ , we claim

\lim_{\varepsilon\rightarrow 0^{+}}\int_{M}\exp\left(a_{n}\frac{\left|u_{\varepsilon}\right|^{\frac{n}{n-1}}}{\left\|\nabla u_{\varepsilon}\right\|_{L^{n}\left(M\right)}^{\frac{n}{n-1}}}\right)d\mu=\infty.

(1.24)

Indeed,

			$\displaystyle\int_{M}\exp\left(a_{n}\frac{\left\|u_{\varepsilon}\right\|^{\frac{n}{n-1}}}{\left\\|\nabla u_{\varepsilon}\right\\|_{L^{n}\left(M\right)}^{\frac{n}{n-1}}}\right)d\mu$
		$\displaystyle\geq$	$\displaystyle\int_{B_{\varepsilon}}\exp\left(a_{n}\frac{\left\|\log\varepsilon-v_{\varepsilon,M}\right\|^{\frac{n}{n-1}}}{\left\\|\nabla v_{\varepsilon}\right\\|_{L^{n}\left(M\right)}^{\frac{n}{n-1}}}\right)dx$
		$\displaystyle=$	$\displaystyle\left\|B_{1}\right\|\varepsilon^{n}\exp\left(n\frac{\left\|\log\frac{1}{\varepsilon}+v_{\varepsilon,M}\right\|^{\frac{n}{n-1}}}{\left(\log\frac{1}{\varepsilon}-\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr\right)^{\frac{1}{n-1}}}\right)$
		$\displaystyle\geq$	$\displaystyle c\left(n\right)\exp\left[n\log\frac{1}{\varepsilon}\cdot\left(1-\frac{c_{1}}{\log\frac{1}{\varepsilon}}\right)\left(1+\frac{1}{n-1}\frac{\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr}{\log\frac{1}{\varepsilon}}\right)-n\log\frac{1}{\varepsilon}\right]$
		$\displaystyle\geq$	$\displaystyle c\left(n\right)\exp\left(\frac{n}{n-1}\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr-c\right)$
		$\displaystyle\geq$	$\displaystyle c\exp\left(\frac{n}{n-1}\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr\right).$

This estimate together with (1.21) implies (1.24). (1.19) follows.

Proof of Theorem 1.1.

For any $v\in W^{1,n}\left(M\right)$ , we know

\left\|v-v_{M}\right\|_{L^{n}}\leq c\left\|\nabla v\right\|_{L^{n}}.

On the other hand,

\left\|v-v_{E}\right\|_{L^{n}}\leq\frac{c}{\mu\left(E\right)^{\frac{1}{n}}}\left\|v\right\|_{L^{n}}\leq c\left(\delta\right)\left\|v\right\|_{L^{n}}.

Replacing $v$ by $v-v_{M}$ , we see

\left\|v-v_{E}\right\|_{L^{n}}\leq c\left(\delta\right)\left\|v-v_{M}\right\|_{L^{n}}\leq c\left(\delta\right)\left\|\nabla v\right\|_{L^{n}}.

If (1.18) is not true, then we can find a sequence $u_{i}\in W^{1,n}\left(M\right)$ , $E_{i}\subset M$ with $\mu\left(E_{i}\right)\geq\delta$ , $u_{i,E_{i}}=0$ , $\left\|\nabla u_{i}\right\|_{L^{n}}=1$ and

\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu\rightarrow\infty

as $i\rightarrow\infty$ . Since

\left\|u_{i}\right\|_{L^{n}}=\left\|u_{i}-u_{i,E_{i}}\right\|_{L^{n}}\leq c\left(\delta\right)\left\|\nabla u_{i}\right\|_{L^{n}}=c\left(\delta\right),

we see $u_{i}$ is bounded in $W^{1,n}\left(M\right)$ . Hence after passing to a subsequence we can find $u\in W^{1,n}\left(M\right)$ such that $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(M\right)$ and a measure on $M$ , $\sigma$ such that

\left|\nabla u_{i}\right|^{n}d\mu\rightarrow\left|\nabla u\right|^{n}d\mu+\sigma

as measure. Note that $\sigma\left(M\right)\leq 1$ . For any $x\in M$ , since

0<\frac{a}{a_{n}}<\sigma\left(\left\{x\right\}\right)^{-\frac{1}{n-1}},

it follows from Proposition 1.1 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(x\right)}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

A covering argument implies

\sup_{i}\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

This contradicts with the choice of $u_{i}$ .

The argument above is flexible and can be applied to other cases as well.

Example 1.1.

Let $M^{n}$ be a $C^{1}$ compact manifold with a $C^{0}$ Riemannian metric $g$ . Denote

\kappa\left(M,g\right)=\inf_{\begin{subarray}{c}u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}\\ u_{M}=0\end{subarray}}\frac{\left\|\nabla u\right\|_{L^{n}}}{\left\|u\right\|_{L^{n}}}.

(1.25)

It follows from Poincare inequality that $\kappa\left(M,g\right)$ is a positive number. Assume $0\leq\kappa<\kappa\left(M,g\right)$ , $0<a<a_{n}$ , $u\in W^{1,n}\left(M\right)$ with $u_{M}=0$ and

\left\|\nabla u\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u\right\|_{L^{n}}^{n}\leq 1,

(1.26)

then we have

\int_{M}e^{a\left|u\right|^{\frac{n}{n-1}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(1.27)

Proof.

If the claim is not true, then we can find a sequence $u_{i}\in W^{1,n}\left(M\right)$ s.t. $u_{i,M}=0$ , $\left\|\nabla u_{i}\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u_{i}\right\|_{L^{n}}^{n}\leq 1$ and $\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu\rightarrow\infty$ as $i\rightarrow\infty$ . By the definition of $\kappa\left(M,g\right)$ , we see

\left(\kappa\left(M,g\right)^{n}-\kappa^{n}\right)\left\|u_{i}\right\|_{L^{n}}^{n}\leq\left\|\nabla u_{i}\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u_{i}\right\|_{L^{n}}^{n}\leq 1,

hence

\left\|u_{i}\right\|_{L^{n}}^{n}\leq\frac{1}{\kappa\left(M,g\right)^{n}-\kappa^{n}}.

This together with $\left\|\nabla u_{i}\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u_{i}\right\|_{L^{n}}^{n}\leq 1$ implies

\left\|\nabla u_{i}\right\|_{L^{n}}^{n}\leq\frac{\kappa\left(M,g\right)^{n}}{\kappa\left(M,g\right)^{n}-\kappa^{n}}.

In other words, $u_{i}$ is bounded in $W^{1,n}\left(M\right)$ . After passing to a subsequence, we can find $u\in W^{1,n}\left(M\right)$ s.t. $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(M\right)$ , $u_{i}\rightarrow u$ in $L^{n}\left(M\right)$ and

\left|\nabla u_{i}\right|^{n}d\mu\rightarrow\left|\nabla u\right|^{n}d\mu+\sigma

as measure. In particular, $u_{M}=0$ . On the other hand,

\left\|\nabla u\right\|_{L^{n}}^{n}+\sigma\left(M\right)-\kappa^{n}\left\|u\right\|_{L^{n}}^{n}\leq 1.

Since $\left\|\nabla u\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u\right\|_{L^{n}}^{n}\geq 0$ , we see $\sigma\left(M\right)\leq 1$ . For any $x\in M$ , since

0<\frac{a}{a_{n}}<\sigma\left(\left\{x\right\}\right)^{-\frac{1}{n-1}},

it follows from Proposition 1.1 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(x\right)}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

This together with compactness of $M$ implies

\sup_{i}\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

It gives us a contradiction by the choice of $u_{i}$ at the beginning.

2. Functions on manifolds with nonempty boundary

In this section, we will study functions on manifolds with boundary under no boundary conditions or Dirichlet boundary condition. At first we need to discuss functions on half ball. Let $R>0$ , we denote

B_{R}^{+}=\left\{x\in B_{R}:x=\left(x^{\prime},x_{n}\right),x^{\prime}\in\mathbb{R}^{n-1},x_{n}\geq 0\right\}.

(2.1)

Lemma 2.1.

Assume $u\in W^{1,n}\left(B_{R}^{+}\right)\backslash\left\{0\right\}$ such that $u\left(x\right)=0$ for $\left|x\right|$ close to $R$ , then

\int_{B_{R}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}\left(B_{R}^{+}\right)}^{\frac{n}{n-1}}}}dx\leq c\left(n\right)R^{n}.

(2.2)

Proof.

For $\left|x\right|<R$ , let

v\left(x\right)=\left\{\begin{array}[]{cc}u\left(x\right),&\text{if }x_{n}>0;\\ u\left(x^{\prime},-x_{n}\right),&\text{if }x_{n}<0\text{.}\end{array}\right.

Then $v\in W_{0}^{1,n}\left(B_{R}\right)$ with $\left\|\nabla v\right\|_{L^{n}\left(B_{R}\right)}^{n}=2\left\|\nabla u\right\|_{L^{n}\left(B_{R}^{+}\right)}^{n}$ . Using (1.4) we see

$\displaystyle\int_{B_{R}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}\frac{\left\|u\right\|^{\frac{n}{n-1}}}{\left\\|\nabla u\right\\|_{L^{n}\left(B_{R}^{+}\right)}^{\frac{n}{n-1}}}}dx$	$\displaystyle=$	$\displaystyle\int_{B_{R}^{+}}e^{a_{n}\frac{\left\|u\right\|^{\frac{n}{n-1}}}{\left\\|\nabla v\right\\|_{L^{n}\left(B_{R}\right)}^{\frac{n}{n-1}}}}dx$
	$\displaystyle\leq$	$\displaystyle\int_{B_{R}}e^{a_{n}\frac{\left\|v\right\|^{\frac{n}{n-1}}}{\left\\|\nabla v\right\\|_{L^{n}\left(B_{R}\right)}^{\frac{n}{n-1}}}}dx$
	$\displaystyle\leq$	$\displaystyle c\left(n\right)R^{n}.$

We also need a qualitative property of Sobolev functions.

Lemma 2.2.

Let $u\in W^{1,n}\left(B_{R}^{+}\right)$ and $a>0$ , then

\int_{B_{R}^{+}}e^{a\left|u\right|^{\frac{n}{n-1}}}dx<\infty.

(2.3)

Proof.

We can find $\widetilde{u}\in W^{1,n}\left(B_{R}\right)$ such that $\left.\widetilde{u}\right|_{B_{R}^{+}}=u$ . Then it follows from Lemma 1.1 that

\int_{B_{R}^{+}}e^{a\left|u\right|^{\frac{n}{n-1}}}dx\leq\int_{B_{R}}e^{a\left|\widetilde{u}\right|^{\frac{n}{n-1}}}dx<\infty.

Proposition 2.1.

Let $0<R\leq 1$ , $g$ be a continuous Riemannian metric on $\overline{B_{R}^{+}}$ . Assume $u_{i}\in W^{1,n}\left(B_{R}^{+}\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(B_{R}^{+}\right)$ and

\left|\nabla_{g}u_{i}\right|_{g}^{n}d\mu\rightarrow\left|\nabla_{g}u\right|_{g}^{n}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{1}{n-1}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}p\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

(2.4)

Here

a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}.

(2.5)

Proof.

By linear changing of variable and shrinking $R$ if necessary we can assume $g=g_{ij}dx_{i}dx_{j}$ with $g_{ij}\left(0\right)=\delta_{ij}$ . Fix $p_{1}\in\left(p,\sigma\left(\left\{0\right\}\right)^{-\frac{1}{n-1}}\right)$ , then

\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}.

(2.6)

We can find $\varepsilon>0$ such that

\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}

(2.7)

and

\left(1+\varepsilon\right)p<p_{1}.

(2.8)

Let $v_{i}=u_{i}-u$ , then $v_{i}\rightharpoonup 0$ weakly in $W^{1,n}\left(B_{R}^{+}\right)$ , $v_{i}\rightarrow 0$ in $L^{n}\left(B_{R}^{+}\right)$ . Let $0<R_{1}<R$ be a small number to be determined. For any $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}^{+}\right)$ , we have

			$\displaystyle\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}^{+}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla v_{i}\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}+\left\\|\varphi\nabla u\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\nabla_{g}u_{i}\right\\|_{L^{n}\left(B_{R}^{+},d\mu\right)}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}.$

Hence

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla_{g}u\right\|_{g}^{n}d\mu+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx.$

Since $\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{n-1}}$ , we can choose $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}^{+}\right)$ such that $\left.\varphi\right|_{B_{r}^{+}}=1$ for some $r>0$ and

\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left|\varphi\right|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left|\varphi\right|^{n}\left|\nabla u\right|^{n}dx<\frac{1}{p_{1}^{n-1}}.

(2.9)

Hence for $i$ large enough, we have

\left\|\nabla\left(\varphi v_{i}\right)\right\|_{L^{n}}^{n}<\frac{1}{p_{1}^{n-1}}.

(2.10)

On the other hand,

\left|u_{i}\right|^{\frac{n}{n-1}}\leq\left(\left|v_{i}\right|+\left|u\right|\right)^{\frac{n}{n-1}}\leq\left(1+\varepsilon\right)\left|v_{i}\right|^{\frac{n}{n-1}}+c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-1}},

hence

e^{2^{-\frac{1}{n-1}}a_{n}\left|u_{i}\right|^{\frac{n}{n-1}}}\leq e^{2^{-\frac{1}{n-1}}\left(1+\varepsilon\right)a_{n}\left|v_{i}\right|^{\frac{n}{n-1}}}e^{c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-1}}}.

It follows from Lemma 2.1 that

$\displaystyle\int_{B_{r}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}p_{1}\left\|v_{i}\right\|^{\frac{n}{n-1}}}d\mu$	$\displaystyle\leq$	$\displaystyle c\int_{B_{R}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}p_{1}\left\|\varphi v_{i}\right\|^{\frac{n}{n-1}}}dx$
	$\displaystyle\leq$	$\displaystyle c\int_{B_{R}^{+}}e^{2^{-\frac{1}{n-1}}a_{n}\frac{\left\|\varphi v_{i}\right\|^{\frac{n}{n-1}}}{\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}}^{\frac{n}{n-1}}}}dx$
	$\displaystyle\leq$	$\displaystyle c.$

Using $p<\frac{p_{1}}{1+\varepsilon}$ and Lemma 2.2, it follows from Holder’s inequality that $e^{2^{-\frac{1}{n-1}}a_{n}\left|u_{i}\right|^{\frac{n}{n-1}}}$ is bounded in $L^{p}\left(B_{r}^{+},d\mu\right)$ .

Theorem 2.1.

Let $M^{n}$ be a $C^{1}$ compact manifold with boundary and $g$ be a continuous Riemannian metric on $M$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ , then for any $0<a<a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}$ ,

\int_{M}e^{2^{-\frac{1}{n-1}}a\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}}^{\frac{n}{n-1}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(2.11)

Proof.

For any $v\in W^{1,n}\left(M\right)$ ,

\left\|v-v_{M}\right\|_{L^{n}}\leq c\left\|\nabla v\right\|_{L^{n}}.

On the other hand,

\left\|v-v_{E}\right\|_{L^{n}}\leq\frac{c}{\mu\left(E\right)^{\frac{1}{n}}}\left\|v\right\|_{L^{n}}\leq c\left(\delta\right)\left\|v\right\|_{L^{n}}.

Replacing $v$ by $v-v_{M}$ , we see

\left\|v-v_{E}\right\|_{L^{2}}\leq c\left(\delta\right)\left\|v-v_{M}\right\|_{L^{n}}\leq c\left(\delta\right)\left\|\nabla v\right\|_{L^{n}}.

If (2.11) is not true, then we can find a sequence $u_{i}\in W^{1,n}\left(M\right)$ , $E_{i}\subset M$ with $\mu\left(E_{i}\right)\geq\delta$ , $u_{i,E_{i}}=0$ , $\left\|\nabla u_{i}\right\|_{L^{n}}=1$ and

\int_{M}e^{2^{-\frac{1}{n-1}}a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu\rightarrow\infty

as $i\rightarrow\infty$ . Since

\left\|u_{i}\right\|_{L^{n}}=\left\|u_{i}-u_{i,E_{i}}\right\|_{L^{n}}\leq c\left(\delta\right)\left\|\nabla u_{i}\right\|_{L^{n}}=c\left(\delta\right),

we see $u_{i}$ is bounded in $W^{1,n}\left(M\right)$ . After passing to a subsequence we can find $u\in W^{1,n}\left(M\right)$ such that $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(M\right)$ and a measure on $M$ , $\sigma$ such that

\left|\nabla u_{i}\right|^{n}d\mu\rightarrow\left|\nabla u\right|^{n}d\mu+\sigma

as measure. Note that $\sigma\left(M\right)\leq 1$ . For any $x\in M\backslash\partial M$ , since

0<\frac{a}{a_{n}}<\sigma\left(\left\{x\right\}\right)^{-\frac{1}{n-1}},

it follows from Proposition 1.1 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(x\right)}e^{a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

Hence

\sup_{i}\int_{B_{r}\left(x\right)}e^{2^{-\frac{1}{n-1}}a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

For $x\in\partial M$ , using $0<\frac{a}{a_{n}}<\sigma\left(\left\{x\right\}\right)^{-\frac{1}{n-1}}$ and Proposition 2.1, we can find $r>0$ such that

\sup_{i}\int_{B_{r}\left(x\right)}e^{2^{-\frac{1}{n-1}}a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

A covering argument implies

\sup_{i}\int_{M}e^{2^{-\frac{1}{n-1}}a\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

This contradicts with the choice of $u_{i}$ .

Corollary 2.1.

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded open domain with $C^{1}$ boundary, $0<a<a_{n}$ , then for any $u\in W^{1,n}\left(\Omega\right)\backslash\left\{0\right\}$ with $\int_{\Omega}udx=0$ , we have

\int_{\Omega}e^{2^{-\frac{1}{n-1}}a\frac{\left|u\right|^{\frac{n}{n-1}}}{\left\|\nabla u\right\|_{L^{n}}^{\frac{n}{n-1}}}}dx\leq c\left(a,\Omega\right)<\infty.

(2.12)

Corollary 2.1 should be compared to [Ci, Theorem 1.1].

Example 2.1.

Let $M^{n}$ be a $C^{1}$ compact manifold with boundary and $g$ be a continuous Riemannian metric on $M$ . Denote

\kappa\left(M,g\right)=\inf_{\begin{subarray}{c}u\in W^{1,n}\left(M\right)\backslash\left\{0\right\}\\ u_{M}=0\end{subarray}}\frac{\left\|\nabla u\right\|_{L^{n}}}{\left\|u\right\|_{L^{n}}}.

(2.13)

Assume $0\leq\kappa<\kappa\left(M,g\right)$ , $0<a<a_{n}$ , $u\in W^{1,n}\left(M\right)$ with $u_{M}=0$ and

\left\|\nabla u\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u\right\|_{L^{n}}^{n}\leq 1,

(2.14)

then we have

\int_{M}e^{2^{-\frac{1}{n-1}}a\left|u\right|^{\frac{n}{n-1}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(2.15)

Since the proof of Example 2.1 is almost identical to the proof of Example 1.1 (using Proposition 1.1 and 2.1 when necessary), we omit it here. Example 2.1 should be compared with [N1].

Next we switch to the zero boundary value case. For $R>0$ , we denote $\Sigma_{R}$ as the base of $B_{R}^{+}$ i.e.

\Sigma_{R}=\left\{\left(x^{\prime},0\right):x^{\prime}\in\mathbb{R}^{n-1},\left|x^{\prime}\right|<R\right\}.

(2.16)

Proposition 2.2.

Let $0<R\leq 1$ , $g$ be a continuous Riemannian metric on $\overline{B_{R}^{+}}$ . Assume $u_{i}\in W^{1,n}\left(B_{R}^{+}\right)$ s.t. $\left.u_{i}\right|_{\Sigma_{R}}=0$ , $u_{i}\rightharpoonup u$ weakly in $W^{1,n}\left(B_{R}^{+}\right)$ and

\left|\nabla_{g}u_{i}\right|_{g}^{n}d\mu\rightarrow\left|\nabla_{g}u\right|_{g}^{n}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{1}{n-1}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{a_{n}p\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

(2.17)

Here

a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}.

(2.18)

Proof.

Let $h$ be a continuous Riemannian metric on $\overline{B_{R}}$ , which is an extension of $g$ . We define

v_{i}\left(x\right)=\left\{\begin{tabular}[]{ll}$u_{i}\left(x\right),$&if $x\in B_{R}^{+}$\\ $0,$&if $x\in B_{R}\backslash B_{R}^{+}$\end{tabular}\right.,\quad v\left(x\right)=\left\{\begin{tabular}[]{ll}$u\left(x\right),$&if $x\in B_{R}^{+}$\\ $0,$&if $x\in B_{R}\backslash B_{R}^{+}$\end{tabular}\right.,

and a measure $\tau$ on $B_{R}$ by $\tau\left(E\right)=\sigma\left(E\cap B_{R}\right)$ for any Borel set $E\subset B_{R}$ . Then $v_{i},v\in W^{1,n}\left(B_{R}\right)$ , $v_{i}\rightharpoonup v$ weakly in $W^{1,n}\left(B_{R}\right)$ and

\left|\nabla_{h}v_{i}\right|_{h}^{n}d\mu_{h}\rightarrow\left|\nabla_{h}v\right|_{h}^{n}d\mu_{h}+\tau\text{ as measure on }B_{R}.

Here $\mu_{h}$ is the measure associated with $h$ . Since $\tau\left(\left\{0\right\}\right)=\sigma\left(\left\{0\right\}\right)$ , it follows from Proposition 2.1 that for some $r>0$ ,

\sup_{i}\int_{B_{r}}e^{a_{n}p\left|v_{i}\right|^{\frac{n}{n-1}}}d\mu_{h}<\infty.

Hence

\sup_{i}\int_{B_{r}^{+}}e^{a_{n}p\left|u_{i}\right|^{\frac{n}{n-1}}}d\mu<\infty.

Theorem 2.2.

Let $M^{n}$ be a $C^{1}$ compact manifold with boundary and $g$ be a continuous Riemannian metric on $M$ . Denote

\kappa_{0}\left(M,g\right)=\inf_{u\in W_{0}^{1,n}\left(M\right)\backslash\left\{0\right\}}\frac{\left\|\nabla u\right\|_{L^{n}}}{\left\|u\right\|_{L^{n}}}.

(2.19)

Assume $0\leq\kappa<\kappa_{0}\left(M,g\right)$ , $0<a<a_{n}=n\left|\mathbb{S}^{n-1}\right|^{\frac{1}{n-1}}$ , $u\in W_{0}^{1,n}\left(M\right)$ and

\left\|\nabla u\right\|_{L^{n}}^{n}-\kappa^{n}\left\|u\right\|_{L^{n}}^{n}\leq 1,

(2.20)

then we have

\int_{M}e^{a\left|u\right|^{\frac{n}{n-1}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(2.21)

Since the proof of Theorem 2.2 is almost identical to the proof of Example 1.1 (using Proposition 1.1 and 2.2 when necessary), we omit it here. Theorem 2.2 should be compared with [AD, N2, T, Y1].

3. Second order Sobolev spaces

Let $n\in\mathbb{N}$ , $n\geq 3$ and $\Omega\subset\mathbb{R}^{n}$ be an open domain. For $1\leq p<\infty$ and $u\in W^{2,p}\left(\Omega\right)$ , we denote

\left\|u\right\|_{W^{2,p}\left(\Omega\right)}=\left[\int_{\Omega}\left(\left|u\right|^{p}+\left|\nabla u\right|^{p}+\left|D^{2}u\right|^{p}\right)dx\right]^{\frac{1}{p}}

(3.1)

and $W_{0}^{2,p}\left(\Omega\right)$ as the closure of $C_{c}^{\infty}\left(\Omega\right)$ in $W^{2,p}\left(\Omega\right)$ . For $u\in W^{2,\infty}\left(\Omega\right)$ , we denote

\left\|u\right\|_{W^{2,\infty}\left(\Omega\right)}=\left\|u\right\|_{L^{\infty}\left(\Omega\right)}+\left\|\nabla u\right\|_{L^{\infty}\left(\Omega\right)}+\left\|D^{2}u\right\|_{L^{\infty}\left(\Omega\right)}.

(3.2)

For $R>0$ , it is shown in [A] that for any $u\in W_{0}^{2,\frac{n}{2}}\left(B_{R}^{n}\right)\backslash\left\{0\right\}$ ,

\int_{B_{R}}\exp\left(a_{2,n}\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{n-2}}}\right)dx\leq c\left(n\right)R^{n}.

(3.3)

Here

a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}

(3.4)

and

\Gamma\left(\alpha\right)=\int_{0}^{\infty}t^{\alpha-1}e^{-t}dt

(3.5)

for $\alpha>0$ .

Similar to Lemma 1.1, we have

Lemma 3.1.

If $u\in W^{2,\frac{n}{2}}\left(B_{R}^{n}\right)$ , then for any $a>0$ ,

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-2}}}dx<\infty.

(3.6)

Proof.

First we assume $u\in W_{0}^{2,\frac{n}{2}}\left(B_{R}\right)$ . Without losing of generality, we can assume $u$ is unbounded. For $\varepsilon>0$ , a tiny number to be determined, we can find $v\in C_{c}^{\infty}\left(B_{R}\right)$ such that

\left\|u-v\right\|_{W^{2,\frac{n}{2}}}<\varepsilon.

Hence

\left\|\Delta u-\Delta v\right\|_{L^{\frac{n}{2}}}\leq c\left(n\right)\varepsilon.

Let $w=u-v$ , then

\left|u\right|=\left|v+w\right|\leq\left\|v\right\|_{L^{\infty}}+\left|w\right|.

Hence

\left|u\right|^{\frac{n}{n-2}}\leq 2^{\frac{2}{n-2}}\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-2}}+2^{\frac{2}{n-2}}\left|w\right|^{\frac{n}{n-2}}.

It follows that

e^{a\left|u\right|^{\frac{n}{n-1}}}\leq e^{2^{\frac{2}{n-2}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-2}}}e^{2^{\frac{2}{n-2}}a\left|w\right|^{\frac{n}{n-2}}}\leq e^{2^{\frac{2}{n-2}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-2}}}e^{a_{2,n}\frac{\left|w\right|^{\frac{n}{n-2}}}{\left\|\Delta w\right\|_{L^{\frac{n}{2}}}^{\frac{n}{n-2}}}}

if $\varepsilon$ is small enough. Hence

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-2}}}dx\leq c\left(n\right)e^{2^{\frac{2}{n-2}}a\left\|v\right\|_{L^{\infty}}^{\frac{n}{n-2}}}R^{n}<\infty.

In general, if $u\in W^{2,\frac{n}{2}}\left(B_{R}\right)$ , then we can find $\widetilde{u}\in W_{0}^{2,\frac{n}{2}}\left(B_{2R}\right)$ such that $\left.\widetilde{u}\right|_{B_{R}}=u$ . Hence

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-2}}}dx\leq\int_{B_{2R}}e^{a\left|\widetilde{u}\right|^{\frac{n}{n-2}}}dx<\infty.

Proposition 3.1.

Let $0<R\leq 1$ , $g$ be a $C^{1}$ Riemannian metric on $\overline{B_{R}^{n}}$ . Assume $u_{i}\in W^{2,\frac{n}{2}}\left(B_{R}^{n}\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}\right)$ and

\left|\Delta_{g}u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu+\sigma\text{ as measure on }B_{R}.

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}$ , then for some $r>0$ ,

\sup_{i}\int_{B_{r}}e^{a_{2,n}p\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

(3.7)

Here

a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}

(3.8)

By a linear changing of variable and shrinking $R$ if necessary we can assume $g=g_{ij}dx_{i}dx_{j}$ with $g_{ij}\left(0\right)=\delta_{ij}$ . Let

G=\det\left[g_{ij}\right]_{1\leq i,j\leq n};\quad\left[g^{ij}\right]_{1\leq i,j\leq n}=\left[g_{ij}\right]_{1\leq i,j\leq n}^{-1}.

If $v$ is a function on $B_{R}$ , then

	$\displaystyle\Delta_{g}v$	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{G}}\partial_{i}\left(\sqrt{G}g^{ij}\partial_{j}v\right)$
		$\displaystyle=$	$\displaystyle g^{ij}\partial_{ij}v+\partial_{i}g^{ij}\partial_{j}v+g^{ij}\partial_{i}\log\sqrt{G}\partial_{j}v.$

For $v\in W_{0}^{2,\frac{n}{2}}\left(B_{R}\right)$ , then standard elliptic theory (see [GiT]) tells us

\left\|v\right\|_{W^{2,\frac{n}{2}}\left(B_{R}\right)}\leq c_{1}\left\|\Delta_{g}v\right\|_{L^{\frac{n}{2}}\left(B_{R}\right)}.

(3.9)

On the other hand, let $0<R_{1}<R$ be a small number and $v\in W_{0}^{2,\frac{n}{2}}\left(B_{R_{1}}\right)$ , we have

	$\displaystyle\Delta v$	$\displaystyle=$	$\displaystyle\left(\delta_{ij}-g^{ij}\right)\partial_{ij}v+g^{ij}\partial_{ij}v$
		$\displaystyle=$	$\displaystyle\Delta_{g}v+\left(\delta_{ij}-g^{ij}\right)\partial_{ij}v-\partial_{i}g^{ij}\partial_{j}v-g^{ij}\partial_{i}\log\sqrt{G}\partial_{j}v.$

It follows that

\left|\Delta v\right|\leq\left|\Delta_{g}v\right|+\varepsilon_{1}\left|D^{2}v\right|+c\left|\nabla v\right|.

(3.10)

Here $\varepsilon_{1}=\varepsilon_{1}\left(R_{1}\right)>0$ . Moreover $\varepsilon_{1}\rightarrow 0$ as $R_{1}\rightarrow 0^{+}$ . We have

			$\displaystyle\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}+\varepsilon_{1}\left\\|D^{2}v\right\\|_{L^{\frac{n}{2}}}+c\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}\right)^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left[\left(1+c_{1}\varepsilon_{1}\right)\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}+c\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}\right]^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+c\varepsilon_{1}\right)\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon_{1}\right)\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}.$

To continue, we fix $p_{1}\in\left(p,\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}\right)$ , then

\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.12)

We can find $\varepsilon>0$ s.t.

\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}

(3.13)

and

\left(1+\varepsilon\right)p<p_{1}.

(3.14)

Let $v_{i}=u_{i}-u$ , then $v_{i}\rightharpoonup 0$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}\right)$ and $v_{i}\rightarrow 0$ in $W^{1,\frac{n}{2}}\left(B_{R}\right)$ . Let $0<R_{1}<R$ be a small number to be determined. For $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ , by the previous discussion we have

			$\displaystyle\left\\|\Delta\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+c\varepsilon_{1}\right)\left\\|\Delta_{g}\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon_{1}\right)\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+c\varepsilon_{1}\right)\left(\left\\|\varphi\Delta_{g}v_{i}\right\\|_{L^{\frac{n}{2}}}+c\left\\|\varphi\right\\|_{W^{2,\infty}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}\right)^{\frac{n}{2}}+c\left(\varepsilon_{1}\right)\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{4}\right)\left\\|\varphi\Delta_{g}v_{i}\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{4}\right)\left(\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}}+\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}}\right)^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}\left(B_{R},d\mu\right)}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}\left(B_{R},d\mu\right)}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$

if $R_{1}$ is small enough. Hence

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\Delta\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{\frac{n}{2}}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{\frac{n}{2}}\left\|\Delta_{g}u\right\|^{\frac{n}{2}}d\mu+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}\left(B_{R},d\mu\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{\frac{n}{2}}d\sigma+c\left(\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{\frac{n}{2}}\left\|\Delta_{g}u\right\|^{\frac{n}{2}}d\mu.$

Since $\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}$ , we can choose $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ such that $\left.\varphi\right|_{B_{r}}=1$ for some $r>0$ and

\left(1+\varepsilon\right)\int_{B_{R}}\left|\varphi\right|^{\frac{n}{2}}d\sigma+c\left(\varepsilon\right)\int_{B_{R}}\left|\varphi\right|^{\frac{n}{2}}\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.15)

Hence for $i$ large enough, we have

\left\|\Delta\left(\varphi v_{i}\right)\right\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{2}}<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.16)

This implies

\left\|\Delta\left(\varphi v_{i}\right)\right\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{n-2}}<\frac{1}{p_{1}}.

(3.17)

On the other hand,

\left|u_{i}\right|^{\frac{n}{n-2}}\leq\left(\left|v_{i}\right|+\left|u\right|\right)^{\frac{n}{n-2}}\leq\left(1+\varepsilon\right)\left|v_{i}\right|^{\frac{n}{n-2}}+c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-2}},

hence

e^{a_{2,n}\left|u_{i}\right|^{\frac{n}{n-2}}}\leq e^{\left(1+\varepsilon\right)a_{2,n}\left|v_{i}\right|^{\frac{n}{n-2}}}e^{c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-2}}}.

It follows from (3.3) that

\int_{B_{r}}e^{a_{2,n}p_{1}\left|v_{i}\right|^{\frac{n}{n-2}}}d\mu\leq c\int_{B_{R}}e^{a_{2,n}p_{1}\left|\varphi v_{i}\right|^{\frac{n}{n-2}}}dx\leq c\int_{B_{R}}e^{a_{2,n}\frac{\left|\varphi v_{i}\right|^{\frac{n}{n-2}}}{\left\|\nabla\left(\varphi v_{i}\right)\right\|_{L^{n}}^{\frac{n}{n-2}}}}dx\leq c.

Using $p<\frac{p_{1}}{1+\varepsilon}$ and Lemma 3.1, it follows from Holder’s inequality that $e^{a_{2,n}\left|u_{i}\right|^{\frac{n}{n-2}}}$ is bounded in $L^{p}\left(B_{r},d\mu\right)$ . This finishes the proof of Proposition 3.1.

Theorem 3.1.

Let $M^{n}$ be a $C^{2}$ compact manifold with a $C^{1}$ Riemannian metric $g$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{2,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ , then for any $0<a<a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}$ ,

\int_{M}e^{a\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{n-2}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(3.18)

Proof.

For any $v\in W^{2,\frac{n}{2}}\left(M\right)$ , standard elliptic theory (see [GiT]) and compactness argument tells us

\left\|v-v_{M}\right\|_{L^{\frac{n}{2}}}\leq c\left\|\Delta v\right\|_{L^{\frac{n}{2}}}.

On the other hand,

\left\|v-v_{E}\right\|_{L^{\frac{n}{2}}}\leq\frac{c}{\mu\left(E\right)^{\frac{2}{n}}}\left\|v\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|v\right\|_{L^{\frac{n}{2}}}.

Replacing $v$ by $v-v_{M}$ , we see

\left\|v-v_{E}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|v-v_{M}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|\Delta v\right\|_{L^{\frac{n}{2}}}.

If (3.18) is not true, then we can find a sequence $u_{i}\in W^{2,\frac{n}{2}}\left(M\right)$ , $E_{i}\subset M$ with $\mu\left(E_{i}\right)\geq\delta$ , $u_{i,E_{i}}=0$ , $\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}=1$ and

\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu\rightarrow\infty

as $i\rightarrow\infty$ . Since

\left\|u_{i}\right\|_{L^{\frac{n}{2}}}=\left\|u_{i}-u_{i,E_{i}}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}=c\left(\delta\right),

standard elliptic estimate (see [GiT]) gives us

\left\|u_{i}\right\|_{W^{2,\frac{n}{2}}}\leq c\left(\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}+\left\|u_{i}\right\|_{L^{\frac{n}{2}}}\right)\leq c\left(\delta\right).

Hence $u_{i}$ is bounded in $W^{2,\frac{n}{2}}\left(M\right)$ . After passing to a subsequence we can find $u\in W^{2,\frac{n}{2}}\left(M\right)$ such that $u_{i}\rightharpoonup u$ weakly in $W^{2,\frac{n}{2}}\left(M\right)$ and a measure on $M$ , $\sigma$ such that

\left|\Delta u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta u\right|^{\frac{n}{2}}d\mu+\sigma

as measure. Note that $\sigma\left(M\right)\leq 1$ . For any $x\in M$ , since

0<\frac{a}{a_{2,n}}<\sigma\left(\left\{x\right\}\right)^{-\frac{2}{n-2}},

it follows from Proposition 3.1 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(x\right)}e^{a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

A covering argument implies

\sup_{i}\int_{M}e^{a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

This contradicts with the choice of $u_{i}$ .

Example 3.1.

Let $M^{n}$ be a $C^{2}$ compact manifold with a $C^{1}$ Riemannian metric $g$ . Denote

\kappa_{2}\left(M,g\right)=\inf_{\begin{subarray}{c}u\in W^{2,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}\\ u_{M}=0\end{subarray}}\frac{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}}{\left\|u\right\|_{L^{\frac{n}{2}}}}.

(3.19)

Here $u_{M}=\frac{1}{\mu\left(M\right)}\int_{M}ud\mu$ . Assume $0\leq\kappa<\kappa_{2}\left(M,g\right)$ , $0<a<a_{2,n}$ , $u\in W^{2,\frac{n}{2}}\left(M\right)$ with $u_{M}=0$ and

\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}-\kappa^{\frac{n}{2}}\left\|u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}\leq 1,

(3.20)

then we have

\int_{M}e^{a\left|u\right|^{\frac{n}{n-2}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(3.21)

Since the proof of Example 3.1 is almost identical to the proof of Example 1.1 (using Proposition 3.1 when necessary), we omit it here.

3.1. Paneitz operator and Q curvature in dimension 4

Let $\left(M^{4},g\right)$ be a smooth compact Riemannian manifold with dimension 4. The Paneitz operator is given by (see [CY2, HY2])

Pu=\Delta^{2}u+2\mathop{\mathrm{d}iv}\left(Rc\left(\nabla u,e_{i}\right)e_{i}\right)-\frac{2}{3}\mathop{\mathrm{d}iv}\left(R\nabla u\right).

(3.22)

Here $e_{1},e_{2},e_{3},e_{4}$ is a local orthonormal frame with respect to $g$ . The associated $Q$ curvature is

Q=-\frac{1}{6}\Delta R-\frac{1}{2}\left|Rc\right|^{2}+\frac{1}{6}R^{2}.

(3.23)

In 4-dimensional conformal geometry, $P$ and $Q$ play the same roles as $-\Delta$ and Gauss curvature in 2-dimensional conformal geometry.

For $u\in C^{\infty}\left(M\right)$ , let

	$\displaystyle E\left(u\right)$	$\displaystyle=$	$\displaystyle\int_{M}Pu\cdot ud\mu$
		$\displaystyle=$	$\displaystyle\int_{M}\left(\left(\Delta u\right)^{2}-2Rc\left(\nabla u,\nabla u\right)+\frac{2}{3}R\left\|\nabla u\right\|^{2}\right)d\mu.$

By this formula, we know $E\left(u\right)$ still makes sense for $u\in H^{2}\left(M\right)=W^{2,2}\left(M\right)$ .

On the standard $\mathbb{S}^{4}$ , $P\geq 0$ and $\ker P=\left\{\text{constant functions}\right\}$ . Moreover in [Gu, GuV], some general criterion for such positivity condition to be valid were derived. If $P\geq 0$ and $\ker P=\left\{\text{constant functions}\right\}$ , then for any $u\in H^{2}\left(M\right)$ with $u_{M}=0$ ,

\left\|u\right\|_{L^{2}}^{2}\leq c\left(M,g\right)E\left(u\right).

(3.25)

On the other hand, standard elliptic theory (see [GiT]) tells us

$\displaystyle\left\\|u\right\\|_{H^{2}}^{2}$	$\displaystyle\leq$	$\displaystyle c\left(M,g\right)\left(\left\\|\Delta u\right\\|_{L^{2}}^{2}+\left\\|u\right\\|_{L^{2}}^{2}\right)$
	$\displaystyle\leq$	$\displaystyle c\left(M,g\right)\left(E\left(u\right)+\left\\|u\right\\|_{H^{1}}^{2}\right)$
	$\displaystyle\leq$	$\displaystyle c\left(M,g\right)E\left(u\right)+\frac{1}{2}\left\\|u\right\\|_{H^{2}}^{2}+c\left(M,g\right)\left\\|u\right\\|_{L^{2}}^{2}$
	$\displaystyle\leq$	$\displaystyle\frac{1}{2}\left\\|u\right\\|_{H^{2}}^{2}+c\left(M,g\right)E\left(u\right).$

We have used the interpolation inequality in between. It follows that

\left\|u\right\|_{H^{2}}^{2}\leq c\left(M,g\right)E\left(u\right).

(3.26)

Lemma 3.2.

If $P\geq 0$ , $\ker P=\left\{\text{constant functions}\right\}$ , $u\in H^{2}\left(M\right)\backslash\left\{0\right\}$ such that $u_{M}=0$ , then for any $a\in\left(0,32\pi^{2}\right)$ ,

\int_{M}e^{a\frac{u^{2}}{E\left(u\right)}}d\mu\leq c\left(a\right)<\infty.

(3.27)

In particular,

\log\int_{M}e^{4u}d\mu\leq\frac{4}{a}E\left(u\right)+c\left(a\right).

(3.28)

Note that $a_{2,4}=32\pi^{2}$ . In [CY2, Lemma 1.6], it is shown that under the same assumption as Lemma 3.2,

\int_{M}e^{32\pi^{2}\frac{u^{2}}{E\left(u\right)}}d\mu\leq c\left(M,g\right)<\infty.

(3.29)

The argument is based on asymptotic expansion formula of Green’s function of $\sqrt{P}$ and modified Adams’ argument [A]. Our approach below is more elementary. Moreover, Lemma 3.2 is sufficient for application to Q curvature equation in [CY2, Theorem 1.2], which is for the case $\left(M,g\right)$ not conformal diffeomorphic to the standard $\mathbb{S}^{4}$ (see [Gu] and [HY1, Proposition 1.3]).

Proof of Lemma 3.2.

If (3.27) is not true, then there exists $u_{i}\in H^{2}\left(M\right)$ such that $u_{i,M}=0$ , $E\left(u_{i}\right)=1$ and

\int_{M}e^{au_{i}^{2}}d\mu\rightarrow\infty

as $i\rightarrow\infty$ . By previous discussion we see

\left\|u_{i}\right\|_{H^{2}}^{2}\leq c\left(M,g\right)E\left(u_{i}\right)=c\left(M,g\right).

Hence after passing to a subsequence, we can find a $u\in H^{2}\left(M\right)$ and a measure $\sigma$ on $M$ such that $u_{i}\rightharpoonup u$ weakly in $H^{2}\left(M\right)$ , $u_{i}\rightarrow u$ in $H^{1}\left(M\right)$ and

\left(\Delta u_{i}\right)^{2}d\mu\rightarrow\left(\Delta u\right)^{2}d\mu+\sigma\text{ as measure.}

Note that

	$\displaystyle E\left(u_{i}\right)$	$\displaystyle=$	$\displaystyle\int_{M}\left(\left(\Delta u_{i}\right)^{2}-2Rc\left(\nabla u_{i},\nabla u_{i}\right)+\frac{2}{3}R\left\|\nabla u_{i}\right\|^{2}\right)d\mu$
		$\displaystyle\rightarrow$	$\displaystyle E\left(u\right)+\sigma\left(M\right),$

we see $E\left(u\right)+\sigma\left(M\right)=1$ . Using the fact $E\left(u\right)\geq 0$ , we see $\sigma\left(M\right)\leq 1$ . For any $x\in M$ , because

0<\frac{a}{32\pi^{2}}<\sigma\left(\left\{x\right\}\right)^{-1},

it follows from Proposition 3.1 that for some $r>0$ ,

\sup_{i}\int_{B_{r}\left(x\right)}e^{au_{i}^{2}}d\mu<\infty.

A covering arguments implies

\sup_{i}\int_{M}e^{au_{i}^{2}}d\mu<\infty.

This contradicts with the choice of $u_{i}$ .

Since

4u\leq a\frac{u^{2}}{E\left(u\right)}+\frac{4E\left(u\right)}{a},

we see

e^{4u}\leq e^{\frac{4E\left(u\right)}{a}}e^{a\frac{u^{2}}{E\left(u\right)}}.

Hence

\int_{M}e^{4u}d\mu\leq c\left(a\right)e^{\frac{4E\left(u\right)}{a}}

and

\log\int_{M}e^{4u}d\mu\leq\frac{4}{a}E\left(u\right)+c\left(a\right).

3.2. Functions on manifolds with nonempty boundary

We will use the same notations as in Section 2.

Lemma 3.3.

Assume $u\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)\backslash\left\{0\right\}$ such that $u\left(x\right)=0$ for $\left|x\right|$ close to $R$ and $\partial_{n}u=0$ on $\Sigma_{R}$ , then

\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{n-2}}}}dx\leq c\left(n\right)R^{n}.

(3.30)

Proof.

For $\left|x\right|<R$ , we define

v\left(x\right)=\left\{\begin{array}[]{cc}u\left(x\right),&\text{if }x_{n}>0;\\ u\left(x^{\prime},-x_{n}\right),&\text{if }x_{n}<0\text{.}\end{array}\right.

Then $v\in W_{0}^{2,\frac{n}{2}}\left(B_{R}\right)$ with $\left\|\Delta v\right\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{2}}=2\left\|\Delta u\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}$ . Hence

$\displaystyle\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}\frac{\left\|u\right\|^{\frac{n}{n-2}}}{\left\\|\Delta u\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{n-2}}}}dx$	$\displaystyle=$	$\displaystyle\int_{B_{R}^{+}}e^{a_{2,n}\frac{\left\|u\right\|^{\frac{n}{n-2}}}{\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{n-2}}}}dx$
	$\displaystyle\leq$	$\displaystyle\int_{B_{R}}e^{a_{2,n}\frac{\left\|v\right\|^{\frac{n}{n-2}}}{\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{n-2}}}}dx$
	$\displaystyle\leq$	$\displaystyle c\left(n\right)R^{n}.$

Lemma 3.4.

Let $u\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ and $a>0$ , then

\int_{B_{R}^{+}}e^{a\left|u\right|^{\frac{n}{n-2}}}dx<\infty.

(3.31)

Proof.

We can find $\widetilde{u}\in W^{2,\frac{n}{2}}\left(B_{R}\right)$ such that $\left.\widetilde{u}\right|_{B_{R}^{+}}=u$ . Then it follows from Lemma 3.1 that

\int_{B_{R}^{+}}e^{a\left|u\right|^{\frac{n}{n-2}}}dx\leq\int_{B_{R}}e^{a\left|\widetilde{u}\right|^{\frac{n}{n-2}}}dx<\infty.

Proposition 3.2.

Let $0<R\leq 1$ , $g$ be a $C^{1}$ Riemannian metric on $\overline{B_{R}^{+}}$ such that $g_{ij}\left(0\right)=\delta_{ij}$ for $1\leq i,j\leq n$ . Assume $u_{i}\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ , $\partial_{n}u_{i}=0$ on $\Sigma_{R}$ , $u_{i}\rightharpoonup u$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}p\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

(3.32)

Here

a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}.

(3.33)

First we observe that if $v\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ , $v\left(x\right)=0$ for $\left|x\right|$ close to $R$ , $\left.\partial_{n}v\right|_{\Sigma_{R}}=0$ , then we have

\left\|v\right\|_{W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)}\leq c\left\|\Delta v\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}.

(3.34)

Indeed, let

\widetilde{v}\left(x\right)=\left\{\begin{array}[]{cc}v\left(x\right),&\text{if }x_{n}>0;\\ v\left(x^{\prime},-x_{n}\right),&\text{if }x_{n}<0\end{array}\right.

for $x\in B_{R}$ , then $\widetilde{v}\in W_{0}^{2,\frac{n}{2}}\left(B_{R}\right)$ . Hence elliptic estimates gives us

\left\|v\right\|_{W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)}\leq\left\|\widetilde{v}\right\|_{W^{2,\frac{n}{2}}\left(B_{R}\right)}\leq c\left\|\Delta\widetilde{v}\right\|_{L^{\frac{n}{2}}\left(B_{R}\right)}\leq c\left\|\Delta v\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}.

To continue, we recall (3.10) that if $0<R_{1}<R$ is a small number and $v\in W^{2,\frac{n}{2}}\left(B_{R_{1}}^{+}\right)$ , $v\left(x\right)=0$ for $\left|x\right|$ close to $R_{1}$ , $\left.\partial_{n}v\right|_{\Sigma_{R_{1}}}=0$ , then

\left|\Delta v\right|\leq\left|\Delta_{g}v\right|+\varepsilon_{1}\left|D^{2}v\right|+c\left|\nabla v\right|,

$\varepsilon_{1}=\varepsilon_{1}\left(R_{1}\right)>0$ with $\varepsilon_{1}\rightarrow 0$ as $R_{1}\rightarrow 0^{+}$ . Hence

			$\displaystyle\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}+\varepsilon_{1}\left\\|D^{2}v\right\\|_{L^{\frac{n}{2}}}+c\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}\right)^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{16}\right)\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\varepsilon_{1}^{\frac{n}{2}}\left\\|D^{2}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{16}\right)\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\varepsilon_{1}^{\frac{n}{2}}\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}.$

It follows that

	$\displaystyle\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}$	$\displaystyle\leq$	$\displaystyle\frac{1+\frac{\varepsilon}{16}}{1-c\left(\varepsilon\right)\varepsilon_{1}^{\frac{n}{2}}}\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+\frac{c\left(\varepsilon\right)}{1-c\left(\varepsilon\right)\varepsilon_{1}^{\frac{n}{2}}}\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{8}\right)\left\\|\Delta_{g}v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\nabla v\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$

if $R_{1}$ is small enough.

Proof of Proposition 3.2.

Fix $p_{1}\in\left(p,\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}\right)$ , then

\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.35)

We can find $\varepsilon>0$ s.t.

\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}

(3.36)

and

\left(1+\varepsilon\right)p<p_{1}.

(3.37)

Let $v_{i}=u_{i}-u$ , then $v_{i}\rightharpoonup 0$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ and $v_{i}\rightarrow 0$ in $W^{1,\frac{n}{2}}\left(B_{R}^{+}\right)$ . Let $0<R_{1}<R$ be small enough. For radial symmetric $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ , we have

			$\displaystyle\left\\|\Delta\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{8}\right)\left\\|\Delta_{g}\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{8}\right)\left(\left\\|\varphi\Delta_{g}v_{i}\right\\|_{L^{\frac{n}{2}}}+c\left\\|\varphi\right\\|_{W^{2,\infty}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}\right)^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{4}\right)\left\\|\varphi\Delta_{g}v_{i}\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{4}\right)\left(\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}}+\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}}\right)^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\Delta_{g}u_{i}\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+},d\mu\right)}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+},d\mu\right)}^{\frac{n}{2}}+c\left(\varepsilon\right)\left\\|\varphi\right\\|_{W^{2,\infty}}^{\frac{n}{2}}\left\\|v_{i}\right\\|_{W^{1,\frac{n}{2}}}^{\frac{n}{2}}.$

Hence

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\Delta\left(\varphi v_{i}\right)\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{\frac{n}{2}}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{\frac{n}{2}}\left\|\Delta_{g}u\right\|^{\frac{n}{2}}d\mu+c\left(\varepsilon\right)\left\\|\varphi\Delta_{g}u\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+},d\mu\right)}^{\frac{n}{2}}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{\frac{n}{2}}d\sigma+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{\frac{n}{2}}\left\|\Delta_{g}u\right\|^{\frac{n}{2}}d\mu.$

Since $\left(1+\varepsilon\right)\sigma\left(\left\{0\right\}\right)<\frac{1}{p_{1}^{\frac{n-2}{2}}}$ , we can choose a radial function $\varphi\in C_{c}^{\infty}\left(B_{R_{1}}\right)$ such that $\left.\varphi\right|_{B_{r}}=1$ for some $r>0$ and

\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left|\varphi\right|^{\frac{n}{2}}d\sigma+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left|\varphi\right|^{\frac{n}{2}}\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.38)

Hence for $i$ large enough, we have

\left\|\Delta\left(\varphi v_{i}\right)\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{2}}<\frac{1}{p_{1}^{\frac{n-2}{2}}}.

(3.39)

This implies

\left\|\Delta\left(\varphi v_{i}\right)\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{n-2}}<\frac{1}{p_{1}}.

(3.40)

On the other hand,

\left|u_{i}\right|^{\frac{n}{n-2}}\leq\left(\left|v_{i}\right|+\left|u\right|\right)^{\frac{n}{n-2}}\leq\left(1+\varepsilon\right)\left|v_{i}\right|^{\frac{n}{n-2}}+c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-2}},

hence

e^{2^{-\frac{2}{n-2}}a_{2,n}\left|u_{i}\right|^{\frac{n}{n-2}}}\leq e^{\left(1+\varepsilon\right)2^{-\frac{2}{n-2}}a_{2,n}\left|v_{i}\right|^{\frac{n}{n-2}}}e^{c\left(\varepsilon\right)\left|u\right|^{\frac{n}{n-2}}}.

It follows from Lemma 3.3 that

$\displaystyle\int_{B_{r}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}p_{1}\left\|v_{i}\right\|^{\frac{n}{n-2}}}d\mu$	$\displaystyle\leq$	$\displaystyle c\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}p_{1}\left\|\varphi v_{i}\right\|^{\frac{n}{n-2}}}dx$
	$\displaystyle\leq$	$\displaystyle c\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}\frac{\left\|\varphi v_{i}\right\|^{\frac{n}{n-2}}}{\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}}^{\frac{n}{n-2}}}}dx$
	$\displaystyle\leq$	$\displaystyle c.$

Using $p<\frac{p_{1}}{1+\varepsilon}$ and Lemma 3.4, it follows from Holder’s inequality that $e^{2^{-\frac{2}{n-2}}a_{2,n}\left|u_{i}\right|^{\frac{n}{n-2}}}$ is bounded in $L^{p}\left(B_{r}^{+},d\mu\right)$ .

Theorem 3.2.

Let $\left(M^{n},g\right)$ be a $C^{4}$ compact Riemannian manifold with boundary and $g$ be a $C^{3}$ Riemannian metric on $M$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{2,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ and $\left.\frac{\partial u}{\partial\nu}\right|_{\partial M}=0$ (here $\nu$ is the unit outer normal direction on $\partial M$ ), then for any $0<a<a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}$ ,

\int_{M}e^{2^{-\frac{2}{n-2}}a\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{n-2}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(3.41)

Proof.

For any $v\in W^{2,\frac{n}{2}}\left(M\right)$ with $\left.\frac{\partial v}{\partial\nu}\right|_{\partial M}=0$ , standard elliptic theory and compactness argument tells us

\left\|v-v_{M}\right\|_{L^{\frac{n}{2}}\left(M\right)}\leq c\left\|\Delta v\right\|_{L^{\frac{n}{2}}\left(M\right)}.

On the other hand,

\left\|v-v_{E}\right\|_{L^{\frac{n}{2}}}\leq\frac{c}{\mu\left(E\right)^{\frac{2}{n}}}\left\|v\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|v\right\|_{L^{\frac{n}{2}}}.

Replacing $v$ by $v-v_{M}$ , we see

\left\|v-v_{E}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|v-v_{M}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|\Delta v\right\|_{L^{\frac{n}{2}}}.

If (3.41) is not true, then we can find a sequence $u_{i}\in W^{2,\frac{n}{2}}\left(M\right)$ , $E_{i}\subset M$ with $\mu\left(E_{i}\right)\geq\delta$ , $u_{i,E_{i}}=0$ , $\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}=1$ , $\left.\frac{\partial u_{i}}{\partial\nu}\right|_{\partial M}=0$ and

\int_{M}e^{2^{-\frac{2}{n-2}}a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu\rightarrow\infty

as $i\rightarrow\infty$ . Since

\left\|u_{i}\right\|_{L^{\frac{n}{2}}}=\left\|u_{i}-u_{i,E_{i}}\right\|_{L^{\frac{n}{2}}}\leq c\left(\delta\right)\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}=c\left(\delta\right),

standard elliptic estimate gives us

\left\|u_{i}\right\|_{W^{2,\frac{n}{2}}}\leq c\left(\left\|\Delta u_{i}\right\|_{L^{\frac{n}{2}}}+\left\|u_{i}\right\|_{L^{\frac{n}{2}}}\right)\leq c\left(\delta\right).

\left|\Delta u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta u\right|^{\frac{n}{2}}d\mu+\sigma

as measure. Note that $\sigma\left(M\right)\leq 1$ . For any $y\in M\backslash\partial M$ , since

0<\frac{a}{a_{2,n}}<\sigma\left(\left\{y\right\}\right)^{-\frac{2}{n-2}},

it follows from Proposition 3.1 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(y\right)}e^{2^{-\frac{2}{n-2}}a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu\leq\sup_{i}\int_{B_{r}\left(y\right)}e^{a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

Next we deal with the case $y\in\partial M$ . Note for $\varepsilon_{0}>0$ small enough, we have a well defined map

\phi:\partial M\times\left[0,\varepsilon_{0}\right)\rightarrow M:\left(\xi,t\right)\mapsto\exp_{\xi}\left(-t\nu\left(\xi\right)\right).

By the compactness of $M$ , we can find $\varepsilon_{0}>0$ such that $\phi\left(\partial M\times\left[0,\varepsilon_{0}\right)\right)$ is open and $\phi$ is a $C^{2}$ diffeomorphism. We write $\phi^{-1}\left(z\right)=\left(\xi\left(z\right),t\left(z\right)\right)$ . Let $s_{1},\cdots,s_{n-1}$ be a coordinate near $y$ on $\partial M$ such that $s_{i}\left(y\right)=0$ and $\left\langle\partial_{s_{i}},\partial_{s_{j}}\right\rangle\left(y\right)=\delta_{ij}$ for $1\leq i,j\leq n-1$ . Then we define a coordinate $x_{1},\cdots,x_{n}$ near $y$ as $x_{i}\left(z\right)=s_{i}\left(\xi\left(z\right)\right)$ for $1\leq i\leq n-1$ and $x_{n}\left(z\right)=t\left(z\right)$ . It is clear that $\left.\partial_{x_{n}}\right|_{x_{n}=0}=-\nu$ , hence we see $\left.\partial_{n}u_{i}\right|_{\Sigma_{R}}=0$ . It follows from Proposition 3.2 that for some $r>0$ , we have

\sup_{i}\int_{B_{r}\left(y\right)}e^{2^{-\frac{2}{n-2}}a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

A covering argument implies

\sup_{i}\int_{M}e^{2^{-\frac{2}{n-2}}a\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

This contradicts with the choice of $u_{i}$ .

Example 3.2.

Let $\left(M^{n},g\right)$ be a $C^{4}$ compact Riemannian manifold with boundary and $g$ be a $C^{3}$ Riemannian metric on $M$ . We define

\kappa_{2,N}\left(M,g\right)=\inf_{\begin{subarray}{c}u\in W^{2,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}\\ u_{M}=0,\left.\frac{\partial u}{\partial\nu}\right|_{\partial M}=0\end{subarray}}\frac{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}}{\left\|u\right\|_{L^{\frac{n}{2}}}},

(3.42)

here $\nu$ is the unit outnormal direction on $\partial M$ . Assume $0\leq\kappa<\kappa_{2,N}\left(M,g\right)$ , $0<a<a_{2,n}$ , $u\in W^{2,\frac{n}{2}}\left(M\right)$ with $u_{M}=0$ , $\left.\frac{\partial u}{\partial\nu}\right|_{\partial M}=0$ and

\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}-\kappa^{\frac{n}{2}}\left\|u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}\leq 1,

(3.43)

then

\int_{M}e^{2^{-\frac{2}{n-2}}a\left|u\right|^{\frac{n}{n-2}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(3.44)

Since the proof of Example 3.2 is almost identical to the proof of Example 1.1 (using Proposition 3.1 and 3.2 when necessary), we omit it here.

Next we move to the second boundary condition.

Lemma 3.5.

Assume $u\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)\backslash\left\{0\right\}$ such that $u\left(x\right)=0$ for $\left|x\right|$ close to $R$ and $\left.u\right|_{\Sigma_{R}}=0$ , then

\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{n-2}}}}dx\leq c\left(n\right)R^{n}.

(3.45)

Proof.

For $\left|x\right|<R$ , we define

v\left(x\right)=\left\{\begin{array}[]{cc}u\left(x\right),&\text{if }x_{n}>0;\\ -u\left(x^{\prime},-x_{n}\right),&\text{if }x_{n}<0\text{.}\end{array}\right.

$\displaystyle\int_{B_{R}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}\frac{\left\|u\right\|^{\frac{n}{n-2}}}{\left\\|\Delta u\right\\|_{L^{\frac{n}{2}}\left(B_{R}^{+}\right)}^{\frac{n}{n-2}}}}dx$	$\displaystyle=$	$\displaystyle\int_{B_{R}^{+}}e^{a_{2,n}\frac{\left\|u\right\|^{\frac{n}{n-2}}}{\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{n-2}}}}dx$
	$\displaystyle\leq$	$\displaystyle\int_{B_{R}}e^{a_{2,n}\frac{\left\|v\right\|^{\frac{n}{n-2}}}{\left\\|\Delta v\right\\|_{L^{\frac{n}{2}}\left(B_{R}\right)}^{\frac{n}{n-2}}}}dx$
	$\displaystyle\leq$	$\displaystyle c\left(n\right)R^{n}.$

Proposition 3.3.

Let $0<R\leq 1$ , $g$ be a $C^{1}$ Riemannian metric on $\overline{B_{R}^{+}}$ . Assume $u_{i}\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ , $\left.u_{i}\right|_{\Sigma_{R}}=0$ , $u_{i}\rightharpoonup u$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{2^{-\frac{2}{n-2}}a_{2,n}p\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

(3.46)

By a linear changing of variable and shrinking $R$ if necessary we can assume $g=g_{ij}dx_{i}dx_{j}$ with $g_{ij}\left(0\right)=\delta_{ij}$ . The remaining proof is almost identical to the proof of Proposition 3.2 (using Lemma 3.5 instead of Lemma 3.3), we omit it here.

Theorem 3.3.

Let $M^{n}$ be a $C^{2}$ compact manifold with boundary and $g$ be a $C^{1}$ Riemannian metric on $M$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{2,\frac{n}{2}}\left(M\right)\cap W_{0}^{1,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ , then for any $0<a<a_{2,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{4\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n-2}{2}\right)}\right)^{\frac{n}{n-2}}$ ,

\int_{M}e^{2^{-\frac{2}{n-2}}a\frac{\left|u\right|^{\frac{n}{n-2}}}{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{n-2}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(3.47)

The proof is almost identical to the proof of Theorem 3.2 (using Proposition 3.3 instead of Proposition 3.2), we omit it here.

Example 3.3.

Let $M^{n}$ be a $C^{2}$ compact manifold with boundary and $g$ be a $C^{1}$ Riemannian metric on $M$ . We denote

\kappa_{2,D}\left(M,g\right)=\inf_{u\in W^{2,\frac{n}{2}}\left(M\right)\cap W_{0}^{1,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}}\frac{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}}{\left\|u\right\|_{L^{\frac{n}{2}}}}.

(3.48)

Assume $0\leq\kappa<\kappa_{2,D}\left(M,g\right)$ , $0<a<a_{2,n}$ , $u\in W^{2,\frac{n}{2}}\left(M\right)\cap W_{0}^{1,\frac{n}{2}}\left(M\right)$ and

\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}-\kappa^{\frac{n}{2}}\left\|u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}\leq 1,

(3.49)

then

\int_{M}e^{2^{-\frac{2}{n-2}}a\left|u\right|^{\frac{n}{n-2}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(3.50)

Since the proof of Example 3.3 is almost identical to the proof of Example 1.1 (using Proposition 3.1 and 3.3 when necessary), we omit it here.

At last we turn to the third boundary condition.

Proposition 3.4.

Let $0<R\leq 1$ , $g$ be a $C^{1}$ Riemannian metric on $\overline{B_{R}^{+}}$ . Assume $u_{i}\in W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ , $\left.u\right|_{\Sigma_{R}}=0$ , $\left.\partial_{n}u\right|_{\Sigma_{R}}=0$ , $u_{i}\rightharpoonup u$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}u_{i}\right|^{\frac{n}{2}}d\mu\rightarrow\left|\Delta_{g}u\right|^{\frac{n}{2}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{a_{2,n}p\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

(3.51)

Proof.

Let $h$ be a $C^{1}$ Riemannian metric on $\overline{B_{R}}$ , which is an extension of $g$ . We define

v_{i}\left(x\right)=\left\{\begin{tabular}[]{ll}$u_{i}\left(x\right),$&if $x\in B_{R}^{+}$\\ $0,$&if $x\in B_{R}\backslash B_{R}^{+}$\end{tabular}\right.,\quad v\left(x\right)=\left\{\begin{tabular}[]{ll}$u\left(x\right),$&if $x\in B_{R}^{+}$\\ $0,$&if $x\in B_{R}\backslash B_{R}^{+}$\end{tabular}\right.,

and a measure $\tau$ on $B_{R}$ by $\tau\left(E\right)=\sigma\left(E\cap B_{R}\right)$ for any Borel set $E\subset B_{R}$ . Then $v_{i},v\in W^{2,\frac{n}{2}}\left(B_{R}\right)$ , $v_{i}\rightharpoonup v$ weakly in $W^{2,\frac{n}{2}}\left(B_{R}\right)$ and

\left|\nabla_{h}v_{i}\right|_{h}^{n}d\mu_{h}\rightarrow\left|\nabla_{h}v\right|_{h}^{n}d\mu_{h}+\tau\text{ as measure on }B_{R}.

Here $\mu_{h}$ is the measure associated with $h$ . Since $\tau\left(\left\{0\right\}\right)=\sigma\left(\left\{0\right\}\right)$ , it follows from Proposition 3.1 that for some $r>0$ ,

\sup_{i}\int_{B_{r}}e^{a_{2,n}p\left|v_{i}\right|^{\frac{n}{n-2}}}d\mu_{h}<\infty.

Hence

\sup_{i}\int_{B_{r}^{+}}e^{a_{2,n}p\left|u_{i}\right|^{\frac{n}{n-2}}}d\mu<\infty.

Theorem 3.4.

Let $M^{n}$ be a $C^{2}$ compact manifold with boundary and $g$ be a $C^{1}$ Riemannian metric on $M$ . We denote

\kappa_{2}\left(M,g\right)=\inf_{u\in W_{0}^{2,\frac{n}{2}}\left(M\right)\backslash\left\{0\right\}}\frac{\left\|\Delta u\right\|_{L^{\frac{n}{2}}}}{\left\|u\right\|_{L^{\frac{n}{2}}}}.

(3.52)

Assume $0\leq\kappa<\kappa_{2}\left(M,g\right)$ , $0<a<a_{2,n}$ , $u\in W_{0}^{2,\frac{n}{2}}\left(M\right)$ and

\left\|\Delta u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}-\kappa^{\frac{n}{2}}\left\|u\right\|_{L^{\frac{n}{2}}}^{\frac{n}{2}}\leq 1,

(3.53)

then

\int_{M}e^{a\left|u\right|^{\frac{n}{n-2}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(3.54)

Since the proof of Theorem 2.2 is almost identical to the proof of Example 1.1 (using Proposition 3.1 and 3.4 when necessary), we omit it here. Theorem 3.4 should be compared to [LY].

4. Higher order Sobolev spaces

In this section, we will apply our approach to higher order Sobolev spaces. Since the arguments are similar to those for first and second order Sobolev spaces, we omit all the proofs here. For future references we list the theorems.

4.1. $W^{m,\frac{n}{m}}\left(M^{n}\right)$ for even $m$

Let $m\in\mathbb{N}$ be an even number strictly less than $n$ . For $R>0$ , $u\in W_{0}^{m,\frac{n}{m}}\left(B_{R}\right)\backslash\left\{0\right\}$ , it is shown in [A] that

\int_{B_{R}}\exp\left(a_{m,n}\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{n-m}}}\right)dx\leq c\left(m,n\right)R^{n}.

(4.1)

Here

a_{m,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{\pi^{\frac{n}{2}}2^{m}\Gamma\left(\frac{m}{2}\right)}{\Gamma\left(\frac{n-m}{2}\right)}\right)^{\frac{n}{n-m}}.

(4.2)

Lemma 4.1.

If $u\in W^{m,\frac{n}{m}}\left(B_{R}^{n}\right)$ , then for any $a>0$ ,

\int_{B_{R}}e^{a\left|u\right|^{\frac{n}{n-m}}}dx<\infty.

(4.3)

Proposition 4.1.

Let $0<R\leq 1$ , $g$ be a $C^{m-1}$ Riemannian metric on $\overline{B_{R}^{n}}$ . Assume $u_{i}\in W^{m,\frac{n}{m}}\left(B_{R}^{n}\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{m,\frac{n}{m}}\left(B_{R}\right)$ and

\left|\Delta_{g}^{\frac{m}{2}}u_{i}\right|^{\frac{n}{m}}d\mu\rightarrow\left|\Delta_{g}^{\frac{m}{2}}u\right|^{\frac{n}{m}}d\mu+\sigma\text{ as measure on }B_{R},

here $\mu$ is the measure associated with the metric $g$ . If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{m}{n-m}}$ , then for some $r>0$ ,

\sup_{i}\int_{B_{r}}e^{a_{m,n}p\left|u_{i}\right|^{\frac{n}{n-m}}}d\mu<\infty.

(4.4)

Theorem 4.1.

Let $M^{n}$ be a $C^{m}$ compact manifold with a $C^{m-1}$ Riemannian metric $g$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{m,\frac{n}{m}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ (here $u_{E}=\frac{1}{\mu\left(E\right)}\int_{E}ud\mu$ ), then for any $0<a<a_{m,n}$ ,

\int_{M}e^{a\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{n-m}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(4.5)

4.1.1. Functions on manifolds with nonempty boundary

Lemma 4.2.

Assume $u\in W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)\backslash\left\{0\right\}$ such that $u\left(x\right)=0$ for $\left|x\right|$ close to $R$ and $\partial_{n}^{k}u=0$ on $\Sigma_{R}$ for odd number $k\in\left[0,m\right)$ , then

\int_{B_{R}^{+}}e^{2^{-\frac{m}{n-m}}a_{m,n}\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}\left(B_{R}^{+}\right)}^{\frac{n}{n-m}}}}dx\leq c\left(n\right)R^{n}.

(4.6)

This can be done by even extension for $x_{n}$ direction as in the proof of Lemma 3.3.

Proposition 4.2.

Let $0<R\leq 1$ , $g$ be a $C^{m-1}$ Riemannian metric on $\overline{B_{R}^{+}}$ such that $g_{ij}\left(0\right)=\delta_{ij}$ for $1\leq i,j\leq n$ . Assume $u_{i}\in W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ , $\partial_{n}^{k}u=0$ on $\Sigma_{R}$ for odd number $k\in\left[0,m\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}^{\frac{m}{2}}u_{i}\right|^{\frac{n}{m}}d\mu\rightarrow\left|\Delta_{g}^{\frac{m}{2}}u\right|^{\frac{n}{m}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{m}{n-m}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{2^{-\frac{m}{n-m}}a_{m,n}p\left|u_{i}\right|^{\frac{n}{n-m}}}d\mu<\infty.

(4.7)

Theorem 4.2.

Let $\left(M^{n},g\right)$ be a $C^{m+2}$ compact Riemannian manifold with boundary and $g$ be a $C^{m+1}$ Riemannian metric on $M$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{m,\frac{n}{m}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ and $\left.D^{k}u\left(\overset{k\text{ times}}{\overbrace{\nu,\cdots,\nu}}\right)\right|_{\partial M}=0$ for odd number $k\in\left[0,m\right)$ (here $\nu$ is the unit outer normal direction on $\partial M$ ), then for any $0<a<a_{m,n}$ ,

\int_{M}e^{2^{-\frac{m}{n-m}}a\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{n-m}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(4.8)

Lemma 4.3.

\int_{B_{R}^{+}}e^{2^{-\frac{m}{n-m}}a_{m,n}\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}\left(B_{R}^{+}\right)}^{\frac{n}{n-m}}}}dx\leq c\left(n\right)R^{n}.

(4.9)

This can be done by odd extension in $x_{n}$ direction as in the proof of Lemma 3.5.

Proposition 4.3.

Let $0<R\leq 1$ , $g$ be a $C^{m-1}$ Riemannian metric on $\overline{B_{R}^{+}}$ such that $g_{ij}\left(0\right)=\delta_{ij}$ for $1\leq i,j\leq n$ . Assume $u_{i}\in W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ , $\partial_{n}^{k}u=0$ on $\Sigma_{R}$ for even number $k\in\left[0,m\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}^{\frac{m}{2}}u_{i}\right|^{\frac{n}{m}}d\mu\rightarrow\left|\Delta_{g}^{\frac{m}{2}}u\right|^{\frac{n}{m}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{m}{n-m}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{2^{-\frac{m}{n-m}}a_{m,n}p\left|u_{i}\right|^{\frac{n}{n-m}}}d\mu<\infty.

(4.10)

Theorem 4.3.

Let $\left(M^{n},g\right)$ be a $C^{m+2}$ compact Riemannian manifold with boundary and $g$ be a $C^{m+1}$ Riemannian metric on $M$ . If $E\subset M$ with $\mu\left(E\right)\geq\delta>0$ , $u\in W^{m,\frac{n}{m}}\left(M\right)\backslash\left\{0\right\}$ with $u_{E}=0$ and $\left.D^{k}u\left(\overset{k\text{ times}}{\overbrace{\nu,\cdots,\nu}}\right)\right|_{\partial M}=0$ for even number $k\in\left[0,m\right)$ (here $\nu$ is the unit outer normal direction on $\partial M$ ), then for any $0<a<a_{m,n}$ ,

\int_{M}e^{2^{-\frac{m}{n-m}}a\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{n-m}}}}d\mu\leq c\left(a,\delta\right)<\infty.

(4.11)

Proposition 4.4.

Let $0<R\leq 1$ , $g$ be a $C^{m-1}$ Riemannian metric on $\overline{B_{R}^{+}}$ . Assume $u_{i}\in W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ , $\left.\partial_{n}^{k}u\right|_{\Sigma_{R}}=0$ for integer $k\in\left[0,m\right)$ , $u_{i}\rightharpoonup u$ weakly in $W^{m,\frac{n}{m}}\left(B_{R}^{+}\right)$ and

\left|\Delta_{g}^{\frac{m}{2}}u_{i}\right|^{\frac{n}{m}}d\mu\rightarrow\left|\Delta_{g}^{\frac{m}{2}}u\right|^{\frac{n}{m}}d\mu+\sigma\text{ as measure on }B_{R}^{+}\text{.}

If $0<p<\sigma\left(\left\{0\right\}\right)^{-\frac{2}{n-2}}$ , then there exists $r>0$ such that

\sup_{i}\int_{B_{r}^{+}}e^{a_{m,n}p\left|u_{i}\right|^{\frac{n}{n-m}}}d\mu<\infty.

(4.12)

Theorem 4.4.

Let $M^{n}$ be a $C^{m}$ compact manifold with boundary and $g$ be a $C^{m-1}$ Riemannian metric on $M$ . We denote

\kappa_{m}\left(M,g\right)=\inf_{u\in W_{0}^{m,\frac{n}{m}}\left(M\right)\backslash\left\{0\right\}}\frac{\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}}{\left\|u\right\|_{L^{\frac{n}{m}}}}.

(4.13)

Assume $0\leq\kappa<\kappa_{m}\left(M,g\right)$ , $0<a<a_{m,n}$ , $u\in W_{0}^{m,\frac{n}{m}}\left(M\right)$ and

\left\|\Delta u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{m}}-\kappa^{\frac{n}{m}}\left\|u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{m}}\leq 1,

(4.14)

then

\int_{M}e^{a\left|u\right|^{\frac{n}{n-m}}}d\mu\leq c\left(\kappa,a\right)<\infty.

(4.15)

4.2. $W^{m,\frac{n}{m}}\left(M^{n}\right)$ for odd $m$

Let $m\in\mathbb{N}$ be an odd number strictly less than $n$ . For $R>0$ , $u\in W_{0}^{m,\frac{n}{m}}\left(B_{R}\right)\backslash\left\{0\right\}$ , it is shown in [A] that

\int_{B_{R}}\exp\left(a_{m,n}\frac{\left|u\right|^{\frac{n}{n-m}}}{\left\|\nabla\Delta^{\frac{m-1}{2}}u\right\|_{L^{\frac{n}{m}}}^{\frac{n}{n-m}}}\right)dx\leq c\left(m,n\right)R^{n}.

(4.16)

Here

a_{m,n}=\frac{n}{\left|\mathbb{S}^{n-1}\right|}\left(\frac{\pi^{\frac{n}{2}}2^{m}\Gamma\left(\frac{m+1}{2}\right)}{\Gamma\left(\frac{n-m+1}{2}\right)}\right)^{\frac{n}{n-m}}.

(4.17)

With (4.16) at hands, we can derive similar results for $m$ odd cases. Indeed if we use $\left\|\nabla\Delta^{\frac{m-1}{2}}u\right\|_{L^{\frac{n}{m}}}$ instead of $\left\|\Delta^{\frac{m}{2}}u\right\|_{L^{\frac{n}{m}}}$ (in the $m$ even case), we can get the corresponding statements. The details are left to interested readers.

References

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[AD] Adimurthi, O. Druet. Blow-up analysis in dimension 2 and a sharp form of Trudinger-Moser inequality. Comm. Partial Differential Equations 29 (2004), no. 1-2, 295–322.
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[Au2] T. Aubin. Meilleures constantes dans le théorème d’inclusion de Sobolev et un théorème de Fredholm non linéaire pour la transformation conforme de la courbure scalaire. (French) J. Functional Analysis 32 (1979), no. 2, 148–174.
[CCH] R. Cerny, A. Cianchi and S. Hencl. Concentration-compactness principles for Moser-Trudinger inequalities: new results and proofs. (English summary) Ann. Mat. Pura Appl. (4) 192 (2013), no. 2, 225–243.
[CH] S. Y. Chang and F. B. Hang. Improved Moser-Trudinger-Onofri inequality under constraints. Comm Pure Appl Math, to appear, arXiv:1909.00431
[CY1] S. Y. Chang and P. C. Yang. Conformal deformation of metrics on $\mathbb{S}^{2}$ . J. Differential Geom. 27 (1988), no. 2, 259–296.
[CY2] S. Y. Chang and P. C. Yang. Extremal metrics of zeta function determinants on 4-manifolds. Ann. of Math. (2) 142 (1995), no. 1, 171–212.
[Ci] A. Cianchi. Moser-Trudinger inequalities without boundary conditions and isoperimetric problems. Indiana Univ. Math. J. 54 (2005), no. 3, 669–705.
[F] L. Fontana. Sharp borderline Sobolev inequalities on compact Riemannian manifolds. Comment. Math. Helv. 68 (1993), no. 3, 415–454.
[GiT] D. Gilbarg and N. S. Trudinger. Elliptic partial differential equations of second order. Reprint of the 1998 edition. Classics in Mathematics. Springer-Verlag, Berlin, 2001. xiv+517 pp. ISBN: 3-540-41160-7
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[HY2] F. B. Hang and P. C. Yang. Lectures on the fourth-order Q curvature equation. Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 31 (2016), 1–33.
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			$\displaystyle\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla v_{i}\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}+\left\\|\varphi\nabla u\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\nabla_{g}u_{i}\right\\|_{L^{n}\left(B_{R},d\mu\right)}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}.$

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla_{g}u\right\|_{g}^{n}d\mu+c\left(\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx.$

$\displaystyle\left\\|\nabla v_{\varepsilon}\right\\|_{L^{n}}^{n}$	$\displaystyle=$	$\displaystyle\int_{\varepsilon<\left\|x\right\|<1}\left(g^{ij}\frac{x_{i}}{\left\|x\right\|^{2}}\frac{x_{j}}{\left\|x\right\|^{2}}\right)^{\frac{n}{2}}dx$
	$\displaystyle=$	$\displaystyle\int_{\varepsilon<\left\|x\right\|<1}\frac{1-\alpha\left(\left\|x\right\|\right)}{\left\|x\right\|^{n}}dx$
	$\displaystyle=$	$\displaystyle\left\|\mathbb{S}^{n-1}\right\|\left(\log\frac{1}{\varepsilon}-\int_{\varepsilon}^{1}\frac{\alpha\left(r\right)}{r}dr\right).$

			$\displaystyle\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}\left(B_{R}^{+}\right)}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla v_{i}\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}+\left\\|\varphi\nabla u\right\\|_{L^{n}}+\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}\right)^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\frac{\varepsilon}{2}\right)\left\\|\varphi\nabla u_{i}\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\left\\|\varphi\nabla_{g}u_{i}\right\\|_{L^{n}\left(B_{R}^{+},d\mu\right)}^{n}+c\left(\varepsilon\right)\left\\|\varphi\nabla u\right\\|_{L^{n}}^{n}+c\left(\varepsilon\right)\left\\|v_{i}\nabla\varphi\right\\|_{L^{n}}^{n}.$

			$\displaystyle\lim\sup_{i\rightarrow\infty}\left\\|\nabla\left(\varphi v_{i}\right)\right\\|_{L^{n}}^{n}$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}d\sigma+\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla_{g}u\right\|_{g}^{n}d\mu+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx$
		$\displaystyle\leq$	$\displaystyle\left(1+\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}d\sigma+c\left(\varepsilon\right)\int_{B_{R}^{+}}\left\|\varphi\right\|^{n}\left\|\nabla u\right\|^{n}dx.$

Aubin type almost sharp Moser-Trudinger inequality revisited

Abstract.

1. Introduction

Lemma 1.1.

Proof.

Proposition 1.1.

Remark 1.1.

Theorem 1.1.

Remark 1.2.

Proof of Theorem 1.1.

Example 1.1.

Proof.

2. Functions on manifolds with nonempty boundary

Lemma 2.1.

Proof.

Lemma 2.2.

Proof.

Proposition 2.1.

Proof.

Theorem 2.1.

Proof.

Corollary 2.1.

Example 2.1.

Proposition 2.2.

Proof.

Theorem 2.2.

3. Second order Sobolev spaces

Lemma 3.1.

Proof.

Proposition 3.1.

Theorem 3.1.

Proof.

Example 3.1.

3.1. Paneitz operator and Q curvature in dimension 4

Lemma 3.2.

Proof of Lemma 3.2.

3.2. Functions on manifolds with nonempty boundary

Lemma 3.3.

Proof.

Lemma 3.4.

Proof.

Proposition 3.2.

Proof of Proposition 3.2.

Theorem 3.2.

Proof.

Example 3.2.

Lemma 3.5.

Proof.

Proposition 3.3.

Theorem 3.3.

Example 3.3.

Proposition 3.4.

Proof.

Theorem 3.4.

4. Higher order Sobolev spaces

4.1. Wm,nm​(Mn)W^{m,\frac{n}{m}}\left(M^{n}\right) for even mm

Lemma 4.1.

Proposition 4.1.

Theorem 4.1.

4.1.1. Functions on manifolds with nonempty boundary

Lemma 4.2.

Proposition 4.2.

Theorem 4.2.

Lemma 4.3.

Proposition 4.3.

Theorem 4.3.

Proposition 4.4.

Theorem 4.4.

4.2. Wm,nm​(Mn)W^{m,\frac{n}{m}}\left(M^{n}\right) for odd mm

References

4.1. $W^{m,\frac{n}{m}}\left(M^{n}\right)$ for even $m$

4.2. $W^{m,\frac{n}{m}}\left(M^{n}\right)$ for odd $m$