A blowup proof of the Aubin theorem in the Yamabe problem

November 8, 2011

A blowup proof of the Aubin theorem in the Yamabe problem

Filed under: PDEs, Riemannian geometry — Tags: Yamabe problem — Ngô Quốc Anh @ 13:29

Yamabe’s approach was to consider first the perturbed functional

$\displaystyle Q_s(u)\doteqdot\frac{\displaystyle\int_M\Big(|\nabla u|^2+\frac{n-2}{4(n-1)}R_gu^2\Big)d\mu_g}{\left(\displaystyle\int_M|u|^sd\mu_g\right)^\frac{2}{s}}$

where

$\displaystyle s\in \left(0,\frac{2n}{n-2} \right] \quad \text{ and } \quad u\in H^1(M)\setminus\{0\}.$

Set

$\displaystyle \lambda_s\doteqdot\inf\big\{Q_s(u):u\in H^{1}(M)\setminus\{0\}\big\}\quad\text{and}\quad Y(M)=\lambda_{2^*}.$

By using a direct minimizing procedure, it can be shown that for $2 < s < 2^*$ , there exists a smooth positive function $u_s$ such that its $L^s$ -norm is equal to one, $Q_s(u_s) = \lambda_s$ , and $u_s$ satisfies the equation

$\displaystyle \Delta_gu_s-\frac{n-2}{4(n-1)}R_gu_s+\lambda_su^{s-1}_s=0,\quad \text{in}\;M.$

The direct method does not work when $s=2^*$ because the Sobolev embedding $H^1(M) \hookrightarrow L^{2^*}(M)$ is continuous but not compact. However, if one can show that $u_s$ is uniformly bounded, i.e. there exists a positive constant $c$ such that $u_s \le c$ in $M$ for $2 < s < 2^*$ , then there exists a sequence $\{s_i\} \subset (2, 2^*)$ such that and $u_{s_i}$ converges to a smooth positive function $u$ which satisfies the Yamabe equation .

We discuss a blow-up argument. Suppose that no such upper bound $c$ exists. It follows that there exist sequences $\{s_k\} \subset (2, 2^*)$ and $\{y_k\} \subset M$ such that

$\displaystyle s_k\to 2^*\quad\text{and}\quad m_k\doteqdot u_{s_k}(y_k)=\max u_{s_k}\to\infty,\quad \text{ as } k\to\infty.$

As $M$ is compact, we may assume that $y_k \to y_0$ as $k \to\infty$ . For a normal coordinate system centered at $y_0$ and with radius $\rho$ , let the coordinates of $y_k$ be $x_k$ , $k \geqslant 1$ . In the local coordinates,

$\displaystyle g_{ij}(x)=\delta_{ij}+O(\rho^2),\quad \det g=1+O(\rho^2).$

From the equation, we know that $u_{k}=u_{s_k}$ satisfies

$\displaystyle \frac1{\sqrt{\det g}}\partial_j\Big(\sqrt{\det g}g^{ij}\partial_iu_{k}\Big)-\frac{n-2}{4(n-1)}R_gu_{k}+\lambda_ku^{{s_k}-1}_{k}=0,\quad \text{ in }\;B_0(\rho).$

The idea here is to consider the normalized function

$\displaystyle v_k(x)\doteqdot\frac{u_k(\delta_kx+x_k)}{m_k}$

where $\delta_k=m_k^{(2-s_k)/2}$ . We have $x_k \to 0$ and $\delta_k \to 0$ as $k \to\infty$ . Here $v_k$ is defined on a ball in $\mathbb{R}^n$ of radius $\rho_k = \frac{\rho-|x_k|}{\delta_k} \to\infty$ as $k\to\infty$ . Obviously,

$\displaystyle \frac{{{m_k}}}{{{\delta _k^2}}}\frac{1}{{\sqrt {\det g} }}{\partial _j}\left( {\sqrt {\det g} {g^{ij}}{\partial _i}{v_k}} \right) = \frac{{\frac{{n - 2}}{{4(n - 1)}}R({\delta _k}x + {x_k})m_k^{2 - {s_k}}{v_k}(x) - {\lambda _k}{v_k}{{(x)}^{{s_k} - 1}}}}{{m_k^{1 - {s_k}}}}.$

Under the choice of $\delta_k$ , we have

$\displaystyle\underbrace {\frac{1}{{\sqrt {\det g} }}}_{ \to 1}{\partial _j}\left( {\sqrt {\det g} {g^{ij}}{\partial _i}{v_k}} \right) =\underbrace {\frac{{n - 2}}{{4(n - 1)}}R({\delta _k}x + {x_k})m_k^{2 - {s_k}}}_{ \to 0}{v_k}(x)- \underbrace {{\lambda _k}}_{ \to \lambda }{v_k}{(x)^{{s_k} - 1}}$

where those limits are taken as $k \to \infty$ . By the argument of diagonal subsequence and the property of normal coordinates, one observes that a subsequence of $\{v_k\}$ converges to a smooth positive function $v$ which is a nonnegative solution of the equation

$\displaystyle \Delta_0 v+\lambda v^{\frac{n+2}{n-2}}=0,\quad\text{in}\;\mathbb{R}^n\hfill (1)$

where $\lambda=\lim_{k\to\infty}\lambda_k$ , and $\Delta_0$ is the standard Laplacian on $\mathbb{R}^n$ . By the strong maximum principle, $v>0$ . It is known that

$\lambda \begin{cases} <\lambda(M),& \text{ if }\lambda(M) < 0,\\ =\lambda(M),& \text{ if }\lambda(M) \ge 0\end{cases}$

where $\lambda (M)$ is an invariant depending only on the conformal class $[g]$ of the metric $g$ . Let $d$ be the diameter of $(M, g)$ . By a change of variables we have

$\displaystyle \int_{|x|<\frac{d}{2\delta_k}}v_k^{s_k}\sqrt{\det g}dx=\delta_k^{\frac{2s_k}{s_k-2}-n}\int_{B_{x_k}(\frac d2)}u_k^{s_k}d\mu_g\le\delta_k^{\frac{2s_k}{s_k-2}-n}\hfill (2)$

where $B_{x_k}(d/2)$ denotes the open ball in $(M, g)$ with center at $x_k$ and radius equal to $\frac{d}{2}$ . we note that

$\displaystyle \frac{2s_k}{s_k-2}-n>0\quad \text{and} \frac{2s_k}{s_k-2}-n \to 0\quad\text{as}\;k\to\infty.$

From (2) the Fatou lemma and $\lim_{k\to\infty}\delta_k\to 0$ , we obtain

$\displaystyle \int_{\mathbb{R}^n}v^{\frac{2n}{n-2}}dx\le 1.\hfill (3)$

A similar argument implies

$\displaystyle \int_{\mathbb{R}^n}|\nabla v|^2dx<\infty.$

Let $\eta\in C^{\infty}_0(\mathbb{R}^n)$ be a cutoff function satisfies

$\eta =\begin{cases} 1,& \text{ in } B_0(d),\\ 0, & \text{ in }\mathbb{R}^n\setminus B_0(2d).\end{cases}$

Defined $v_R(x)=\eta{\frac xR}v(x)$ , then

$\displaystyle \int_{\mathbb{R}^n}(|\nabla(v-v_R)|^2+|v-v_R|^{2^*})dx\to0,\quad \text{as}\;R\to\infty.\hfill (4)$

Multiplies (1) by $v_R$ and integration by parts, we obtain

$\displaystyle \int_{\mathbb{R}^n}\nabla v_R\nabla vdx=\lambda\int_{\mathbb{R}^n}v^{2^*-1}v_Rdx.$

Taking $R\to\infty$ in above equation and thanks to (4) we get

$\displaystyle \int_{\mathbb{R}^n}|\nabla v|^2dx=\lambda\int_{\mathbb{R}^n}v^{2^*}dx.\hfill(5)$

If $\lambda\le 0$ , then $v=\text{constant}$ , and (3) implies $v\equiv 0$ , which is a contradiction with $v>0$ .
If $\lambda>0$ , $\lambda=\lambda (M)$ . (2) (5) and the best Sobolev imbedding implies

$\displaystyle \Lambda\Big(\int_{\mathbb{R}^n}v^{2^*}dx\Big)^{2/2^*}\le\int_{\mathbb{R}^n}|\nabla v|^2dx=\lambda(M)\int_{\mathbb{R}^n}v^{2^*}dx.$

Thus

$\displaystyle \Lambda\le\lambda(M)\Big(\int_{\mathbb{R}^n}v^{2^*}dx\Big)^{n/2}\le\lambda(M).$

We are led to the contradiction with

$\lambda(M)<\lambda(\mathbb{S}^n)=\Lambda.$

Therefore, $u_s$ is uniformly bounded.

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Ngô Quốc Anh

November 8, 2011