## A blowup proof of Aubin’s theorem in the Yamabe problem

There is a classic blowup analysis proof for Aubin’s theorem, due to Uhlenbeck’s renormalization method described below to give another proof that the Yamabe problem. See the chapter5 at Schoen & Yau              ‘ Lectures on differential geometry’.

Yamabe’s approach was to consider first the perturbed functional

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

where $s\in(0,\frac{2n}{n-2}]$ and $u\in H^1(M)\setminus\{0\}$. Set

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

By using a direct minimizing procedure, it can be shown that for $2 < s < 2^*$, there exists a smooth positive function us such that its $L^s$-norm is equal to one, $Q_s(u_s) = \lambda_s$ and us 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 $H^1(M)\to L^{2^*}(M)$  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 $\{s_i\} \in 2 < s < 2^*$ 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

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

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 $x_k, k = 1, 2, ....$.

In the local coordinates,

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

$u_{k}=u_{s_k}$ satisfies the equation

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

The idea here is to consider the normalized function

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

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 $\rho_k = (\rho-|x_k|)/\delta_k$ and $\rho_k\to\infty$ 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 $\lambda=\lim\limits_{k\to\infty}\lambda_k$,and $\Delta_0$ is the standard Laplacian on $\mathbb{R}^n$.

By the strong maximum principle, $v>0$. It is known that $\lambda<\lambda(M)$ if $\lambda(M) < 0$; and $\lambda=\lambda(M)$ if $\lambda(M) \ge 0$ . Let $d$ be the diameter of $(M, g)$. By a change of variables we have

$\displaystyle \int_{|x|<\frac d2\delta_k^{-1}}v_k^{s_k}\sqrt{\det g}dx=\delta_k^{\frac{2s_k}{s_k-2}-n}\int_{B_{x_k}(\frac d2)}u_k^{s_k}d\mu_g\le\delta_k^{\frac{2s_k}{s_k-2}-n}\hfill (2)$

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

$\displaystyle \frac{2s_k}{s_k-2}-n>0\quad \text{and}\;\to0\quad\text{as}\;k\to\infty.$

From (2) the Fatou lemma and $\lim\limits_{k\to\infty}\delta_k\to0$ , we obtain

$\displaystyle \int_{\mathbb{R}^n}v^{\frac{2n}{n-2}}dx\le1\hfill(3)$

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 $\eta =1$ in $B_0(d)$ and  $\eta =0$ in $\mathbb{R}^n\setminus B_0(2d)$

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 $\lambda\le0$, then $v=\text{constant}$, and (2)  implies $v\equiv0$, 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.

### 2 Responses to A blowup proof of Aubin’s theorem in the Yamabe problem

1. Ngô Quốc Anh说道：

It seems there is a typo in your calculation. After introducing the normalized function, I think the constant $\delta_k$ should be $m_k^{(2-s_k)}$. If I am not wrong, your equation of $u_k$ can be transformed to the following

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

which forces $\delta_k=m_k^{(2-s_k)}$. Please clarify.

• Ngô Quốc Anh说道：

Oh you are right, in fact we have

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