# Ngô Quốc Anh

## March 28, 2014

### Lower bound for conformal factor in terms of upper bound of L^4-norm of the scalar curvature in the negative case

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 18:01

In this note, we talk about the following interesting result obtained by X. Xu in Nagoya Mathematical Journal in 1995, here.

To state his result, let us fix $(M,g_0)$ a compact manifold of dimension $4$. We shall denote by $\text{Scal}_g$ the scalar curvature computed with respect to the metric $g$. It is clear that under the conformal change $g=u^2g_0$, $\text{Scal}_{g_0}$ and $\text{Scal}_g$ are related by the following rule

$\displaystyle \text{Scal}_g = u^{-3}\big(u\text{Scal}_{g_0}-6\Delta_{g_0}u \big).$

Furthermore, we may also assume that $\text{Scal}_{g_0}$ is a negative constant. We are now able to state his result.

Theorem. If $\text{Scal}_{g_0} < 0$ and $\int_M |\text{Scal}_g|^4 dv_g< C_o$, then there exists a constant $C_1 > 0$ such that $u > C_1$ where $g = u^2 g_0$.

Proof. Using the conformal change rule shown above, we obtain

$\begin{array}{lcl}{C_0} &\ge& \displaystyle\int_M {|\text{Scal}_g{|^4}d{v_g}} \\ &=& \displaystyle\int_M {{{[{u^{ - 3}}(u \text{Scal}_{g_0} - 6{\Delta _{{g_0}}}u)]}^4}({u^4}d{v_{{g_0}}})} \\ &=& \displaystyle\int_M {{{[{u^{ - 1}}\text{Scal}_{g_0} - 6{u^{ - 2}}{\Delta _{{g_0}}}u]}^4}d{v_{{g_0}}}} \\ &=& \displaystyle\int_M {{u^{ - 4}}|\text{Scal}_{g_0}|^4 d{v_{{g_0}}}} + {\rm{1296}}\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^4}d{v_{{g_0}}}} + \\&& \displaystyle + {\rm{216}}\int_M {{u^{ - 6}}|\text{Scal}_{g_0}|^2{{({\Delta _{{g_0}}}u)}^2}d{v_{{g_0}}}} - 24\int_M {{u^{ - 5}}|\text{Scal}_{g_0}|^3{\Delta _{{g_0}}}ud{v_{{g_0}}}} \\ &&\displaystyle - 864\int_M {{u^{ - 7}} \text{Scal}_{g_0} {{({\Delta _{{g_0}}}u)}^3}d{v_{{g_0}}}} \\ &=& \displaystyle |\text{Scal}_{g_0}|^4\int_M {{u^{ - 4}}d{v_{{g_0}}}} - 24|\text{Scal}_{g_0}|^3 \int_M {{u^{ - 5}}{\Delta _{{g_0}}}ud{v_{{g_0}}}} \\ &&\displaystyle +432\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^4}d{v_{{g_0}}}} + 216\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^2}{{[\text{Scal}_{g_0} - 2{\Delta _{{g_0}}}u]}^2}d{v_{{g_0}}}} .\end{array}$

Since $\text{Scal}_{g_0}<0$, $\text{Scal}_{g_0}^3<0$. Also, using integration by parts,

$\displaystyle \int_M {{u^{ - 5}}{\Delta _{{g_0}}}ud{v_{{g_0}}}} = 5\int_M {{u^{ - 6}}|\nabla u|_{{g_0}}^2d{v_{{g_0}}}} > 0.$

Therefore, each term on the right hand side of the preceding inequality is positive, hence we obtain

$\displaystyle |\text{Scal}_{g_0}|^4\int_M {{u^{ - 4}}d{v_{{g_0}}}} \le {C_0}$

and

$\displaystyle\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^4}d{v_{{g_0}}}} \le {C_0}.$

Now from the identity

$\displaystyle {\Delta _{{g_0}}}(\frac{1}{u}) = - \frac{{{\Delta _{{g_0}}}u}}{{{u^2}}} + \frac{{2|\nabla u{|^2}}}{{{u^3}}}$

it is easy to see that for any point $p \in M$

$\displaystyle \frac{1}{{u(p)}} - \frac{1}{{\text{vol}(M,{g_0})}}\int_M {\frac{1}{u}d{v_{{g_0}}}} = - \int_M {G(p,q)\left[ - \frac{{{\Delta _{{g_0}}}u}}{{{u^2}}} + \frac{{2|\nabla u|^2}}{{{u^3}}} \right](q)d{v_{{g_0}}}}$

where $G(p,q)$ is the Green function on $M$ which can be assumed to be positive everywhere on $M$. The well known fact is that the Green function on a $4$-dimensional manifold is $L^\alpha$ integrable for $\alpha < 2$. Thus using the Holder inequality, it is clear that $u$ is bounded from below.