# Ngô Quốc Anh

## November 6, 2013

### The prescribed scalar curvature: An inf x sup inequality in the positive Yamabe invariant

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 0:56

Today, we talk about an inequality of the form $\inf \times \sup \geqslant c$ for solutions of the so-called prescribed scalar curvature problem, i.e. $\displaystyle -\frac{4(n-1)}{n-2}\Delta_g u + \text{Scal}_g u = V u^{2^\star -1}$

where the scalar curvature $\text{Scal}_g>0$ is fixed but the function $V$ which satisfies $0

for any $x \in M$. Here $2^\star = 2n/(n-2)$ is the Sobolev critical exponent and $n \geqslant 3$ is the dimension of $M$. In addition, the manifold $M$ is compact and has no boundary.

Theorem (Bahoura). There exists a positive constant $c$ depending on $a,b,M$ such that $\displaystyle \sup_M u \times \inf_M u \geqslant c$

for any solution $u$ of the PDE.

We prove this theorem using contradiction: There exists a sequence of solutions $u_i$ of $\displaystyle -\frac{4(n-1)}{n-2}\Delta_g u_i + \text{Scal}_g u_i = V_i u_i^{2^\star -1}$

such that $\displaystyle \sup_M u_i \times \inf_M u_i \to 0$

as $i \to \infty$.

Step 1. There holds $\|u_i\|_{2^\star} \to 0$ as $i \to \infty$.

To conclude this step, we make use of the Green function $\mathbb G$ associated to the operator $\displaystyle L := \Delta_g -\frac{n-2}{4(n-1)}\text{Scal}_g.$

It is worth noticing that $\mathbb G_L$ is invertible and $\displaystyle 0

for some positive constants $m$ and $c$.  Using $\mathbb G_L$, we have the following representation $\displaystyle u_i(x) = \frac{n-2}{4(n-1)}\int_M \mathbb G_L (x,y) V_i(y)u_i(y)^{2^\star-1} dv_g.$

WLOG, we may assume that $\inf_M u_i = u_i(x_i)$. Thanks to $V_i \geqslant a$, for any $y\in M$, there holds $\displaystyle \frac{4(n-1)}{n-2} \min_M u_i = \int_M \mathbb G_L (x_i,y) V_i(y)u_i(y)^{2^\star-1} dv_g \geqslant am \int_M u_i(y)^{2^\star-1}dv_g.$

Hence, $\displaystyle \|u_i\|_{2^\star}^{2^\star} \leqslant \sup_M u_i \int_M u_i^{2^\star -1} dv_g \leqslant \frac{4(n-1)}{am(n-2)} \sup_M u_i \times \inf_M u_i \to 0.$

as claimed.

Step 2. There holds $\sup_M u_i \to 0$ as $i \to \infty$.

For $k>1$, we multiply the PDE by $u_i^{2k-1}$ $\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}\Delta {u_i} + \text{Scal}_g {u_i}} \right)u_i^{2k - 1}d{v_g}} = \int_M {{V_i}u_i^{{2^ \star } + 2k - 2}d{v_g}}$

and integrate by parts to get $\displaystyle\frac{{4(n - 1)}}{{n - 2}}(2k - 1)\int_M {u_i^{2k - 2}|\nabla {u_i}{|^2}d{v_g}} + \int_M {\text{Scal}_g u_i^{2k}d{v_g}} = \int_M {{V_i}u_i^{{2^ \star } + 2k - 2}d{v_g}} .$

Thanks to $|\nabla u_i^k{|^2} = {k^2}u_i^{2k - 2}|\nabla {u_i}{|^2}$ we eventually get $\displaystyle\int_M {|\nabla u_i^k{|^2}d{v_g}} = \frac{{n - 2}}{{4(n - 1)}}\frac{{{k^2}}}{2k - 1}\left( { - \int_M {\text{Scal}_g u_i^{2k}d{v_g}} + \int_M {{V_i}u_i^{{2^ \star } + 2k - 2}d{v_g}} } \right).$

The positivity of $\text{Scal}_g$ plus the bounds for $V$ imply $\displaystyle\int_M {|\nabla u_i^k{|^2}d{v_g}} \leqslant \frac{{n - 2}}{{4(n - 1)}}\frac{{b{k^2}}}{{2k - 1}}\int_M {u_i^{{2^ \star } + 2k - 2}d{v_g}} .$

Using the Holder inequality, we know that $\displaystyle\int_M {u_i^{{2^ \star } + 2k - 2}d{v_g}} = \int_M {u_i^{2k}u_i^{{2^ \star } - 2}d{v_g}} \leqslant {\left( {\int_M {u_i^{k{2^ \star }}d{v_g}} } \right)^{2/{2^ \star }}}{\left( {\int_M {u_i^{{2^ \star }}d{v_g}} } \right)^{({2^ \star } - 2)/{2^ \star }}}.$

Hence we have proved that $\displaystyle\int_M {|\nabla u_i^k{|^2}d{v_g}} \leqslant \frac{{n - 2}}{{4(n - 1)}}\frac{{b{k^2}}}{{2k - 1}}{\left( {\int_M {u_i^{k{2^ \star }}d{v_g}} } \right)^{2/{2^ \star }}}{\left( {\int_M {u_i^{{2^ \star }}d{v_g}} } \right)^{({2^ \star } - 2)/{2^ \star }}},$

i.e. $\displaystyle\left\| {\nabla u_i^k} \right\|_2^2 \leqslant \frac{{n - 2}}{{4(n - 1)}}\frac{{b{k^2}}}{{2k - 1}}\left\| {{u_i}} \right\|_{{2^ \star }}^2\left\| {u_i^k} \right\|_{{2^ \star }}^2.$

Using the Sobolev inequality, that is $\displaystyle\left\| {u_i^k} \right\|_{{2^ \star }}^2 \leqslant K(n,2)\left\| {\nabla u_i^k} \right\|_2^2 + B\left\| {u_i^k} \right\|_2^2$

we obtain $\displaystyle\left\| {u_i^k} \right\|_{{2^ \star }}^2\left( {1 - K(n,2)\frac{{n - 2}}{{4(n - 1)}}\frac{{b{k^2}}}{{2k - 1}}\left\| {{u_i}} \right\|_{{2^ \star }}^{{2^ \star } - 2}} \right) \leqslant B\left\| {u_i^k} \right\|_2^2.$

Thanks to Step 1, for $i$ large $\displaystyle 1 - K(n,2)\frac{{n - 2}}{{4(n - 1)}}\frac{{b{k^2}}}{{2k - 1}}\left\| {{u_i}} \right\|_{{2^ \star }}^{{2^ \star } - 2} \leqslant \frac{1}{2}$

holds. Therefore, $\displaystyle\left\| {u_i^k} \right\|_{{2^ \star }}^2 \leqslant 2B\left\| {u_i^k} \right\|_2^2.$

Repeatedly using $k=(2^\star/2)^l$ for $l=1,2,...$ we can conclude from Step 1 that $u_i \to 0$ strongly in any $L^p(M)$ with $p>1$. Then the presentation using the Green function plus the Holder inequality immediately imply that $u_i \to 0$ uniformly in $M$. This certainly proves Step 2.

Step 3. A contradiction.

Again, WLOG, we assume $\sup_M u_i = u_i(y_i)$. Then $\displaystyle u_i(y_i) = \frac{n-2}{4(n-1)} \int_M \mathbb G_L (x,y_i)V_i(x)u_i(x)^{2^\star-1} dv_g.$

Since the Green function $\mathbb G_L$ verifies $\displaystyle 0 < m \text{vol}(M) \leqslant \int_M {{\mathbb G_L}(x,{y_i})d{v_g}} \leqslant c\int_M {\frac{{d{v_g}}}{{{d_g}{{(x,{y_i})}^{n - 2}}}}} \leqslant C(M),$

we obtain $\displaystyle \sup_M u_i \leqslant \frac{n-2}{4(n-1)}b C(M) (\sup_M u_i)^{2^\star-1}.$

Thus we have a contradiction in view of Step 2.

Note that when $n=2$, the $\sup \times \inf$ inequality we have just discussed become the inequality $\sup+\inf$. This type of result is bascially due to Brezis-Li-Shafrir.

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