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<a \leqslant V(x) \leqslant b

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<m \leqslant \mathbb G_L (x,y) \leqslant c d_g(x,y)^{2-n}

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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