A simple proof of Aubin’s theorem in Yamabe problem

Theorem(Aubin) Let (M, g) be a compact Riemannian manifold with Y(M, g) < Y (S_n), where Y(M, g) is called the Yamabe invariant and defined by

\displaystyle  Y(M,g)=\inf_{u\in C^{\infty}(M)}Q_g(u)=\inf_{ u\in C^{\infty}(M)}\frac{\int_M(|\nabla u|^2+\frac{n-2}{4(n-1)}R_gu^2)d\mu_g}{\|u\|^2_{L^{2^*}}}.

Then the infimum of the functional Q_g(u) is attained. Namely, the Yamabe problem can be solved.

The original proof of Theorem used the subcritical equations. There is another simple proof by Druet and Hebey using the  Brezis and Lieb’s lemma.

Proof: After passing to a subsequence, we may also assume that there exists u\in H^1(M) such that u_i\rightharpoonup u weakly in H^1(M), applying the imbedding theorem, we obtain u_i\to u strongly in L^2(M), furthermore u_i\to u almost everywhere as i\to\infty. In particular, u is nonnegative. It easy form the weakly convergence that

\displaystyle  \|\nabla u_i\|^2_{L^2}=\|\nabla(u_i-u)\|_{L^2}^2+\|\nabla u\|^2_{L^2}+o(1)

for all i, where o(1)\to0 as i\to\infty. We also have that(p=2^*=\frac{2n}{n-2}) by  Brezis and Lieb’s lemma

\displaystyle  \|u_i\|^p_{L^p}=\|u_i-u\|_{L^p}^p+\|u\|_{L^p}^p+o(1).

Thanks to the sharp Sobolev inequality of Hebey and Vaugon, there exists B>0 such that for any i,

\displaystyle  \|u_i-u\|^2_{L^p}\le K_n^2\|\nabla(u_i-u)\|^2_{L^2}+B\|u_i-u\|^2_{L^2}.

Since u_i\in \mathcal{H}=\{u\in H^1(M):\int_M|u|^pd\mu_g=1\}, it follows that

\displaystyle  (1-\|u\|^p_{L^p})^{\frac 2p}\le K_n^2\left(\|\nabla u_i\|^2_{L^2}-\|\nabla u\|^2_{L^2}\right)+o(1).

Since Q_g(u_i)\to Y_g(M), and since u_i\to u in L^2(M), we also have that

\displaystyle\begin{gathered}  K_n^2\left(\|\nabla u_i\|^2_{L^2}-\|\nabla u\|^2_{L^2}\right)\hfill\\  \qquad=K_n^2Y(M)-K_n^2\left(\int_M|\nabla u|^2d\mu+\int_MaRu^2d\mu\right)+o(1)\hfill\\  \qquad\le K_n^2Y(M)-K_n^2Y(M)\|u\|^2_{L^p}+o(1)\hfill\\  \end{gathered}

Hence,

\displaystyle  (1-\|u\|^p_{L^p})^{\frac 2p}\le K_n^2Y_g(M)(1-\|u\|^2_{L^p})+o(1).

We assume that Y_g(M)<Y_g(S^n)=1/K_n^2 and note that

\displaystyle  1-\|u\|^2_{L^p}\le (1-\|u\|^p_{L^p})^{\frac 2p}

this implies that

\displaystyle  \|u\|^2_{L^p}=1.

Then,

\displaystyle  \|\nabla u_i\|_{L^2}\to \|\nabla u\|_{L^2}

as i\to\infty. and since

\displaystyle  \|\nabla u_i\|^2_{L^2}=\|\nabla(u_i-u)\|_{L^2}^2+\|\nabla u\|^2_{L^2}+o(1)

We obtain that u_i\to u strongly in H^1 as i\to\infty. In particular, u is a minimizer for Y_g(M) and u is a weak nonnegative solution of the Yamabe equation

\displaystyle  \Delta_gu-\frac{n-2}{4(n-1)}R_gu=Y_g(M)u^{2^*-1}.

Regularity argument and the maximum principle then give that u is smooth and positive. This prove the Theorem.

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2 Responses to A simple proof of Aubin’s theorem in Yamabe problem

  1. Ngô Quốc Anh说道:

    The terminology Yamabe invariant has just been generalized by Aubin’s student in his PhD dissertation, 2010 I guess. If I am not wrong, this solved the Hebey–Vaugon conjecture. It is great if you could reprocedure this stuff here.

    • mathsnail说道:

      Thank you for your reminding, i know that paper in Arxiv, but not reading carefully, maybe i will write down here in summer vocation. I’m interested in your blog, which also enlighten me. moreover, you are the first commenter of my blog.

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