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I am reading the paper of Convergence of the Yamabe flow for arbitrary initial energy

I am stuck by one part of the paper. Suppose $u_\infty>0$ is a smooth function on $(M, g_0)$ and $$L_0=\frac{4(n-1)}{n-2}\Delta_{g_0}-R_{g_0}$$ is the conformal laplacian, where $R_{g_0}$ is the scalar curvature. $u_\infty$ satisfies $$L_0u_\infty+r_\infty u_\infty^{\frac{n+2}{n-2}}=0$$ consider the linear operator $$\psi\mapsto T\psi=u_\infty^{-\frac{4}{n-2}}L_0\psi$$ which is symmetric with respect to the inner product $$(\psi_1,\psi_2)_\infty=\int_M u_\infty^{\frac{4}{n-2}}\psi_1\psi_2d\mu_0$$ on $L^2(M)$. By the spectral theorem, eigenvalue and eigenvector of $T$, suppose they are $\{-\lambda_i,i\in\mathbb{N}\}$ and $\{\psi_i, i\in\mathbb{N}\}$, satisfies $$L_0\psi_i+\lambda_i u_\infty^{\frac{4}{n-2}}\psi_i=0$$ $$\int_M u_\infty^{\frac{4}{n-2}}\psi_i\psi_jd\mu_0=\begin{cases}1\quad i=j\\0\quad i\neq j\end{cases}$$ $$\lambda_i\to\infty \text{ as } i\to \infty$$ Let $A=\{i|\lambda_i\leq \frac{n+2}{n-2}r_\infty\}$. He constructs the projection operator $\Pi$ which is $$\Pi f=\sum_{i\not\in A}\left(\int_M \psi_ifd\mu_0\right)u_\infty^{\frac{4}{n-2}}\psi_i=f-\sum_{i\in A}\left(\int_M \psi_ifd\mu_0\right)u_\infty^{\frac{4}{n-2}}\psi_i$$ He said that by using the implicit function theorem one can get the following significance: for every vecrtor $z\in \mathbb{R}^A$ sufficiently small, there exists a smooth function $\bar{u}_z$ such that $$\int_M u^{\frac{4}{n-2}}(\bar{u}_z-u_\infty)\psi_id\mu_0=z_i \quad \forall\,i\in A$$ $$\Pi\left(L_0\bar{u}_z+r_\infty\bar{u}_z^\frac{n+2}{n-2}\right)=0$$ futhermore the map $z\mapsto \bar{u}_z$ is real analytic.

Here comes my questions, how can we apply and justify the assumption of implicit function theorem? I can not get the motivation of the projection operator either.

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  • $\begingroup$ Have you read this paper? @Otis Chodosh $\endgroup$
    – Slm2004
    May 29, 2014 at 20:20

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Consider the map $F:C^{2,\alpha}\to \mathbb{R}^A \times (C^{0,\alpha}\cap Im \Pi)$ $$ F: f \mapsto \left( \left(\int u_\infty^{\frac{4}{n-2}}(f-u_\infty)\psi_a\right)_{a\in A},\Pi\left( \frac{4(n-1)}{n-2} \Delta f - Rf +r f^\frac{n+2}{n-2}\right)\right). $$ Compute $DF(u_\infty)$: $$ DF(u_\infty)f = \left( \left(\int u_\infty^{\frac{4}{n-2}}f\psi_a\right)_{a\in A},\Pi\left( \frac{4(n-1)}{n-2} \Delta f - Rf + \frac{n+2}{n-2} r u_\infty^\frac{4}{n-2} f\right)\right). $$ By definition of $A$, this is clearly invertible. (This a place where we use the projection operator, $\psi_a$ for $a \in A$ contains elements that would be in the kernel. But, we've projected away from these directions). Thus, we can use the implicit function theorem to define $u_z \in C^{2,\alpha}$ as claimed. The fact that $u_z$ depends analytically on $z$ is because $F$ is easily seen to be an analytic map, and the implicit function theorem yields an analytic function in this case.

So, it remains to prove that $u_z$ is smooth. This follows from standard elliptic regularity: we have that $$ \frac{4(n-1)}{n-2} \Delta u_z - Ru_z +r u_z^\frac{n+2}{n-2} \in Span_{a\in A}\{ \psi_a\}. $$ Hence, $u_z$ satisfies an elliptic PDE with smooth coefficients, so it is smooth.

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  • $\begingroup$ I have come up this $F$ before, but not as clear as you think. Thank you very much for your explanation. @otis chodosh $\endgroup$
    – Slm2004
    May 29, 2014 at 23:52
  • $\begingroup$ The projection map is still not natural for me. Why we can not project like this way: $\Pi f=\sum_{a\not\in A}\left(\int_Mu_\infty^{\frac{4}{n-2}}\psi_a fd\mu_0\right)\psi_a$, which is the projection with respect to the inner product $()_\infty$? I guess the answer is if we project this way, we get nothing $\endgroup$
    – Slm2004
    May 30, 2014 at 0:02
  • $\begingroup$ I could be wrong, but I think that at this stage, either choice would be fine. I'd guess there is just some reason that the other way is more convenient. $\endgroup$ May 30, 2014 at 0:58

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