I am reading the paper of [Convergence of the Yamabe flow for arbitrary initial energy](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-69/issue-2/Convergence-of-the-Yamabe-flow-for-arbitrary-initial-energy/10.4310/jdg/1121449107.full)


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.



Implicit function theorem for operator
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. 


