Lee-Parker Yamabe problem proposition 4.6

Question

I believe there may be a gap towards the end of the proof of proposition 4.6 in the Bulletin of the AMS paper The Yamabe Problem by Lee and Parker : https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-17/issue-1/The-Yamabe-problem/bams/1183553962.full. They are trying to produce a minimizer to the Yamabe quotient on the standard round sphere. They proceed by considering the corresponding sub-critical problem for which it is standard to produce a smooth minimizer. This produces a family $\{\varphi_{s}\}$, which (since we are on the sphere), may not stay uniformly bounded as $s \to p= \frac{2n}{n-2}. $ They proceed by renormalizing the sequence using a family of chosen conformal diffeomorphisms, and establish that this renormalized sequence converges weakly in $ W^{1,2} $ and $C^2_{loc}(S^n)\setminus\{P\} $ (P denotes the north pole) to a function $\psi$, which they want to show is a minimizer. However towards the end of page 56, they claim that $$ \|\psi_{s}\|_{p}^p \geq Vol(S^n)^{1-\frac{s}{p}}\|\varphi_{s}\|_{s}^p ,$$ implies $ \|\psi\|_{p} \geq 1.$ It is this claim which I have problems with. The control they have on $\psi_{s}$ near the north pole is not enough to rule out the mass "escaping out to the infinity". Indeed one can consider a family of Aubin-Talenti bubbles on the sphere which would show this claim cannot be true, unless they are using something else about $\psi_{s}$ I'm not seeing. Hope someone can clarify this issue.

Answer 1

You are forgetting that, as part of the renormalization, the functions $\psi_s$ all satisfy $\psi_s(-P)=1$ (here $-P$ is the south pole). In particular, since $\psi$ is $C^2$ away from the north pole, $\lVert \psi \rVert_p > 0$. You can get $\lVert \psi \rVert_p^p \geq 1$ using the Brezis-Lieb lemma:

$$ \lVert \psi \rVert_p^p = \lim_{s \to p} \left( \lVert \psi_s \rVert_p^p - \lVert \psi - \psi_s \rVert_p^p \right) . $$

Set $u_s := \psi_s - \psi$. By the Brezis-Lieb lemma, the normalization of $\psi_s$, and the fact $\lVert \psi \rVert_p \in (0,1]$,

$$ \tag{$\ast$}\label{bl}\lim_{s \to p} \lVert u_s \rVert_p^p = \lim_{s \to p} \lVert \psi_s \rVert_p^p - \lVert \psi \rVert_p^p \geq 1 - \lVert \psi \rVert_p^2 . $$

Now recall that, by construction, $u_s \rightharpoonup 0$ in $W^{1,2}$. Then $u_s \to 0$ in $L^2$. Therefore

\begin{align*} \lim_{s \to p} \Lambda\lVert u_s \rVert_p^2 & \leq \lim_{s \to p} \int_{S^n} (\psi - \psi_s)\Box (\psi - \psi_s) && \text{(defn $\Lambda$)} \\ & = \lim_{s \to p} \int_{S^n} \left( -\psi\Box\psi + 2\psi\Box(\psi-\psi_s) + \psi_s\Box\psi_s \right) && \text{(expand)} \\ & = -\int_{S^n} \psi\Box\psi + \lim_{s \to p} \int_{S^n} \psi_s \Box \psi_s && \text{(weak convergence)} \\ & \leq -\Lambda\lVert\psi\rVert_p^2 + \Lambda && \text{(defn $\Lambda$ and $\psi_s$)} . \end{align*}

Combining this with \eqref{bl} yields $\lVert\psi\rVert_p^2 \leq \lVert\psi\rVert_p^p$. Since $\lVert \psi \rVert_p > 0$, we conclude that $\lVert\psi\rVert_p \geq 1$.