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, $\lVert \psi \rVert_p > 0$. Combining this with Fatou and the normalization $\lVert\varphi_s\rVert_s=1$ yields $\lVert\psi\rVert_p \in (0,1]$. 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 $\lVert \psi \rVert_p \in (0,1]$,

$$ \tag{$\ast$}\label{bl}\lim_{s \to p} \lVert u_s \rVert_p^p = 1 - \lVert \psi \rVert_p^p \geq 1 - \lVert \psi \rVert_p^2 , $$ with equality if and only if $\lVert\psi\rVert_p=1$.

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} implies that $\lVert\psi\rVert_p=1$.

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 $\lVert\psi\rVert_p \geq 1$.

You are forgetting that, as part of the renormalization, the functions $\psi_s$ are uniformly bounded above by one. This together with the convergence in $C_{loc}^2(S^n \setminus \{ P \})$ implies that $\lVert \psi_s \rVert_p^p \to \lVert \psi \rVert_p^p$. Since $\lVert \varphi_s \rVert_s = 1$ by construction of $\varphi_s$, the conclusion follows.

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, $\lVert \psi \rVert_p > 0$. Combining this with Fatou and the normalization $\lVert\varphi_s\rVert_s=1$ yields $\lVert\psi\rVert_p \in (0,1]$. 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 $\lVert \psi \rVert_p \in (0,1]$,

$$ \tag{$\ast$}\label{bl}\lim_{s \to p} \lVert u_s \rVert_p^p = 1 - \lVert \psi \rVert_p^p \geq 1 - \lVert \psi \rVert_p^2 , $$ with equality if and only if $\lVert\psi\rVert_p=1$.

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} implies that $\lVert\psi\rVert_p=1$.