How to calculate the infimum of Yamabe functional on upper hemisphere We introduce the following functional to study Yamabe problem with boundary.
$$
Q_g(\varphi)=\frac{\int_M\left(|\nabla \varphi|_g^2+\frac{n-2}{4(n-1)} R_g \varphi^2\right) d v+\frac{n-2}{2} \int_{\partial M} h_g \varphi^2 d \sigma}{\left(\int_M|\varphi|^{2 n /(n-2)} d v\right)^{(n-2) / 2}}
$$
$$
Q(M)=\inf \left\{Q_g(\varphi): \varphi \in C^1(\bar{M}), \varphi \neq 0\right\}
$$
In Escobar's The Yamabe Problem On Manifolds With Boundary,he says if $Q(M)<Q(S^{+}_{n})$,the functional has a critical point which is the solution of Yamabe Problem.
He also gives the formula of $Q(S_{n}^{+})$
$$
Q\left(S_{+}^n\right)=\frac{\int_{\mathbb{R}_{+}^n}\left|\nabla u_{\varepsilon}\right|^2}{\left(\int_{\mathbb{R}_{+}^n} u_{\varepsilon}^{2 n /(n-2)}\right)^{(n-2) / n}}=n(n-2)\left(\int_{\mathbb{R}_{+}^n} u_{\varepsilon}^{\frac{2 n}{n-2}}\right)^{2 / n},
$$
$$
u_{\varepsilon}\left(x, x_n\right)=\left(\frac{\varepsilon}{\varepsilon^2+|x|^2+x_n^2}\right)^{(n-2) / 2}
$$
I wonder how this is derived.
When studying the Yamabe problem on compact manifolds,we use conformal change to calculate $Q(S_{n})$.Since the functional is conformally invariant, we use conformal change to transfer the problem to $R^{n}$,which becomes the problem of best Sobolev constant.
How can we do this when studying the problem on manifolds with boundary?Any help will be thanked.
 A: The first observation is that $Q_g(\phi)$ is conformally invariant.
Define
\begin{align*}
 L_gu & := -\Delta u + \frac{n-2}{4(n-1)}Ru , \\
 B_gu & := \partial_\nu u + \frac{n-2}{2}Hu ,
\end{align*}
where $\nu$ is the outward-pointing unit normal and everything is defined with respect to the metric $g$.
On the one hand, integration by parts gives
\begin{equation*}
 Q_g(u) = \frac{\int_M u\,Lu\,dv_g + \oint_{\partial M} u\,Bu\,d\sigma_g}{\left( \int_M \lvert u \rvert^{\frac{2n}{n-2}} \, dv_g \right)^{\frac{n-2}{n}}} .
\end{equation*}
On the other hand, if $\hat g = w^{\frac{4}{n-2}}g$, then
\begin{align*}
 L_{\hat g}u & = w^{-\frac{n+2}{n-2}}L_g(wu) , \\
 B_{\hat g}u & = w^{-\frac{n}{n-2}}B_g(wu) .
\end{align*}
Therefore
\begin{equation*}
 Q_{\hat g}(u) = Q_g(wu) .
\end{equation*}
It follows from stereographic projection that $Q(S_+^n) = Q(\mathbb{R}_+^n)$, where $\mathbb{R}_+^n$ is the upper half plane $\{ x_n \geq 0 \}$.
The next observation is that $u_\varepsilon$ really minimizes $Q(\mathbb{R}_+^n)$.
One argument goes as follows:
By the Concentration Compactness Principle, a minimizer $u$ exists.
Moreover, the usual argument using elliptic regularity and the maximum principle implies that a minimizer is smooth.
By an adaptation of the Obata argument, if $u$ is a minimizer, then $u^{\frac{4n}{n-2}}dx^2$ is an Einstein metric on $S_+^n$ with respect to which the boundary is minimal.
This requirement implies that $u$ takes the form $u_\varepsilon$.
Note that the latter argument proceeds by using stereographic projection again to study the minimizer $u$ on $S_+^n$.
