# 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),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.

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 $$S_+^n$$.
• May I ask why $Q(S^{n}_{+})=Q(R^{n}_{+})?$Are they conformally equivalent?By stereographic projection I can only show $S^{n}_{+}$ and $D_{n}$(unit disk) are conformally equivalent. Commented Oct 15, 2022 at 15:10
• They are equivalent by stereographic projection when you choose a point on $\partial S_+^n$ to go to infinity. Commented Oct 15, 2022 at 23:31