Further Results on the Fractional Yamabe Problem: The Umbilic Case

Abstract

We prove some existence results for the fractional Yamabe problem in the case that the boundary manifold is umbilic, thus covering some of the cases not considered by González and Qing. These are inspired by the work of Coda-Marques on the boundary Yamabe problem but, in addition, a careful understanding of the behavior at infinity for asymptotically hyperbolic metrics is required.

Appendix

Lemma 5.1

The Fourier transform of the function

$$\begin{aligned} w(x)=\left( \frac{1}{1+|x|^2}\right) ^{\frac{n-2\gamma }{2}}, \quad x\in {\mathbb {R}}^n, \end{aligned}$$

is given by

$$\begin{aligned} {\hat{w}}(\zeta )=C_0 |\zeta |^{-\gamma }K_\gamma (|\zeta |), \end{aligned}$$

for some constant $C_0=C_0(n,\gamma )$, and $K_\gamma$ the modified Bessel function from Lemma 3.1.

Proof

In the following, all the equalities will be so up to multiplicative constant that may change from line to line. Since w is a radial function, its Fourier transform will be radial too, and we can choose coordinate axes such that $\zeta =|\zeta |e_1$. Then, expanding in spherical coordinates,

$$\begin{aligned} \hat{w}(\zeta )=\int _{\mathbb R^n} e^{-i x\cdot \zeta }w(x)\,dx =\int _0^\infty \int _0^\pi e^{-i|\zeta |\cos \theta _1}(1+r^2)^{-\mu }r^{n-1}\sin ^{n-2}\theta _1\, d\theta _1\,dr. \end{aligned}$$

It is well known ([20, page 48]) that

$$\begin{aligned} J_{\frac{n}{2}-1}(ar)=(ar)^{\frac{n}{2}-1} \int _0^\pi e^{ia r\cos \theta }\sin ^{n-2}\theta \,d\theta , \end{aligned}$$

and this function is real. Thus

$$\begin{aligned} {\hat{w}}(\zeta )=|\zeta |^{-\frac{n}{2}+1}\int _0^\infty r^{\frac{n}{2}} J_{\frac{n}{2}-1}(|\zeta |r)(1+r^2)^{-\mu }\mathrm{d}r. \end{aligned}$$

Finally, we recall (11.4.44 in [1]) that

$$\begin{aligned} \int _0^\infty \frac{r^{\nu +1} J_\nu (ar)}{(1+r^2)^\mu }\,\mathrm{d}r=a^{\mu -1} K_{\nu -\mu +1}(a), \end{aligned}$$

so

$$\begin{aligned} {\hat{w}}(\zeta )=|\zeta |^{-\frac{n}{2}+\mu }K_{\frac{n}{2}-\mu }(|\zeta |), \end{aligned}$$

as desired. $\square$