Stationary phase in spherical integral

Question

I'm trying to find asymptotics for an oscillatory integral on $\mathbb{S}^{n-1}$, for which my advisor said I should use stationary phase arguments. The particular, he claims that:

If $\lambda\gg 1$, then $$I(\lambda,x) = \int_{\mathbb{S}^{n-1}}(x\cdot y)e^{i\lambda(x\cdot y)}\,d\sigma(y),$$ is $\mathcal{O}(|x|(\lambda |x|)^{-\frac{n-1}{2}})$ when $|x|\geq \lambda^{-1}$.

When I hear stationary phase, I think of working with operators of the form $$ L = \frac{\nabla_y(x\cdot y)}{i\lambda |\nabla_y (x\cdot y)|^2} \cdot \nabla_y, $$ since then $L^N[e^{i\lambda (x\cdot y)}] = e^{i\lambda (x\cdot y)}$, for any $N \geq 1$, and I can use integration by parts to move these operators over to the $(x\cdot y)$ term. However, wouldn't that require that the integral be defined over an $n$-dimensional region, rather than an $(n-1)$-dimensional surface?

I wouldn't be struggling so much if the gradient $\nabla_y$ could be taken in Cartesian coordinates. But we have a surface integral in $d\sigma(y)$, meaning that we would need to parametrize our surface with $n-1$ parameters, say $\omega = (\omega_1, \ldots, \omega_{n-1}) \in \Omega \subset \mathbb{R}^{n-1}$, with any derivates now being taken in these new variables. Specifically $\nabla_y$ would generate an $n$-vector, while $\nabla_{\omega}$ would generate an $n-1$-vector. So how should I go about applying the same kind of stationary phase arguments to this new integral? $$ I(\lambda, x) = \int_{\Omega} (x \cdot y(\omega)) e^{i\lambda (x\cdot y(\omega))} \,dV(\omega) $$ Now that $y$ depends on $\omega$, the gradients $\nabla_\omega (x\cdot y(\omega))$ become much trickier to get a grasp on. I'm particularly struggling trying to argue how we can find regions where $|\nabla_\omega (x\cdot y(\omega))| > 0$, so that our $L$ type operators are properly defined.

Am I going about this all wrong? If I can get the big-oh asymptotics I mentioned above in some other way, it doesn't really matter. I just need to prove these results as a lemma to something bigger. Any help is much appreciated!

Answer 1

You have $ I(\lambda, x)=x\cdot\int_{\mathbb S^{n-1}} ye^{i \lambda x\cdot y} d\sigma(y)=x\cdot J(x,\lambda) $ and you claim that for $\vert x\vert \lambda \ge 1$, you have $$ J(x,\lambda)=O((\vert x\vert \lambda)^{-\frac{n-1}{2}}). $$ Indeed, using coordinate charts and a finite partition of unity, you are reduced to the case where $$ J(x,\lambda)=\int_{\mathbb R^{n-1}} a(z) e^{i\lambda (x'\cdot z+ x_n\sqrt{1- \vert z\vert^2})}dz, \quad\text{$a\in C^\infty_0(\mathbb R^{n-1})$ supported near $0$, $x=(x', x_n)\in \mathbb R^{n-1}\times \mathbb R$}. $$ Let us set $\phi(x,z)=x'\cdot z+ x_n\sqrt{1- \vert z\vert^2}$. We have $$ \frac{\partial \phi}{\partial z}= x'-(1- \vert z\vert^2)^{-1/2} z x_n, $$ which vanishes at $z=0$ when $x'=0$. Then you calculate the Hessian of $\phi$ at $z=0$ and get $$ \phi''_{zz}=-x_n. $$ The stationary phase in $n-1$ dimensions gives the sought estimate with $O((\vert x_n\vert \lambda)^{-\frac{n-1}{2}})$ (note that you know that $\vert x_n\vert \lambda \ge 1$, since you are near $x'=0$).